CBSE Chemistry Notes for Class 11 Chemical Properties Of Group-13 Elements

Chemical Properties Of Group-13 Elements

The members of group-13 elements have three electrons in their valence shells. Except the last member Tl, all other members use these electrons to form three bonds and thus exhibit an oxidation state of +3. In +3 oxidation state.

The members of the boron family are expected to form covalent bonds for the following reasons:

  1. Small size and high charge (+3) cause high polarisation of the anions leading to the formation of covalent bonds.
  2. The large value of the sum of the first three ionisation enthalpies, ΔiH1< ΔiH2 < ΔiH3 of these elements also suggests that the bonds will be largely covalent.
  3. The difference in electronegativity between the elements of group 13 and those ofthe higher groups is not very. This fact also agrees with the formation of covalent bonds.

Because of its small size and high ionisation enthalpies, it is not possible for boron to form B3+ ions by losing its three valence electrons. Therefore, Boron does not form ionic compounds. In fact, it always forms covalent compounds by sharing its valence electrons. The sum of the first three ionisation enthalpies, ΔiH1< ΔiH2 < ΔiH3 of Al is also higher

But less than that of B. So Al also has a strong tendency to form covalent compounds,

For example:

AlCl3, AlBrv and AlI3. Like Al, compounds of the rest of the members such as GaCl3, InCl3 etc. are covalent when anhydrous. However, all the members except B form metal ions in solution.

This change from covalent to ionic nature may be explained by the fact that in aqueous solutions these ions undergo hydration and the amount of hydration enthalpy exceeds the ionisation enthalpy. Ga, In and Tl show two oxidation states of +1 and +3 due to the inert pair effect. The compounds in the +1 oxidation state are more ionic than the compounds in the +3 oxidation state.

In a trivalent state, the number of electrons in the valence shell of the central atom in a molecule of these elements is only six (two electrons less than the octet) and therefore, such electron-deficient molecules behave as Lewis acids. For example, BCl3 (Lewis acid) readily accepts an unshared pair of electrons from ammonia (Lewis base) to form the adduct, BCl3 -NH3

1. Reaction with dioxygen or air

1. All the members of group 13 react with dioxygen at higher temperatures to form trioxides of the general formula M2O3. Tl forms both T1203 and some amount of Tl3O

P Block Elements Reaction With Dioxygen

The reaction of Al with O2 is known as a thermite reaction which is highly exothermic (ΔH0 =-1670kJ mol-1 ). A very strong affinity of Al for oxygen is used in the extraction of other metals from their oxides (thermite process). For example, Mn and Cr can be extracted from Mn304 and Cr2O3 respectively by this process.

2. The reactivity of group-13 elements towards dioxygen increases on moving down the group. Pure crystalline boron is almost unreactive towards air at ordinary temperature. Al does not react with dry air. However, it gets tarnished readily in moist air even at ordinary temperatures due to the formation of a thin oxide (Al2O3) layer on the surface which prevents the metal from further reaction. When amorphous boron and aluminium metal are heated in air, they form boron trioxide and aluminium trioxide (Al2O3) respectively.

P Block Elements Aluminium Trioxide

Ga and In are not affected by air but Tl forms an oxide on its surface in the presence of air

3. B and Al react with dinitrogen at high temperatures to form the corresponding nitrides

P Block Elements Nitrides

Ga, In and Tl do not react with N2 to form the corresponding nitrides.

4. Boron nitride is a white slippery solid which melts under pressure at 3246K. It is chemically inert towards the air, oxygen, hydrogen, chlorine, etc. even on heating. The total number of valence electrons of one B and one N- atom is equal to the number of valence electrons of two C-atoms.

5. Therefore, the structure of boron nitride is almost the same as that of graphite having a layer lattice. In each layer, alternate B and N-atoms (both sp² -hybridised) form a planar hexagon  The layers are stacked over one another in such a way that the N-atom of one layer is directed over the B-atom of another layer. Because of its structural similarity with graphite, boron nitride is also called inorganic graphite

P Block Elements Structure Of Boron Nitride

When boron nitride is heated at 1800°C under very high pressure, it gets converted to a cubic form comparable to diamond. This extremely hard variety known as borazon is used for cutting diamonds.

The acid-base character of oxides and hydroxides

1. Trioxides of the elements of the boron family react with water to form their corresponding hydroxides.

M2O3 + 3H2O → 2M(OH)3

2. The nature of these oxides and hydroxides changes on moving down the group. Both B203 and B(OH)3 are weakly acidic. They dissolve in alkali to form metal borates.

B2O3, + 2NadH → 2NaBO2(sodium metaborate) +H2O

B(OH)3+ 3NaOH → Na3BO3(sodium borate) + 3H2O

Aluminium oxide and hydroxide are amphotericin in nature. Both of them dissolve in alkalies as well as acids.

Al2O3(s) + 3H2SO4(aq)→ Al2(SO4)3(aq) + 3H2O(l)

P Block Elements Sodium Aluminate

Similarly, Al(OH)3(s) + NaOH(s)→Na[Al(OH)4](aq)

Al(OH)3(aq) + 3HCl(aq)→AlCl3 (aq) + 3H2O(l)

The oxide and hydroxide of Ga are also amphoteric while those of Tl are basic.

Therefore, the basic character of oxides and hydroxides increases down the group

P Block Elements Hydroxides Increases

3. Thallium forms two types of hydroxides:

Thallic hydroxide [Tl(OH)3] and thallous hydroxide (TlOH). Tl(OH)3 is insoluble in H2O but TlOH is soluble and is a strong base like alkali metalhydroxides.

Hydroxides Explanation:

On moving down the group, the magnitude of ionisation enthalpy decreases. As a result, the strength of the M—O bond also decreases and therefore, its cleavage becomes progressively easier resulting in the increased basic strength down the group.

An extremely hard crystalline form of aluminium oxide called corundum is used as an abrasive.It can be made by heating amorphous aluminium oxide at about 2000K. Aluminium forms a series of mixed oxides with other j metals, some of them occurring naturally as semi-precious stones. These include ruby (Cr2+) and blue sapphire (CO2+, Fe2+, Tl4+)

2. Reaction with hydrogen

1. Group-13 elements form hydrides ofthe type MH3. The members of the boron family do not combine directly with hydrogen. However several hydrides are known which can be prepared indirectly. Boron forms several stable covalent hydrides which are collectively called boranes.

The two most important types of boranes are as follows:

The members of the boron family do not combine directly with hydrogen. However, several hydrides are known which can be prepared indirectly. Boron forms several stable covalent hydrides which are collectively called boranes.

The two most important types of boranes are as follows:

1. Boranes with general formula BnHn+4  are called nido – boranes .

Example: Diborane (B2H6)  pentaborane-9 (B5H9).

Boranes with general formula BnHn+6  are called arachno boranes

Example: Tetraborane (B4H10),pentaborane-11 (B5H11).

2. The simplest hydride is diborane (B2H6) which is prepared by the reaction of BF3 with lithium hydride industrially.

P Block Elements Lithium Hydride

3. The other members of group 13 also form several hydrides which are polymeric.

For example: (AlH3)R, (GaH3)n, (InH3)n.

The stability of these hydrides decreases down the group and thallium hydride is quite unstable.

Boron, aluminium and gallium also form complex anionic hydrides such as NaBH4 (sodium borohydride), LiAlH4 (lithium aluminium hydride) and LiGaH4 (lithium gallium hydride). These complex hydrides act as powerful reducing agents.

P Block Elements Hydrides As Power Full Reducing Agents

4. The hydrides are weak Lewis acids and readily form adducts with strong Lewis bases to form compounds ofthe type MH3:B (B = base).

For example: AlH4 :NMe3, GaH4

NMe etc. NMe3 + AlH3→ [Me4N: → AlH3]

3. Reaction with acids and alkalies

The action of acids:

1. Boron remains inert in the presence of non-oxidising acids such as HCl. However, it undergoes oxidation by strong oxidising acids such as a mixture of hot concentrated H3SO4 and HNO3 (2: 1) to form boric acid (H3BO3) at very high temperature

P Block Elements Boric Acids

2. The remaining elements of this group react with both oxidising and non-oxidising acids. For example, Al dissolves in dilute HCl and liberates dihydrogen

2Al(s) + 6HCl(aq) → 2Al2+(aq) + 6Cl(aq) + 3H2(g)

3. Concentrated nitric acid renders aluminium passive by forming a protective layer of its oxide (Al2O33) on the surface of the metal. Thus aluminium vessels can be used to store concentrated HNO3

2A1 + 6HNO3→ Al2O3 + 6NO2 + 2H2>O

Ga, In and TI react with dilute acids to liberate H3

Action of alkalies:

1. When boron is fused with alkalies (NaOH or KOH) at a temperature greater than 775K, it forms borates and liberates dihydrogen

P Block Elements Borates And Liberates Dihydrogen

2. Boron dissolves in a fused mixture of Na2CO3 and NaNO3 at 1123K to produce borate and nitrite salt and liberates carbon dioxide

P Block Elements Borate And Nitrite Salt And Liberates Carbon Dioxide

3. Al and Ga also react with aqueous alkalies with the evolution of dihydrogen.

P Block Elements Aqueous Alkalies With The Evolution Of Dihydrogen

In and Tl do not react with alkalis

4. Reaction with halogens

Elements of group-13 react with halogens at high temperatures to produce trihalides of the general formula, MX3. However, thallium (III) iodide doesn’t exist

P Block Elements Reaction With Halogens

Trihalides of boron:

Due to its small atomic size and high, ionisation enthalpy, boron forms covalent trihalides

P Block Elements Structure Of Boron Trihalides

BX3. BF3 is a gas, BCl3 and BBr3 are liquids and BI3 is a solid. All these are trigonal planar molecules in which the central B -atom is sp² -hybridised. The three unpaired electrons of p -orbitals of three halogen atoms overlap with the three sp2 -orbitals of boron to form three sp²-p, B—X,  σ -bonds. The unhybridised empty p-orbital remains perpendicular to the plane of the molecules

Since there are only six electrons in the valence shell ofthe central boron atom in boron trihalides, they can accept two more electrons to acquire a stable octet, i.e., boron trihalides can behave as Lewis acids. The Lewis acid character, however, decreases in the order:

BI > BBr3 > BCl3 > BF3.

