NCERT Class 11 Chemistry Chapter 6 Chemical Thermodynamics Short Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics Short Question And Answers

Question 1. What is an adiabatic system? Is this an isolated system?
Answer:

An adiabatic system Is a closed system that can exchange several forms of energy (for example Work) but not heat with its surroundings. It is not an isolated system as an isolated system cannot exchange either matter or energy with its surroundings

Question 2. A closed container with impermeable diathermal walls contains some amount of gas. If the gas is considered to be a system, what type of system will it be X is a state function of a thermodynamic system. How are its infinite and infinitesimal changes denoted?
Answer:

As the walls of the container are impermeable, flow of matter into or out ofthe system is not possible. Again, the walls are diathermal. Thus, the exchange of heat between the gas and the surroundings is possible. Hence, the gas is a closed system.

Question 3. A closed system participates in the following process: A→B→C. In step A→B heat absorbed by the system = q cal and in step B→C, heat released by the system = qcal. Therefore, in this process, the sum of the heat absorbed and heat released by the system is zero. Is this an adiabatic process? Give reason.
Answer:

In an adiabatic process, the system does not exchange heat with its surroundings at any step in the process. In the given process, the system absorbs heat in one step and releases heat in the other step. So, this process cannot be regarded as an adiabatic process.

Question 4. One mole of an ideal gas participates in a cyclic reversible process as described. Indicate the type of processes the system undergoes in the steps AB, BC and CA. Assume T₂>T₁.
Answer: AB:

It is an isochoric process as the volume of the system remains unaltered in this step. BC: It is given that T₁<T₂– Again, the given indicates V₂> V₁. This means that the volume of the system increases with a decrease in temperature. This happens in case of an adiabatic expansion of a gas. Therefore, the BC step indicates an adiabatic process.

Question 5. Calculate the work done in the following process which an
ideal gas undergoes
Answer:

⇒ \(\text { 1st step: } w_1=-n R T_1 \ln \frac{V_2}{V_1}\)

⇒  \( 2 \text { nd step: } w_2=-n R T_1 \ln \frac{V_1}{V_2}\)

So, total work \(w_1+w_2=-n R T_1 \ln \frac{V_2}{V_1}-n R T_1 \ln \frac{V_1}{V_2}\)

⇒ \(=-n R T \ln \frac{V_2}{V_1}+n R T \ln \frac{V_2}{V_1}=0\)

Question 6. In which of the following reactions is the work done zero? Assign the sign of w (+ ve or- ve)for the cases in which work is involved.
Answer:

In a chemical reaction, pressure-volume work, ω = -P_exΔV = -ΔnRT; where Δn = total number of moles of gaseous products – total number of moles of gaseous reactants.

In reaction 3 Δn = 0Δ, So, w =0

In reaction 1 Δn = 2-(1+2)=-1

So, w = -ΔnRT = RT, i.e., w> 0

In reaction Δn=l. So, ω= -ΔnRT =-RT, i.e., w < 0.

Question 7. Among the following processes identify those In which the change in internal energy (Δ U) Is zero: Isothermal compression of ideal gas Adiabatic expansion of ideal gas Free adiabatic expansion of an ideal gas Reversible cyclic process Irreversible cyclic process.
Answer:

The change in internal energy of an ideal gas in its isothermal compression is zero, When an ideal gas undergoes an adiabatic expansion, its internal energy decreases. In the adiabatic free expansion of an ideal gas, the internal energy of the gas remains the same, Since U is a state function, its change in any cyclic process (reversible or irreversible) will be zero.

Question 8. Write down the mathematical form of the first law of thermodynamics for an infinitesimal change that involves only pressure-volume work. Write down the form of this equation ifthe above change occurs reversibly.
Answer:

In case of an infinitesimal change, the mathematical form of the first law of thermodynamics is: dU= δq + δw; where δq = heat absorbed by the system, δw = work done on the system and dU is the change In Internal energy of tyre system.

