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.

Leave a Comment