Trihalides of boron Explanation:

This order of relative Lewis acid strength of boron trihalides, which is just the reverse of what may be expected based on the electronegativities of the halogen atoms, can well be explained based on the tendency of the halogen atom to donate its lone pair of electrons to the boron atom through pn-pn back bonding.

Since the vacant 2p -orbital of B and the 2p-orbital of Fatom containing a lone pair of electrons are equal in size, therefore, the tendency of the F -atom to donate the unshared pair by pn-pn back bonding is maximum.

BF3 can well be represented as a resonance hybrid of four resonating structures As a result of resonance involving pn-pn back bonding, the electron density on the boron atom increases effectively and so its strength as a Lewis acid decreases considerably.

P Block Elements Resonance Of Boron Trihalides Molecuke Involving Back Bonding

As the size of the halogen atom increases on going from Cl to I, the extent of overlap between the 2p -orbital of boron and a large p -orbital of halogen (3p of Cl, 4p of Br and 5p of I] decreases. As a consequence, the electron deficiency strength decreases on going from BF3 to BI3

Halides of aluminium:

The halides of aluminium in the vapour state as well as in an inert solvent such as benzene exist ns dimers. For example, AlCl3 exists as Al2Cl6.

Halides of aluminium Explanation:

In AlCl3, there are six electrons (two electrons less than the tyre octet) around the central Al-atom. In the dimeric structure, each Al completes its octet by accepting a lone pair of electrons from the Cl-atom of another AlCl3 molecule. The dimeric form exists lit vapour state at < 473K. However, at higher temperatures, it dissociates to a trigonal planar AlCl3 molecule.

In polar solvents such as water, the dimer dissociates and it is the high hydration enthalpy which helps this dissociation leading to the formation of Al3+ion.

P Block Elements Dimeric Structure Of Aluminium Chloride

Al3Clg + 6H2O ⇌   2[Al(H2O)6]3++(aq) + 6Cl(aq)

Therefore, anhydrous AlCl3 is covalent but, hydrated aluminium chloride is ionic.

Unlike aluminium halides, boron halides exist as monomers and this is because the boron atom is so small that cannot accommodate four large-sized halogen atoms
around it.

CBSE Chemistry Notes for Class 11 Enthalpy

Class 11 Chemistry Concept Of Enthalpy

The word enthalpy is based on the Greek word ‘enthalpos’ which means to put heat into. It is denoted by the symbol H.

Concept Of Enthalpy Definition: (H) of a system is the total energy, which includes the internal energy (If) associated with the system and PV energy i.e., energy due to pressure-volume work.

Mathematical expression: The enthalpy of a system is expressed by the relation: H=U=Pv, where H, U, P, and V are the enthalpy, internal energy, pressure, and volume of the system, respectively. Since both the internal energy (U) and PV term have energy units, so does the enthalpy (H).

If the unit of P is expressed in atm and that of Vin L, then the unit of PV will be L-atm. Again, lL-atm=101.3J. Therefore, PV expresses an energy term.

Importance of the concept of enthalpy: According to the first law of thermodynamics, AU = q + w =q- PA V (considering only P-V work). At constant volume (AV = 0), AU = qy. Thus at constant volume heat absorbed or released (qy) by the system, is equal to the change in internal energy of the system.

So, we can determine the change in internal energy in a chemical reaction at constant volume by determining the amount of heat that is exchanged in the reaction. However, reactions are seldom carried out at constant volume as it is difficult to maintain constant volume during most reactions.

Chemical reactions are usually carried out In an open container (r, beaker, test tube, etc.) under atmospheric pressure. Under this condition, the pressure of the systems is the same as that of the surroundings t.e., P the first law of thermodynamics, we have Alt U=q+w=q-pv (considering only P-V work)

∴ \(q_p=\Delta(U+P V) \quad \text { or, } q_p=\Delta H\)

Thus at constant pressure heat absorbed or released by a system (r/;>) Is equal to the change in enthalpy of the system. Since r/;) can be measured easily and most of the physical and chemical changes are carried out at constant pressure, the change in enthalpy rather than the change in internal energy is more important in thermodynamics.

Class 11 Chemistry Characteristics of enthalpy

Enthalpy Is a state function: According to the definition of enthalpy, H = U + PV; where U, P, and V are the internal energy, pressure, and volume of the system respectively. All three quantities viz., U, P, and V are state functions. So, H is also a state function.

The change in enthalpy (A) of a system does not depend upon how the process has occurred: Since enthalpy is a state function, its change in a process depends only on the initial and final states of the system in the process. Its change does not depend on the path of the process.

The absolute value of enthalpy of a system cannot be determined experimentally: According to the definition, H = U + PV. Since the absolute value of internal energy cannot be determined experimentally, the absolute value of enthalpy cannot also be determined. However, the change in enthalpy of a system in any physical or chemical transformation can be determined experimentally.

Enthalpy is an extensive property: Enthalpy of a system depends upon the amount of substance present in the system. Thus, the enthalpy of a system is an extensive property.

Change in enthalpy of a system in a process

The change in enthalpy of a system in a process = final enthalpy- initial enthalpy 0 of the system in the process.

Let us consider a chemical reaction: aA + bB -+ cC + dD.

The initial enthalpy of the system is equal to the total enthalpy of the reactants, and the final enthalpy of the system is equal to the total enthalpy of the products.

Thus, the change in enthalpy in the above reaction:

⇒ \(\Delta H_{\text {reaction }}=\sum H_{\text {products }}-\sum H_{\text {reactants }}\)

Where, \(\sum H_{\text {products }} \text { and } \sum H_{\text {reactants }}\) are the values of the total enthalpies of the products and reactants, respectively.

∴ Where, \(\bar{H}_A, \bar{H}_B, \bar{H}_C \text { and } \bar{H}_D\) are the molar enthalpies (enthalpy per mole) of A, B, C, and D respectively

If in a chemical reaction \(\sum H_{\text {products }}>\sum H_{\text {reactants }}\) hen AH = +ve. This type of reaction is called endothermic reaction.

If in a chemical reaction \(\sum H_{\text {products }}<\sum H_{\text {reactants }}\) then H= -ve. This type of reaction is called exothermic reaction.

Definition Of change in Enthalpy: The change in enthalpy in a process that a closed system undergoes at constant pressure and is associated with only pressure-volume work may be defined as the amount of heat that the system absorbs or releases in the process.

Expression for the change in enthalpy of a system in a process occurring at a constant pressure

Suppose, a closed system with a volume of V1 and internal energy of U1 undergoes an isobaric process. Also, the volume and tire internal energy of the system after the process are V2 and U2, respectively.

So, the enthalpy of the system in the initial state is, H1 = UX + PVX, and that in the final state is, H2 = U2 + PV2 Therefore, the change in enthalpy (AH) of the system in the given process is

⇒ \(\begin{aligned}
\Delta H & =H_2-H_1=\left(U_2+P V_2\right)-\left(U_1+P V_1\right) \\
& =\left(U_2-U_1\right)+P\left(V_2-V_1\right) \\
\Delta H & =\Delta U+P \Delta V
\end{aligned}\)

If the system performs only P-V work, then w = -PA V [in case of expansion, V2 > V x and w = -ve; in case of compression, V2 < Vx and w =+ve] Applying the first law of thermodynamics, AH = qp+w = qp-PAV

where AH and qp represent the change in internal energy and amount of heat transfer respectively, during the process. From equations [1] and [2], we have

⇒ \(\Delta H=\Delta U+P \Delta V=q_P-P \Delta V+P \Delta V=q_P\)

∴ \(\Delta H=q_P \quad \cdots[3] \quad \text { or, }-\Delta H=-q_p \cdots[4]\)

[when the process occurs at a constant pressure and only P- V work is performed] According to equation [3], heat absorbed by a closed system (capable of doing only P-V work) at a constant pressure is equal to the increase in enthalpy ofthe system. Similarly, equation [4] states that heat lost by a closed system (capable of doing only P- V work) at constant pressure is equal to the decrease in enthalpy of the system

In an isothermal expansion or compression, the change in enthalpy of an ideal gas (system) is zero: in an isothermal process, if the pressure and volume of an ideal gas change from P1 to P2 and V1 to V2, respectively, then the change in enthalpy of the gas, \(\Delta H=\Delta(U+P V)=\Delta U+\Delta(P V)=\Delta U+\left(P_2 V_2-P_1 V_1\right) \quad \cdots[1]\) For an ideal gas in an isothermal process, PÿV j = P2 V2- Thus from equation [1] we obtain, AH = AH. In an isothermal process, since the change in internal energy of ideal gas is zero, i.e., AU = 0, so AH = 0