For an infinitesimal change involving only P-V work, δw=-P_exdV. So, for an infinitesimal change Involving only P-V work, the form of the first law of thermodynamics will be, Δw=-P_exdV

Question 9. A closed system undergoes a process A→B. If it occurs reversibly, then the system absorbs qy amount of heat and performs a amount of work. However, if it occurs irreversibly, then the system absorbs the q₂ amount of heal and does the w₂ amount of work. Is (q₂ + w₁) greater than, less than or equal to (q₂+ w₂)?
Answer:

For a reversible process:

ΔU₁ = q₁ + w₁ and for the irreversible process: ΔU₂ = q₂+w₂.

In both processes, the initial state (A) and final state (B) of the system are identical. Since U is a state function, its change in a process depends only on the initial and final states ofthe system, and not on the nature of the process.

As the initial and final states in both processes are identical, the change in internal energy in both cases will be the same. Therefore, ΔU₁ = ΔU₂ and q₁+ ω₁ = q₂ + w₂

Question 10. C_p-C_y = x J-g^-1.K^-1 and C_p-C_y = x J-g^-1.K^-1 ] mol^-1. K^-1 for an ideal gas. The molecular mass of the gas is M then establishes a relation among x, X and M.
Answer:

For a substance, molar heat capacity = specific heat capacity x molar mass.

Therefore, \(C_{P, m}=C_P \times M \text { and } C_{V, m}=C_V \times M\)

Given: Cp,m- Cv,m = XJ mol^-1. K^-1

∴ CpxM-CvxM = X

or, (Cp-CV)M = X; hence, X = Mx

Question 11. Why is the sign of ΔH negative for an exothermic reaction and why is it positive for an endothermic reaction?
Answer:

In a chemical reaction, the change in enthalpy, AH = sum of the total enthalpies of products – Sum of the total enthalpies of-reactants

= \(\Sigma H_P-\Sigma H_R\)

In case of an exothermic reaction

⇒  \(\Sigma H_P-\Sigma H_R\), and hence ΔH<0;

While for an endothermic reaction \(\Sigma H_P-\Sigma H_R\) resulting ΔH>0.

Question 12. Mention the standard stales of the following elements at 25°C and later: carbon, bromine, Iodine, sulphur, oxygen, calcium, chlorine, fluorine and nitrogen.
Answer: At 25C; and 1 atm, the standard states of the given elements are—

1. Carbon: C(s, graphite);
2. Bromine: Br₂(Z);
3. Iodine: l₂(s):
4. Sulphur: S(s, rhombic);
5. Oxygen: O₂(g);
6. Calcium: Ca(s)
7. Chlorine: Cl₂(g);
8. Pluorine: P₂(g) ;
9. Nitrogen: N₂(g)

Question 13. Give an example of a physical change for each of the following relations between
ΔH And ΔU:

1. ΔH > ΔU
2. ΔH < ΔU
3.  ΔH≈ ΔU

Answer:

The equation ΔH= ΔH + ΔnRT can be used in case of a process involving phase change ofa substance.

1.  \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(g) \text {. Here, } \Delta n=+1 \text {. So, } \Delta H>\Delta U \text {. }\)

2. \(\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \text {. Here, } \Delta n=-1 \text {. So, } \Delta H<\Delta U \text {. }\)

Question 14. Which element in each of the following pairs has the standard heat of formation to zero?

1. [O₂(g), O₃(g)] 
2. [Cl₂(g), Cl(g) ]
3. [S (s, rhombic), S (s, monoclinic)]?

Answer:

The enthalpy of formation of an element in its standard state is zero, at 25 the standard state of oxygen, chlorine and sulphur are

O₂(g)> Cl₂(g) and S (s, rhombic). So, at 25 °C, the standard heats of formation of O₂(g), CI₂(g) and S(s, rhombic) will be zero.