Class 11 Chemistry Enthalpy Numerical Examples

Question 1. 1 mol of a non-ideal gas undergoes the given change: (2 atm, 3 L, 95 K)-(4 atm, 5 L, 245 K). In this process, if the increase in internal energy of the gas is 30 L-atm, then what will be its change in enthalpy?
Answer: The change in enthalpy,

⇒ \(\Delta H=\Delta(U+P V)=\Delta U+\Delta(P V)=\Delta U+\left(P_2 V_2-P_1 V_1\right)\)

Given: AH = 30L-atm, Px = 2 atm, P2 = 4 atm V1 = 3L and V2 =5l

∴ AH = 30 + [(4 x 5)- (2 x 3)] L.atm = 44 L .atm

Question 2. Determine the change in enthalpy and internal energy when I mole of water completely vaporizes at 1OO°C and 1 atm picture. [latent heat of vaporization of water 530 cal – g 1]
Answer: Here, water vaporizes at a constant pressure of 1 atm. Therefore, the heat required for the vaporization will be equal to the change in enthalpy of the system. 1 mole of water = ( 1 X 10) = IB g of water. So, the heat required for the vaporization of IBg of water

⇒ \(\begin{array}{l|l}
=(18 \times 536) \mathrm{cal} & \begin{array}{l}
\text { The latent heat of vaporisation } \\
\text { of water is } 536 \mathrm{cal} \cdot \mathrm{g}^{-1}
\end{array}
\end{array}\)

At a constant pressure, AH = qp

So, the change in enthalpy = 40.5kJ [since 1 cal=4.2]

We know, AH = A(U + PV) = AH + A(PV)2 AU + PAVV [at constant pressure]

For Vapourisation \(P \Delta V=P\left(V_{\text {vapour }}-V_{\text {water }}\right)=P V_{\text {vapour }}\)

\(\left\lfloor V_{\text {vapour }} \gg V_{\text {water }}\right\rfloor\)

or, PAV = RT [considering ideal behaviour of water vapour] or, PAV = 8.314 X (273 + 100) = 3.10kJ

∴ So, A11 = A11 -PAV = (40.5- 3.1 )kj = 37.4 kj .

Therefore, in the process of vaporization of 1 mole of water at 100 °C and 1 atm pressure, the change in enthalpy, AH = 40.5 kj, and the change in internal energy, AH = 37.4 kj.

CBSE Chemistry Notes for Class 11 Some P Block Elements Group-14 Elements

Class 11 Chemistry Some P Block Elements Group-14 Elements (Carbon Family) Introduction

The valence shell electronic configuration of the elements of group 14 is ns²np², where n = 2-6.

It becomes clear from these electronic configurations (given in the table below) that carbon and silicon have noble gas cores, germanium and tin have noble gas plus 10 d-electron cores and lead has a noble gas core in addition to 14 f and 10 d-electron cores.

Thus, the electronic configurations of group-14 elements are similar to that of group-13 elements. However, they contain one more p-electron as compared to group-13 elements.

Electronic configurations of the group- 14 elements:

P Block Elements Electronic Configurations Of Group Of Elements

Occurrence Of Group-14 Elements

The members of group 14 are carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb).

Carbon is the seventeenth most abundant element by mass in the earth’s crust It is widely distributed in nature in free as well as in combined states. In a free state, it is found to occur in coal, graphite and diamond. These are the main allotropes of carbon.

  • Carbon in the form of coal and coke is used mainly as fuel. In a combined state, it is present widely as metal carbonates, hydrocarbons (petroleum), carbohydrates and carbon dioxide (0.03%) in air. Gases like propane and butane are the major constituents of LPG.
  • Moreover, the main constituent of all organic compounds is carbon. Two stable isotopes of carbon are present in nature namely 12Cand 13C6. Another isotope of carbon (14C6) is radioactive. The age of antique articles is determined by the ratio of 12Cand 13C6. present in them. This process is called radiocarbon dating.
  • Silicon is the second (about 27.7% by mass) most abundant element (next to oxygen) in the earth’s crust and is present in the form of silica and silicates. Germanium occurs only in traces (1.5 ppm). Both germanium and silicon in very pure form find applications as semiconductors.
  • The natural abundance of tin and lead is very low (2 and 13ppm respectively). The principal ore of tin is tinstone or cassiterite (SnO2) and that of lead is galena (PbS). Both tin and lead form several alloys. Tin is also used for tin plating while some lead-containing compounds are used as the constituents of paints.
  • The first two elements of this group, carbon and silicon are non-metals, germanium is a semi-metal (metalloid) while tin and lead are metals

Allotropic Forms Of Carbon

The phenomenon of the existence of an element in two or more forms having different physical but similar chemical properties i? called allotropy and the different forms are called allotropes. Carbon exists in some allotropic forms which may be classified as

  1. Crystalline and
  2. Amorphous: ‘

The four crystalline allotropic forms of carbon are:

  • Diamond
  • Graphite
  • Fullerene
  • Carbon nanotubes.

The four amorphous allotropic forms of carbon are:

  1. Charcoal
  2. Soot or lamp black,
  3. Coke and
  4. Gas carbon.

Amorphous carbon is not pure and remains mixed with various elements and compounds. Finer X-ray studies have shown that the amorphous varieties ofc arbon are composed of very minute crystalline units like graphite which are distributed throughout their masses in a most disordered fashion. A synopsis of various is allotropes of carbon is given in the following chart

P Block Elements A Synopsis Of Various Allotropes Of Carbon

Crystalline Carbon

1. Diamond

Diamond is a very precious substance which is available in South Africa, New South Wales, Brazil, the Ural mountains and at Golconda in India. Two varieties of natural diamonds are available. One is the lustrous and colourless (or slightly coloured) variety which is generally used as precious gemstones and the other is the black or deep-coloured opaque variety, known as carbonado orbort.

The weight of precious diamond is expressed in carats (1 carat = 200mg). Some well-known diamonds are:

  • Cullinan (3032 carats), Kohinoor (present weight 106 carats),
  • Pitt (136.25 carats), Regent (193 carats),
  • Orloff (193 carats) and Great Mogul (186 carats).
  • Diamond can also be prepared artificially but because of high cost and poor quality is seldom made artificially

Diamond Physical properties

  • Diamond is transparent, lustrous and crystalline.It may be colourless or slightly yellow coloured although some black or dark-coloured varieties are also available.
  • It is the hardest naturally occurring substance known and it has a very high melting point (3843 K).
  • It is the heaviest among all the allotropic forms of carbon; its density is 3.51 gem-3.
  • Diamond has a very high refractive index (2.417) and thus light passing through it, suffers total internal reflection innumerable times.
  • For this reason, the diamond appears to be extremely bright and lustrous.0It is transparent to X-rays and this property helps to distinguish a real diamond from an artificial one (made of glass).
  • Diamond is a non-conductor of electricity but a good conductor of heat

Diamond Chemical properties:

At ordinary temperatures, diamond is chemically inert. It does not react with acids, alkalies, chlorine, potassium chlorate etc. However, it reacts with certain substances at much higher temperatures.

1. It is oxidised by oxygen at 800-900°C to produce pure carbon dioxide:

C + O2  → CO2

2.  Diamond is converted into graphite in the absence of air at much higher temperatures

P Block Elements High Temperature

Molten diamond can be converted into graphite by applying heat but graphite cannot be converted into diamond by heating to a very high temperature. The change is, therefore, unidirectional and this is because graphite is thermodynamically more stable than diamond. This type of allotropy is known as monotropy.

3. When a diamond is reacted with molten sodium carbonate, sodium monoxide and carbon monoxide are produced

C + Na2CO3→ Na2O + 2CO

4. It undergoes oxidation by fluorine at 700°C to form carbon tetrafluoride C+ 2F CF

5. At about 250°C, diamond gets oxidised by a mixture of K2Cr2O7 and concentrated sulphuric acid (i.e., chromic acid) to CO2

P Block Elements Sulphuric Acids

6. Diamond reacts with sulphur vapour at 1000°C to form carbon disulphide: C + 2S→ CS2

Structure of diamond

  • In diamond, each C-atom is sp³ – hybridised and linked to four other C-atoms tetrahedrally by covalent bonds.
  • The value of each C—C—C bond angle is 109°28′ and each C —C bond distance is 1.54A
  • An innumerable number of such tetrahedral units are linked together to form a three-dimensional giant molecule containing very strong bonds extended in all directions. Because of such a three-dimensional network of strong covalent bonds, diamond is extremely hard.
  • Since a huge amount of thermal energy is required to break a large number of strong covalent bonds, its melting point is very high.
  • All 4 valence electrons of each sp³ -hybridised C-atom in a diamond crystal participate in forming covalent bonds and there is no free electron on any carbon atom. Thus, a diamond is an anon-conductor of electricity.
  • Diamond has the highest known thermal conductivity because its structure distributes thermal motion in three dimensions very effectively.
  • Unlike graphite in which the C-atoms are arranged in different distant layers, the C-atoms in diamond are placed at a covalent bond distance (1.54A). Because of this, the density of diamond is higher than that of graphite

P Block Elements Structure Of Dimond

Diamond Uses:

  • Because of its transparency, dazzling lustre and beauty, diamond is extensively used as precious gemstone.
  • Because of its extreme hardness,it is used for cutting glass, polishing hard surfaces and drilling purposes. Black or dark-coloured diamonds are generally used for this purpose.

2. Graphite

Graphite is available as minerals in Sri Lanka, Mexico, Italy, California (U.S.A), Siberia, Korea, Spain and India. The word graphite originates from the Greek word ‘graphic’ which

Preparation of artificial graphite:

Acheson process in this process, coke dust mixed with silica is heated to a temperature of 3000-3500°C with the help of electrodes in an electric furnace made offire-bricks for 25-30 hours. The mixture is kept covered by sand.

  • In the first stage of the reaction, silica reacts with carbon to form silicon carbide (SiC) and carbon monoxide (CO).
  • Silicon carbide thus formed decomposes to yield graphite and silicon. Q At higher temperatures, silicon, on being vapourised, escapes from the furnace and graphite is left.

SiO2 + 3C→SiC + 2CO↑

SiC→Si + C [graphite]

Graphite Physical Properties

  •  Graphite is a dark greyish-coloured opaque, soft and slippery crystalline substance possessing metallic lustre.
  • It is lighter (density 2.25 g-cm-3 ) than diamond.
  • It is a good conductor of heat and electricity.

Graphite Chemical properties

Graphite is more reactive than diamond.

1. When graphite is heated in air at 700°C, it is oxidised to carbon dioxide:  C + O2→CO2

2. At 500°C, fluorine reacts with graphite to produce carbon tetrafluoride (CF4).  The compound is a non-conductor of electricity.It is also called graphite fluoride.

C+ 2F2 → CF4

3. Graphite is not attacked by dilute acid or alkali. However, when it is subjected to react with molten sodium carbonate, carbon monoxide is formed.

4. Graphite is oxidised to carbon dioxide with a mixture of K2Cr2O and concentrated H2SO4 (chromic acid).

P Block Elements Chromic Acid

When graphite is heated in the presence of a mixture of cones. nitric acid and sulphuric acid containing a small amount of potassium chlorate, greenish-yellow-coloured solid graphitic acid (CnH4O5) is obtained.

Its exact structural formula is not known yet. Graphite can be identified by this test (diamond does not respond to this test).

On complete combustion, graphite produces mellitic acid [C6(COOH)6].

Structure of graphite

  •  Each carbon atom in graphite is sp² -hybridised and is linked to three other carbon atoms directly in the same plane forming a network of planar hexagons and these two-dimensional layers exist in different parallel planes.
  • In each layer, the C—C bond length is 1.42A and the distance between two adjacent layers is 3.35A which is greater than the C—C covalent bond distance. So, the layers are supposed to be held together by relatively weak van der Waals forces of attraction
  • As the distance between two parallel layers is sufficiently large, graphite is less dense than diamond.
  • Since the layers are weakly held together, on the application of pressure, one layer easily slides over the other. Thus, graphite is found to be soft and lubricating in nature.
  • In the formation of hexagons in a layer of graphite, only three of its four valence electrons are used to form three sigma bonds (Csp²-Csp²). The remaining electrons of each carbon atom present in an unhybridised p -orbital is utilised to form n -n-bonds.
  • The -π-electrons are mobile and can move freely through the graphite crystal. Because of the presence of free mobile electrons, graphite is a good conductor of electricity and heat. Of all the crystalline allotropes of carbon, graphite is thermodynamically the most stable one. Its standard enthalpy of formation (AfH°) is taken as zero.

P Block Elements Structure Of Graphite

Graphite Uses

  • Graphite is largely used for lining and making electrodes for electric furnaces.
  • When mixed with oil and water, graphite is used as a lubricant in machinery.
  • It finds use in making crucibles resistant to high temperatures.
  • By mixing with desired quantities of wax or clay, graphite is used for making cores of lead pencils.
  • Graphite is used as a moderator in nuclear reactors

3. Fullerens

Fullerenes or Buckminsterfullerene (named after the famous American designer of the geodesic dome, Robert Buckminster Fuller) is the latest allotrope of carbon discovered in 1985 collectively by three scientists namely R. E. Smalley, R. F. Curl and H. W. Kroto.It is a crystalline allotrope of carbon in which the carbon atoms exist in a cluster form.

It is also known to be the purest form of carbon because unlike diamond and graphite does not have surface bonds that are to be attracted by other atoms.

1. Structure of fullerenes C60:

  • Fullerenes are expressed by the general formula Cn, where n is an even number between 30-600, for example, C60, C70, C84 ….etc.
  • All these are cage-like spheroidal molecules having polyhedral geometry containing pentagonal and hexagonal planes. Number of hexagons in a Cn molecule =(n/2- 10),
  • For example,  in fullerene C60 molecule, number of hexagons =(60/2-10) = 20. Structure of fullerene C60; The C60 molecule consists of twenty-six-membered rings and twelve five-membered rings of
  • sp² -hybridised carbon atoms fused into each other. Each carbon atom forms three or -bonds with the other three carbon atoms and the remaining electron on each carbon is involved in the formation of n -bond and as a result, the system is expected to be aromatic.
  • However, it is not aromatic because the molecule is not planar and it does not have (4n + 2) electrons. It is a non¬ aromatic system. This fusion pattern provides a marvellous symmetry to the structure in which the fused ring system bends around and closes to form a soccer ball-shaped molecule (“buckyball”). Of all the fullerenes, C60 is the most stable one.

P Block Elements Buckminsterfullerene

2. Structure of fullerene C70:

The molecule acquires the shape of a rugby ball. It consists of twelve five-membered rings and twenty-five six-membered rings and their arrangement is the same as that of a C60 molecule.

Fullerens Preparation

  • The preparation of fullerenes involves heating of graphite in an electric arc in the presence of an inert gas such as He or Ar.
  • A sooty material is recovered which consists mainly of C60 with a small amount of C70 and traces of other with an even number of C atoms up to 350 and above. The C60 and C70 fullerenes can further be separated from the sooty material by extraction with benzene or toluene followed by chromatographic separation using alumina (Al2O3) as the adsorbent.
  • In Russia, America, Canada and New Zealand C60 and C70 fullerenes are isolated from natural sources. Fullerenes of this type are formed in the red giant star Antares.

Fullerenes Properties and Applications:

  • Fullerenes are solids with high melting points.
  • Being covalent, they are soluble in organic solvents.
  • They react with alkali metals to form solid compounds such as K3C60.
  • This compound acts as a superconductor even at temperatures of the order of 10-40K.
  • Because of their spherical shapes, they exhibit wonderful lubricating property

4. Carbon nanotubes

Carbon nanotubes are crystalline allotropes of carbon with cylindrical nanostructure. This allotrope was discovered by Sumio Iijima (Japan) in 1991. A carbon nanotube consists of a two-dimensional array of hexagonal rings of carbon just as in a layer of graphite or a chicken wire. The layer is then rolled. Cylinder and capped at each end with half of a C60. fullerene

P Block Elements Carbon Nanotubes

Carbon nanotubes Properties and applications:

  • The nanotubes are approximately 50000 times thinner than a human hair. These are very tough, about 100 times as strong as steel. They are electrically conducting along the length of the tube.
  • These cylindrical carbon molecules having unusual properties are valuable for nanotechnology, electronics, optics and other fields of material science and technology.
  • They are also being used as probe tips for the analysis of DNA and proteins by atomic force microscopy (AFM). Many other applications have been envisioned for them as well, including molecular-size test tubes or capsules for drug delivery

 

CBSE Chemistry Notes for Class 11 Thermodynamic Process and Their Types

Thermodynamic Process

Thermodynamic Process Definition: A system is said to undergo a thermodynamic process when it goes from one equilibrium state to another.

Path Of A process: The sequence of states through which a system passes when it undergoes a process is called the path of the process.

A system can undergo a change from an equilibrium state to another through a variety of processes or paths. In most of these processes, one or more of the properties of the system are held constant. Some of the common thermodynamic processes encountered in chemical thermodynamics arc are discussed below.

Cyclic process

Cyclic process Definition: A system is said to have undergone a cyclic process if it returns to Its initial state after a series of successive.

Cyclic process Explanation: Let us consider a process, in which the initial state of a gaseous system is A (Pv Tx). The system returns to its initial state after undergoing consecutive processes IB, BC, and CA. Since the system returns to its initial state through three successive operations, it is a cyclic process.

Cyclic process Example: The following change indicates a cyclic process because 1 mol of water (system) returns to its initial state again through successive changes

Chemical Thermodynamics Cyclic Process

The changes in state functions are zero in a cyclic process. The value of a state function depends upon the present state of the system. Since the initial and the final states of the system are the same in a cyclic process, the rallies of the state functions in these two states are also the same. So die change in state function (AP, AV, A7′, A77.A/7, etc.) becomes zero for a cyclic process.

Isothermal Process

Isothermal Process Definition: If the temperature of a thermodynamic system remains constant throughout a process, then the process Is said to be an Isothermal process.

At the time of conducting this process, the system Is kept lit contact with Constant temperature heat (f.u, thermostat) with a high heat capacity. Such a heat hath Is capable of gaining or losing heat without changing Its temperature.

Condition(s) for the isothermal process: During an Isothermal process, the temperature of the system remains constant. So we can write, 7′[system] = constant and dV[system] or AT [system] = 0.

Example: The boiling of a liquid at its boiling point Is an isothermal process. This Is because the temperature of the liquid remains constant until the entire liquid converts to vapor. Thus, the boiling of water at 100°C and l atm is an isothermal process

The temperature of the system remains fixed in the isothermal process. U does not mean that heat is not absorbed or liberated by the system during this process.

Chemical Thermodynamics Work Done On The System By Its Surroundings And Work By The system On Surroundings

Isobaric Process

Isobaric Process Definition: A process in which the pressure of the system remains fixed at each step of the process is called an isobaric process.

Condition(s)during constant for this and isobaric Process [system] process: remains APAs [system]theconstant,pressure= Pof. [system]the system=

Example: The vaporization of any liquid in an open container occurs under atmospheric pressure. If the atmospheric pressure remains fixed, then the process of vaporization is said to be an isobaric process.

Isochoric Process

Isochoric Process Definition: A process in Which The Volume Of The System remains constant throughout the process is called an isochoric process

Condition(s) for an isochoric process: The volume of the system remains constant during this process. Hence, V[system] = constant and dV[system] or ΔV [system] =0.

Example: The combustion of a substance In a bomb calorimeter. For a closed system consisting of an ideal gas. the plots of p vs V for O Isothermal Isobarle and Isochoric changes are given below.

Chemical Thermodynamics Isothermal, Isobaric, Isochoric Changes

Adiabatic process

Adiabatic process Definition: A process in which no exchange of heat takes place between the system and Its surroundings at any stage of the process is called an adiabatic process. This process requires the system to be covered with a perfect thermal insulator. But in reality, no such material is available and hence the process never becomes one hundred percent adiabatic.

Condition(s) for the adiabatic process: No heat is exchanged between a system and its surroundings In an adiabatic process. So, for such type of process q – 0; where q = heat absorbed or released by the system.

Example: The sudden expansion or compression of a gas is considered to be an adiabatic process because when a gas (system) undergoes such a process, it does not get a chance to exchange heat with its surroundings.

As a result, the sudden compression of a gas results in an increase in the temperature of the gas, while its sudden expansion leads to a decrease in temperature. For example, when the valve ofa bicycle or car tire is removed, the air coming out of the tire undergoes very fast expansion and gets cold.

The temperature of a system does not remain constant in an adiabatic process (except in the case of an adiabatic expansion of an Ideal gas against zero pressure).

For a process involving more than one step, the algebraic sum of heat absorbed or released in different steps may be zero qx + q2 + q3 + … = o, but it does not mean that the process is an adiabatic one.

Reversible process

Reversible process Definition: A process It salt to be reversible If It Is carried out Infinitesimally slowly so that In each step the thermodynamic equilibrium of the system Is maintained, and any Infinitesimal change In conditions can reverse the process to restore both the system and Its surroundings lo their Initial states.

Explanation: Lot ns consider that gas (system) Is enclosed lit a cylinder lilted with a weightless and frictionless piston and the cylinder Is kept In a constant temperature hath (surroundings). So, any change occurring In the lire system would be Isothermal.

Since the temperature of the system and surroundings tiro the same, the system will remain In thermal equilibrium. Let the applied external pressure acting on the piston Is the same as that of the gas. So, the system will remain In mechanical equilibrium. Hence, In this condition, the system (the gas) Is In a state of thermodynamic equilibrium and the properties of the system remain unchanged.

Chemical Thermodynamics Reversible Expansion Of A Gas

Now, if the external pressure is decreased by an infinitesimal amount of dP, the volume of the gas will increase very slowly by an infinitesimal amount until the pressure of the gas becomes equal to that of the external pressure. Let the infinitesimal increase of volume = dV.

If the external pressure is further decreased by an Infinitesimal amount of dP, the volume of the gas will also increase by an infinitesimal amount of dV. In this way, if the gas is allowed to expand very slowly in an infinite number of steps by decreasing an infinitesimal amount (dP) of the external pressure at each step, then the expansion ofthe gas will reversibly take place.

In a step, if the external pressure is increased by an infinitesimal amount (dP), then the volume of the gas will decrease by an infinitesimal amount dV. Hence, by decreasing or increasing the external pressure by an infinitesimal amount, the direction of the process can be reversed.

Characteristics of a reversible process:

The driving force of a reversible process is infinite simally greater than the opposing force and by increasing or decreasing the driving force by an infinitesimal amount, the direction of the process can be changed.

In this process, the system remains in thermodynamic equilibrium at every intermediate step. This process is extremely slow. From the theoretical point of view, any reversible process requires infinite time for its completion.

After the completion of a reversible process, if the system is made to return to its initial state by traversing the forward sequence of steps in the reverse order, then both the system and its surroundings are restored to their initial states.

The work done by a system in a reversible process is always the maximum. The reversible process is extremely slow and infinite time is required to complete the process.

A true reversible process is a hypothetical concept. In practice, no process can be carried out reversibly. All processes occurring in nature (i.e., real processes) are irreversible. Nevertheless, the concept of reversibility has immense importance in thermodynamics Examples of some reversible processes:

The vaporization of a liquid at a particular temperature in a closed container can be considered a reversible process. Let us consider a certain amount of water is in equilibrium with its vapor at T K temperature in a closed container. Here water and water vapour together constitute a system.

If the temperature of the system is increased by an infinitesimal amount of dT, then a very small amount of water willvaporise, and a new equilibrium will be established. Consequently, the vapor pressure of water will be increased by an infinitesimal amount of DP.

If the temperature of the system is decreased by an infinitesimal amount of dT, the same amount of water vapor will condense to establish equilibrium. As a result, the vapor pressure of water is also decreased by an infinitesimal amount of DP.

So, by increasing or decreasing the driving force (i.e., by increasing or decreasing temperature) the direction of the process can reversed. Thus, the vaporization of a liquid at a particular temperature in a closed vessel approximates a reversible process.

The reaction occurring in a galvanic cell is reversible. If an external potential applied between the two electrodes is slightly less in magnitude but opposite in sign than the electromotive force (EMF) of the cell, then the direction of flow of the current and the cell reaction remains unaltered.

But, if the externally applied potential slightly exceeds the EMF ofthe cell in magnitude, then the direction of the cell reaction and the direction of the current are found to be reversed. Therefore, by slightly increasing or decreasing the external potential (for EMF of the cell), the direction of the cell reaction can be changed. Thus, the reaction occurring in a galvanic cell approximates a reversible process.

Irreversible process

Process which occurs at a finite rate finite changes in properties of the system, and at any stage during the process, the system cannot get a chance to remain in thermodynamic equilibrium is called an irreversible process

All natural processes are irreversible

Explanation: Let us consider that a certain amount of gas is enclosed in a cylinder fitted with a weightless and frictionless piston and the cylinder is kept in a constant temperature heat bath (thermostat). Therefore, the temperature of the system (gas) becomes equal to that of the 5(system) surroundings (thermostat).

Let the external pressure applied on the piston (P) and Urnmibk the pressure of the gas be the same. Now, if the external pressure is suddenly reduced to P’, then the gas will expand at a finite rate and this will be irreversible because during expansion the system does not maintain thermodynamic equilibrium.

Chemical Thermodynamics Irreversible Expansion Of A Gas

Characteristics of an Irreversible Process

  • In an irreversible process, there is an appreciable difference between the driving force and the opposing force (actually, the driving force is greater than the opposing force). As a result, such a process takes place at a finite rate, although sometimes a very slow process may also be irreversible.
  • In an irreversible process, the system does not remain in thermodynamic equilibrium at any stage during the process.
  • If an irreversible process is reversed and the system is made to go back to its initial state, then the work done in the backward direction will not be the same as that in the forward direction.

After the completion of an irreversible process, although the system can be brought back to Its initial state, its surroundings cannot be.

  1. Irreversible processes complete in a finite time
  2. The flow of heat from an object of higher temperature to an object of lower temperature.
  3. The downward flow of water from a mountain.
  4. The expansion of a gas against zero pressure.
  5. The formation of curd from milk

Chemical Thermodynamic Difference Between The reversible and irreversible process

Chemical Thermodynamic Difference Between The Isothermal And Adiabtic Process

CBSE Chemistry Notes for Class 11 First Law Of Thermodynamics

The First Law Of Thermodynamics

The first law of thermodynamics is an extension of the law of conservation of energy. The law of conservation of energy states that energy can neither be created nor destroyed. It can only be converted from one form to another in an equivalent amount.

The first law of thermodynamics can be stated in different ways: The total energy of the universe is always constant. or, Energy can neither be created nor be destroyed. It can only be transformed from one form to another in an equivalent quantity.

Whenever a definite quantity of one form of energy disappears, an exactly equivalent amount of another form ofenergy appears. or, The energy of an isolated system is always constant.

Justification Of The First Law Of Thermodynamics

There is no theoretical proof of the first law of thermodynamics. Human experience of the macroscopic behavior of a large number of systems is indeed the basis of the law. The law, when applied to various fields, is valid.

The following observations justify the law—

Equivalence of work and heat: From different experiments it has been observed that heat and work are always equivalent. Irrespective of the method by which work is performed, a definite quantity of heat is obtained in exchange for a definite quantity of work. Scientist Joule proved that 1 calorie of heat is produced in exchange for 4.1843 J of work.

So 4.184J of work is equivalent to 1 cal of heat. Therefore, creation or destruction of energy is not possible. When one form of energy disappears exactly an equivalent amount of another form ofenergy is obtained.

Perpetual motion machine: The impossibility of constructing a perpetual motion machine of the first kind supports the first law A perpetual motion Is a type that can work In a cyclic process and produce work without consuming any energy or producing more work by consuming lost energy.

All efforts In making such machines are In vain. Hence, it can be concluded that energy can neither be created nor be destroyed, but one form of energy can be converted Into an equivalent quantity of another form of energy.

The total energy of a system Is constant: Since an Isolated system does not Interact with Its surroundings, it cannot exchange either matter or energy with Its surroundings. This is why the energy of the system always remains constant.

The mathematical form of the first law of thermodynamics

A closed system changes its state if it exchanges heat or work or both with its surroundings. Consequently, the internal energy of the system also changes. Suppose, a closed system has an internal energy of Ux. Let q amount of heat be added to the system.

This increases the internal energy of the system to a value of ( Ux + q). Now, if an amount of work w is done on the system, then the internal energy of the system will further increase and will become (Ul+q + w). Suppose, the internal energy of the system at the final state is U2. So, U2 = Ux + q + w or, U2– t1 = q + w

∴ AU=q+w

[AU = U2-U1 = the change in internal energy of the system] vi i.e., the change in internal energy of a closed system = heat absorbed by the system + work done on the system. The equation is the mathematical formulation of the first law of thermodynamics. Important points regarding the equation, AU = q + wi]

Unit of AU, q, or w: In this equation, AU, q, and w must be expressed in the same unit such as joule or calorie.

IUPAC sign convention of q and w: w = positive: work is done on the system. w = negative: work is done by the system. q = positive: heat is absorbed by the system. q = negative: heat is lost by the system

Illustration: In a process, If a system absorbs 140 1 of heat and performs 80 J of work, then q = +140 J and w = -80 I.

In this case the change in internal energy, AU = q + w – 140- 80 = 60 J.0 In a process, if a system rejects 2001 heat and performs 60 J of work, then q = – 200 I and w = – 60 J. In this case, the change in internal energy, AU = q+w = -200-60 = -260 J.

Although q and w are not state functions, the sum of these two quantities does not depend upon the path. This Is because the sum of q and w is equal to AU, which is path-independent.

The mathematical expression of the first law of thermodynamics for an infinitesimal change: According to the first law of thermodynamics, for a closed system, AU = q + w; where q = heat absorbed by the system, w = work done on the system, A U = change in internal energy.

In case of an infinitesimal change, the above equation can be written as: dU = 6q + 5w where 5q = infinitesimal amount of heat absorbed by the system, 8w = infinitesimal amount of work done on the system, dU = infinitesimal change in the internal energy. Equation [1] represents the mathematical form of the first law of thermodynamics for infinitesimal changes.

The mathematical form of the first law of thermodynamics in case of a process that involves only P-work: According to the first law, AU = q+w. In this equation, w represents the sum of all types of work (mechanical, electrical, magnetic work, etc.). If only P-Vwork is performed, then \(w=-\int_{V_1}^{V_2} P_{e x} d V \quad\left[P_{e x}=\text { external pressure on the system }\right]\)

V1 and V2 are the initial and final volumes of the system respectively. When only P-V work is performed, the mathematical form ofthe first law of thermodynamics is

⇒ \(\Delta U=q-\int_{V_1}^{V_2} P_{e x} d V\)

If the volume of the system increases, then V2 > V1 (expansion). In this case, work is done by the system of the surroundings against the external pressure (Pex).

If the volume of the system is decreased by the external pressure (Pa) then V2<VX (compression). In this case, work is done on the system by the surroundings.

If the die volume of the system increases from V1 to V2 against a constant external pressure (PfX). then AU = q-Pex{Vs-V{)=q-PexV <Va>

If the volume of the system is decreased from V1 to V2 bv a constant external pressure (Ppv), then A U = q-P1V2-V2)=q-PexAV(V2< V2) isolated system and the first law of thermodynamics: According to the first law, AU = q + w. An isolated system does not interact with its surroundings. So such a system does not exchange heat (q) or work (w) with its surroundings and hence, q = 0 and w = 0. Therefore, the internal energy of an isolated system remains constant (AC/ = 0)

Chemical Thermodynamic Applications Of The First Law Of Theromodynamics In Different Proceses

CBSE Exemplar Solutions For Class 11 Chemistry Chemical Thermodynamics

Class 11 Chemistry Chemical Thermodynamics  Numerical Examples

Question 1. The volume and temperature of 2mol of an ideal gas 10L & arc, respectively. The gas is allowed to expand in an isothermal reversible process to attain a final volume of 25 L. Calculate the maximum work done
Answer: We know that maximum work is obtained in a reversible isothermal expansion
Now, work done by an ideal gas in an isothermal reversible expansion, \(w=-n R T \ln \frac{V_2}{V_1}\) Given: n = 2, T = (273 + 27)=300K. V1= 10 L, V2=25L

∴ \(w=\left[-2 \times 8.314 \times 300 \ln \frac{25}{10}\right] \mathrm{J}=-4570.82 \mathrm{~J}\)

∴ The maximum work done bythe gas = 4570.82 J

Question 2. A 3 mol sample of an ideal gas at STP expands in an isothermal reversible process to attain a final volume of 100L. Calculate the work done by the gas.
Answer: The volume of1 mol of an ideal gas at STP – 22.4 L. So, at STP the volume of 3 mol of an ideal gas =3×22.4 = 67.2 L

Now, work done by an ideal gas in an isothermal reversible expansion process, \(w=-n R T \ln \frac{V_2}{V_1}\)

Given: n = 3, T= (273 + 0) =273K, V1 = 67.2L, V2 =100L

∴ \(w=-3 \times 8.314 \times 273 \ln \frac{100}{\bar{\sigma} .2}=-2706.62 \mathrm{~J}\)

∴ Work done bythe gas = 2706.62 J.

Question 3. A 23 mol sample of an ideal gas is compressed in i reversible isothermal process from a volume of 12 to a volume of 5 L at 27°C. Calculate the work done on the gas.
Answer: In a reversible isothermal compression of an ideal gas, the work done on the gas, w \(=-n R T \ln \frac{V_2}{V_1}\left[\text { where } V_2<V_1\right]\)

Given: n = 2.5, T= (273 + 27) = 300K, V1=121., V2=5L

∴ \(w=-2.5 \times 8.314 \times 300 \ln \frac{5}{12}=5458.98 \mathrm{~J}\)

∴ Work done on the gas = 5458.98J.

Question 4. A gas is compressed by an external pressure of atm. The work done in the process is 1034 J. How much volume of the gas is reduced?
Answer: Workdone on the gas due to compression

⇒ \(w=-P_{e x} \Delta V=-P_{e x}\left(V_2-V_1\right)=P_{e x}\left(V_1-V_2\right)\)

Where Pex = external pressure; V1 and V2 are the initial and final volumes ofthe gas respectively.

As per the given data, \(w=1034 \mathrm{~J}=\frac{1034}{101.3}=10.2073 \mathrm{~L} \cdot \mathrm{atm}\)

and Pex = 5 atm ∴ 1 L-atm = 101.3 J

∴ 10.2073 = 5 X ( V1- V2)

∴ V1-V2 = 2.04 L

∴ The decrease in volume of the gas = 2.04l.

Question 5. Work done by 3 mol of an ideal gas in an isothermal reversible expansion at 30°C is 9.5 kJ. If the initial volume of the gas is 20 L, then what will be the final volume of the gas?
Answer: Work done by an ideal gas in an isothermal reversible expansion \(w=-n R T \ln \frac{V_2}{V_1}\) As per given data, n vi = (273 + 30)K = 303K and w = -9.5 kj = -9500 J. Initial volume ofthe gas, V1 = 20 L; Finalvoiume, V2=?

∴ \(-9500=-3 \times 8.314 \times 303 \ln \frac{V_2}{20} \text { or, } V_2=70.3 \mathrm{~L}\)

∴ After expansion, the final volume ofthe gas = 703 L.

Question 6. The pressure of 3 mol of an ideal gas is 10 atm at 27°C. Calculate work done by the gas when it is expanded isothermally against an external pressure of1 atm
Answer: As per the given data, the initial pressure of the gas, P1= 10 atm, and the gas is expanded against an external pressure of 1 atm. Therefore, the final pressure, P2 = 1 atm. Since the external pressure is very much Jess than the initial pressure, the gas will expand irreversibly and after expansion, the pressure of the gas will be equal to the external pressure i.e., 1 atm. In an isothermal irreversible expansion

⇒ \(w=-n R T\left(1-\frac{P_2}{P_1}\right)\)

Here, n = 3, T = (273 + 27)K = 300K

∴ \(w=-3 \times 8.314 \times 300\left(1-\frac{1}{10}\right)=-6734.34 \mathrm{~J}\)

Therefore, work done by the gas in this isothermal irreversible expansion = 6734.34 J.

Question 7. The pressure of 6 mol N2 gas kept in a cylinder is 30 atm at 30°C. Suddenly the gas comes out of the cylinder due to leakage. If the atmospheric pressure and temperature are 1 atm and 30°C, then calculate the work done by the gas. Assume that the gas behaves ideally.
Answer: The atmospheric pressure (1 atm) is very much less than the initial pressure of the gas in the cylinder (30 atm) and the temperature of the gas is the same as the atmospheric temperature. So, the gas will expand isothermally and irreversibly against an external pressure of1 atm.

In an isothermal irreversible expansion, \(w=-n R T\left(1-\frac{P_2}{P_1}\right)\)

As per given data, P1 = 30 atm, P2 = 1 atm, n = 6 and T = (273 + 30) = 303K

∴ \(w=-6 \times 8.314 \times 303\left(1-\frac{1}{30}\right)=-14611.02 \mathrm{~J} .\)

Question 8. A 5 mol sample of an ideal gas is compressed isothermally and irreversibly from 1.5 atm to 15 atm at 27 °C. Calculate the work done on the gas in the calorie unit.
Answer: Work done in an isothermal irreversible expansion, \(w=-n R T\left(1-\frac{P_2}{P_1}\right)\)

As per given data, n = 5 , Px = ,1.5atm, P2 = 15 atm, T = (273 + 27)K = 300K

∴ \(w=-5 \times 1.987 \times 300 \times\left(1-\frac{15}{1.5}\right)=26824.5 \mathrm{cal}\)

∴ The work done on the gas = 26.82 kcal.

Question 9. What amount of work is done if an ideal gas expands from 10L to 20L at 2 atm pressure?
Answer: We know, w = -Pex(V2-v1).

Given: Pex = 2 atm, = 10 L and V2 = 20 L

∴ w = -2(20-10) L-atm= -20 L-atm

=-20 x 101.3J = -2026J

since 1 L-atm = 1013 J

A negative value of w indicates work is done by the gas (system). Therefore, the magnitude of work done by the gas (system) = 2026 J.

Question 10. What amount of work is done if an ideal gas is compressed from 0.5 to 0.25L under 0.1 atm pressure?
Answer: We know, w = —Pex(.V2 — Y{) [In case of compression V2<V1]

Given: Pgx = 0.1 atm, V1 = 0.5 L and V2 = 0.25

∴ w =-0.1(0.25- 0.5) = 0.025 L-atm

= 0.025 X 101.3J = 2.53 J

Therefore, work done on the gas = 2.53 J

Question 11. Calculate the work done when 1 mol of water vaporizes at 100°C and atm pressure. Assume water vapor behaves like an ideal gas
Answer: 1 mol water,100C , 1 atm – 1 mol water vapour 100C, 1 atm

Pex = external pressure = 1 atm, V1 and V2 are volumes of1 mol of water and 1 mol of water vapor, respectively. The volume of1 mol of water vapor is very much greater than that of1 mol water. Thus, (V2-V1) and it makes(V2– V1) = V2.

∴ Work done, w = -PexV2

If water vapour behaves like an ideal gas, then PexV2 = PT amount of water vapour = 1 mol

∴ w = -PexV2 = -RT

Here, T = (273 + 100)K = 373K

W = -RT = -8.314 X 373 = -3101.12

Therefore, the amount of work done by the system = 3101.12 J.

Question 12. ron reacts with dilute HCI quantitatively to form \(\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_2(a q)+\mathrm{H}_2(g)\) 56g of iron is allowed to react completely with dil HCI at 25 °C. If this reaction is carried out separately in a closed container of fixed volume and an open breaker, then calculate the work done in each case. Assume If2 gas behaves like an ideal gas.
Answer: Work done, w \(-P_{e x} \Delta \mathrm{V}=-P_{e x}\left(V_2-V_1\right)\)

Pex – Constant external pressure, V1 and V2 are the initial and final volumes, respectively.

When the reaction occurs in a closed container of fixed volume, the change in volume of the system, AV = 0, hence w = 0.

In this case, the volume of the system increases against atmospheric pressure due to the formation of hydrogen gas in the reaction.

⇒ \(\begin{aligned}
& \text { So, } w=-P_{e x} \Delta V=-P_{e x} V_{\mathrm{H}_2}=-n_{\mathrm{H}_2} R T \\
& \text { Reaction: } \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_2(a q)+\mathrm{H}_2(g) \\
& 56 \mathrm{~g}(1 \mathrm{~mol}) \quad 1 \mathrm{~mol}
\end{aligned}\)

∴ nH2 = 1. Given: T = (273 + 25) = 298K

∴ w = -1 X RT = -8.314 x 298 = -2.477 kj

Therefore, work done by the system = 2.477 kj.

CBSE Chemistry Notes for Class 11 Quantum Number

Quantum Number

Quantum Number Definition: A set of four numbers that provide complete information about any electron in an atom are known as quantum numbers.

Quantum Number Classification: The four quantum numbers are—

  1. Principal Quantum Number (N)
  2. Azimuthal Or Subsidiary quantum number (l)
  3. Magnetic quantum number (m or m1)
  4. Spin quantum number (5 or mg ).
  5. To specify an electron in an atom, the following four quantum numbers should be mentioned.
  6. Principal quantum number [n]

Quantum Number Origin:

  1. From Bohr’s postulates, it is known that each electronic orbit surrounding the nucleus in an atom represents an energy level.
  2. The average energy of the electrons revolving in a particular orbit is fixed. So, these orbits are called principal energy levels or principal quantum levels.
  3. Depending on their distance from the nucleus, these orbits or principal energy levels are designated by the numbers 1,2,3, 4… etc. These numbers 1,2,3,4… etc. are called principal quantum numbers.

Quantum Number Designation: The principal quantum number is denoted by the letter ‘n ’. For AT-shell n = 1, for L -shell n = 2, for Mshell n = 3 and so on.

Information obtained:

  1. The higher the value of n, the greater the distance of the orbit from the nucleus, and hence, the greater the size ofthe orbit. Thus, r1<r2<r3< r4< …
  2. The higher the value of ‘ n,’ the greater will be the electronic energy associated with the orbit.
  3. Thus, El<E2<E3<E4<………..
  4. A maximum number of electrons that can be accommodated in a principal quantum level n is given by the formula 2n2.
  5. Limitations of 2β2 The maximum number of electrons in any orbit can never be more than 32 even if the value of n exceeds 4.
  6. The outermost electronic shell does not contain more than 8 electrons.
  7. The penultimate shell (i,e., the shell just preceding the outermost shell) does not contain more than 18 electrons.

Azimuthal or subsidiary quantum number

Azimuthal or subsidiary quantum number Origin: A spectrograph with high resolving [/]power has revealed that each bright line in the spectrum of atomic hydrogen consists of some closely spaced finer lines.

This fact suggests that each orbit or energy level in an atom is composed of subshells. Electrons occupying these subshells within the same -shell, exhibit slight differences in energy.

In order to explain the formation of the fine structure of spectral lines, Sommerfeld proposed
the existence of elliptical orbits, besides Bohr’s circular
orbits.

To specify the shape of the elliptical orbit, another supplementary quantum number is necessary.

This supplementary quantum number which indicates the captivity of the electronic orbit is called azimuthal or subsidiary quantum number denoted by the letter.

If the principal quantum number of any orbit is n, then the total number of subshells incorporated in that orbit will also be n.

Class 11 Chemistry Sturcture Of Atoms Circular And Elliptical Orbits Of Electrons

Magnitude:

  1. As per quantum mechanical calculations, the angular momentum of a moving electron in an elliptical path is given by, L = Jl(l+ 1) X.
  2. This is often called orbital angular momentum.
  3. The value of l determines the shape of the path. So, with the help of the principal quantum number and azimuthal quantum number, a precise idea about the size and shape of the electronic path can be obtained.
  4. If n stands for the principal quantum number of an electronic orbit, the values of l will be from to (n- 1) i.e., with respect to the value of principal quantum number n, the azimuthal quantum number / may assume n number of different values including zero, e.g., for n = 4, 1=0, 1, 2 and 3.
  5. To indicate the subshells within a shell, spectroscopic symbols are used instead ofthe numbers 0, 1, 2, 3 etc.
  6. The symbols s, p, d,f, etc., (spectroscopic coinage) are merely the first letters ofthe words sharp, principal, diffuse, and fundamental, used extensively in spectral analysis.
  7. To express the position of an electron in the atom, the principal quantum number should be written first followed by the symbol of the azimuthal quantum number to its right side, e.g., the subshells included in K, L, M, and N-shells are represented as

Class 11 Chemistry Sturcture Of Atoms Symbol of subshells

Class 11 Chemistry Sturcture Of Atoms K,L,M,N-shells

Class 11 Chemistry Sturcture Of Atoms m and n shells

  • The ratio of the major axis to the minor axis of an elliptical path is given by = (/ + 1)/n .
  • An elliptical path for which l = (rc- 1), becomes circular e.g., in the case of 4-orbit if 1 = 3, then that orbit becomes circular. The greater the difference between the values of l and n, the larger the ellipticity of that path.
  • The penetrating power and screening effect of an elliptical orbit increases on increasing the ellipticity of the orbit.
  • So the penetrating and screening powers of different subshells within the same shell follow the sequence: s> p> d>f.
  • Due to the difference in the internal energies of the subshells [s, p, d, f, etc.), the electrons moving in those subshells also possess different energies. Energy associated with the subshells in a particular orbit increases in the following order: s <p<d<f.

Magnetic quantum number (m or mt)

Magnetic quantum number Origin:

  1. Zeeman in 1896 observed that each fine line in atomic spectra splits further into finer lines in the presence of the highly powerful magnetic field.
  2. In the absence of a magnetic field, such finer splitting i.e., hyperfine splitting disappears. This phenomenon is called the Zeeman effect. To explain the Zeeman effect, a third type of quantum number, known as a magnetic quantum number was introduced.

Magnetic quantum number Discussion:

  1. Due to the angular motion of electrons around the nucleus, a magnetic field is produced, which interacts with the external magnetic field.
  2. As a result subshells of definite energy split into three-dimensional spatial regions called orbitals.
  3. Magnetic quantum number (MI) signifies the orientation of the orbitals in space in which the electron exists.
  4. The value of m depends on the azimuthal quantum number l.
  5. For a certain value of l, m has an o total of (2Z +1 ) different values. These values may be any whole number starting from -Z to +1 (including zero).
  6. For s- subshell, l = 1 and m – 1. This subshell, l = 0 and m – 0. This orbital (i.e., s-orbital). Z = 1 denotes p -subshell consisting of three orbitals which are directed along three axes.
  7. These are marked as px, py, and pz orbitals which have the respective values of m = -1, 0, and + 1 . Similarly, d and /-subshells contain 5 and 7 orbitals respectively.
  8. The negative values of the magnetic quantum number signify that these orbitals are inclined in the direction opposite to the magnetic field and the positive values indicate that these orbitals are inclined in the direction ofthe magnetic field.
  9. shows the different directions of the d -d-subshell (Z = 2) in the magnetic field.

Orientation of different orbitals of ZV-shell (n = 4] under the influence of magnetic field.

Class 11 Chemistry Sturcture Of Atoms Orientation Of DIfferent Obritals Of N-shell Under The Influence Of Magnetic Feild

Values of magnetic quantum number (m] for different values of azimuthal quantum number [l]

Spin quantum number [s or ms]

Uhlenbeck and Goudsmit introduced a fourth quantum number called the spin quantum number.

This is because the other three quantum numbers were not able to give sufficient explanation to the hyperfine structure of the atomic spectra.

% Just like the earth, an electron while moving around the nucleus also spins about its own axis either in a clockwise or in an anti-clockwise direction,

Each type of spin can give rise to characteristic spectral lines with the formation of a hyperfine spectrum in the spectral series.

The spin quantum number denoted by the symbol ‘s’ expresses two opposite types of spinning motions of each electron.

The spin quantum number ‘s’ can have only two values, \(+\frac{1}{2} \text { and }-\frac{1}{2}\) The positive and negative signs represent two opposite directions of spinning motion of any spinning motion of any spinning motion of electron are very often represented by two arrows pointing in opposite directions,| and.

Q A spinning electron behaves like a tiny magnet with a definite magnetic moment. The angular mentum associated with the spinning electron is given by the mathematical expression.

Class 11 Chemistry Sturcture Of Atoms Spinning Of Electron About Its Own Axis

\(s=\sqrt{s(s+1)} \times \frac{h}{2 \pi}\)

Spin Quantum number (s) signifies the mode of Electron Spin (Clockwise or Anti-clockwise).

Class 11 Chemistry Sturcture Of Atoms Significance Of The Quantum Numbers

Shapes Of Orbitals From Wave Function

It has been stated earlier that the three-dimensional space around the nucleus in which the probability of finding an electron is maximum is called an orbital.

In order to obtain a clear idea about the shapes of orbitals, we will first discuss the variation of—

  1. The radial part of the wave function,
  2. Square of the radial wave function, and
  3. Radial distribution function with an increase in distance from the nucleus.

Variation Of Radial Part Of Wave Function With Distance From The Nucleus

Schrodinger wave equation for the electron in a one-electron atom (H-atom) can be solved to get different expressions for wave function \((\psi)\) for different orbitals.

The orbital wave function for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.

However, for different orbitals the plots of the radial part of the corresponding wave functions as a function of r (distance from the nucleus) are different. depicts such plots for Is, 2s, 2p and 3s orbitals.

For is -orbital, the radial part of the wave function [ψ(r) or R] decreases sharply with increasing distance, r, from the dying nucleus.

For 2s -orbital ψ (r) or R, decreases sharply in the beginning, becomes zero at a particular distance, and then becomes negative as r increases.

Class 11 Chemistry Sturcture Of Atoms Plot Of DIfference From The Nucleus (r) Verus

FM 3s-orbilal \(\psi\). decreases sharply in the Beginning with an Increase In r, becomes zero At A Particcular Distance, And Then Becomes negative. On Further Increases In \(r, \psi(r)\) again becomes zero and finally becomes positive.

For 2p -orbital if \(\psi(r)\) rises from zero to a maximum, then decreases with increasing distance (r) from the nucleus. On further increase in distance, ifr(r) approaches almost zero.

For 3p -orbital,\(\psi(r)\) rises from zero and attains a maximum value. On further increase in ψ(r) begins to decrease and becomes zero at a particular distance. Then it becomes negative with a further increase in r.

Characteristic features observed in the plots of r vs Ψ(r):

  1. The radial part of the wave functions for 2s, 3s, 3p, etc. orbitals can be positive or negative depending upon the distance (r) from the nucleus. These are not related to the positive and negative charges.
  2. For each orbital, the radial part of the wave function Ψ(r) approaches zero as r→∞.
  3. For 2s, 3s, and 3p -orbitals, one common feature for the variation of wave function Ψ(r)) with distance is that Ψ(r) becomes zero at a finite distance from the nucleus. However, for different orbitals if Ψ(r) becomes zero at different distances Ψ(r).
  4. The distance Ψ(r) at which becomes zero is called a nodal point radial node or simply node. At the nodal point, the radial wave function if Ψ(r) changes its sign from positive to negative or vice versa.
  5. For different orbitals, the number of radial nodes =(n-1-1).
  6. This indicates that the number of radial nodes is determined by the values of the principal quantum number ( n) and azimuthal quantum number (Z) of the orbital under consideration.

Class 11 Chemistry Sturcture Of Atoms Number of radial nodes

There is no relation between the positive and negative values of the wave function with the positive and negative charges.

Radial probability density [Ψ²(r) Or R²] graphs variation Of the square Of Radical Wave Function With Distance From The nucleus (r)

The square of the radial wave function, Ψ²(r) or R2 for an orbital gives the radial density.

According to the German physicist, Max Bom, the radial density, Ψ²(r) at a point gives the probability density of the electron at that point along a particular radial line.

The variation of Ψ²(r) as a function of r for different orbitals is given in the figure. The nature of these curves is different for different orbitals.

Class 11 Chemistry Sturcture Of Atoms Graph Of Electron Probability

For Is -orbital, probability density is maximum near the nucleus (r≈0) and decreases sharply as we move from it.

For 2s -orbital the probability density is maximum near the nucleus (r≈0).

With increasing distance, Ψ2(r) first decreases sharply to zero and starts increasing again. After reaching a small maxima it decreases again and approaches zero as the value of r increases further.

The intermediate region (a spherical shell) where this probability density reduces to zero is called the nodal surface or simply node.

In general ns -orbital has (n- 1) nodes. Thus, the number of nodes for 2s -orbital is one, two for 3s and so on, i.e., the number of nodes increases with an increase of principal quantum number n.

The probability density variation for Is and 2s orbitals can be visualized in terms of charge cloud diagrams. In these diagrams, the density of the dots in a region represents the electron probability density in that region.

For 2p- Orbital Probability Density Is zero at r = 0. However, with increasing distance, it begins to increase and reaches a maximum and then decreases gradually as the distance (r) from the nucleus increases and ultimately approaches zero.

From similar plots of various orbitals, it has been found that all orbitals except s, have zero electron density at r = 0.

Radial probability distribution curve: Variation of radial distribution function (RDF) with distance from the nucleus (r)

The plot of Ψ²(r) versus r gives the probability density for the electron around the nucleus. However, in order to determine the total probability in an infinitesimally small region, we have to multiply probability density if Ψ2(r) by the volume of the region i.e., probability = Ψ²(r) x dv [where dv = volume of the region].

Since the atoms have spherical symmetry, it is more useful to discuss the probability of finding the electron in a spherical shell between the spheres of radii r and (r + dr).

The volume of such a shell of extremely small thickness, dr, is 4nr2dr. So we have, Probability = R2 x 4rrr2dr = 47tr2Ψ²(r)(r)dr [since R = Ψ²(r))].

This gives the probability of finding the electron at a particular distance (r) from the nucleus. This is called radial distribution function (RDF).

Radial distribution function (RDF) = 4πr2ψ2(r)dr

Class 11 Chemistry Sturcture Of Atoms 2p- orbital

Important information obtained from the plots of RDF vs r:

  1. For all orbitals, the probability is zero at the nucleus.
  2. If the point r = 0 is neglected, then it can be seen that,
  3. The number of radial nodes for any orbital -n-l- 1,
  4. The number of maxima (peak) for any orbital =(n-l- 1) +1 = (n-/). The peak in any curve gives the distance from the nucleus to that point where the probability of finding the electron is maximum. This is called the radius of maximum probability.
  5. All the s -orbitals, except the first one (Is), have a shell-like structure, rather like an onion, or a hailstone, consisting of concentric layers of electron density. Similarly, all but the first p -p-orbital (2p) and the first dorbital (3d) have shell-like structures.
  6. The first s -s-orbital (Is), first p -p-p-orbital (2p) and first orbital (3d) have two important characteristics—
  7. they do not contain radial nodes and contain only one maxima.
  8. Examination of the plots for Is, 2s, and 3s -orbitals shows that the most probable distance of maximum probability density increases markedly as the principal quantum number increases.
  9. Furthermore, by comparing the plots for 2s and 2p, or 3s, 3p, and 3d -orbitals it is seen that the most probable radius decreases slightly as the azimuthal quantum number increases.