Question 15. “At 25°C the standard heat offormation ofliquid benzene is + 49.0 kj.mol^-1 What does it mean
Answer:

At 25°C, the standard enthalpy of formation of liquid benzene is +49.0 kj.mol^-1. This means that at 25 °C and 1 atm when 1 mol liquid benzene forms from its constituent elements, the enthalpy change that occurs is +49.0 kj. In other words, at 25 °C and atm pressure, the change in enthalpy in the following reaction is +49.0 kj

6C(s, graphite) + 3H₂(g)→C₆H₆(l)

Question 16. The standard heat of sublimation of sodium metal is 108.4 kj.mol^-1. What is its standard heat of atomization?
Answer:

At 25 °C, the standard state of sodium is Na(s). The sublimation process of Na(s) is Na(s)→Na(g). At 25 °C the enthalpy change process is equal to the sublimation enthalpy of Na-ihetal. Again, in the above process, I mol of Na(g) is formed from Na(s). So, at 25°C, the enthalpy change in this process is equal to the standard enthalpy of atomisation of sodium. Thus, the standard enthalpy of atomisation of sodium is 108.4 kj. mol^-1

Question 17. What will be the sign of ASsys (+ve or -ve ) in the process of—

1. The vaporisation of a liquid
2. Condensation of a vapour
3. Sublimation of a solid.

Answer:

In the vaporisation of a liquid (liquid → vapour) \(\Delta S_{\text {system }} \text { is +ve. }\) This is because a substance in its vapour state possesses greater entropy than its liquid state.

Question 18. Consider the reaction, A → 2B, if the free energy per mole of A is GA and that of B is GB then what will the relation be between GA and GB when reaction 1 occurs spontaneously and 2 is at equilibrium?
Answer:

For a spontaneous reaction at a given temperature and pressure, ΔG < 0. Therefore, 2G_B< G_A

At a given temperature and pressure, for a reaction at equilibrium, AG = 0

Therefore, ΔG = 2G_B-G_A = 0 or, G_A = 2G_B

Question 19. “The amount of heat present in hot water is greater than that in cold water”—explain whether the statement is correct or not.
Answer:

1. The statement is wrong because heat can never be stored in any system as it is a form of energy In transit.
2. During a process beat appears at the boundary of a system. Heat does not exist before and after the process.

Question 20. Give examples of two processes involving only P-V work, where the system does not perform any work.
Answer:

1. During the expansion of a gas against zero external pressure, work done is zero.
2. If a process, involving only pressure-volume work, is carried out at constant volume, then work done in the process will be zero. For example, in the case of the vaporization of water in a closed container of fixed volume, the work done is zero

Question 21. A plant is growing. What do you think of the entropy changes of the plant and its surroundings?
Answer:

The entropy decreases during the growth of the plant (i.e., system) because the ordered structure of the plant is formed during 1(8 growth. However, the entropy of the surroundings increases during the process. The increase in entropy of the surroundings is much greater than the decrease in entropy of the system. As a result, the net entropy change of the system and its surroundings is always positive during the growth of a plant.

Question 22. When does an adiabatic process become isentropic?
Answer:

1. In a process, if the entropy of a system remains unchanged, then the process is called isoentropic.
2. In a reversible adiabatic process \(\delta q_{r e v}\) =0 So the entropy change \(d S=\frac{\delta q_{r e \nu}}{T}=0\) Therefore, a reversible adiabatic process is isentropic.

Question 23. mol of an ideal gas is freely expanded at a constant temperature. In this process, which of the quantities or quantities among w, q, AU, and AH are 0 or >0 or <0?
Answer:

1. During the isothermal free expansion of a gas, the work done is zero. So w = 0. Again internal energy and enthalpy remain the same during the isothermal expansion of an ideal gas. Therefore, ΔU = 0 and ΔH = 0.
2. According to the first law of thermodynamics, ΔU = q + w. For the given process ΔH = 0 and w = 0.
3. Therefore, q = 0. Thus, for isothermal free expansion of lmol of an illegal gas q = 0, w = 0, ΔU = 0, ΔH = 0.

Question 24. In process A →B →C → D, the heat absorbed by the system in steps A → B and B C are q₁ and q₂, respectively, and the heat released by the system in step C→ D is q₃. If q₁ + q₂ + q₃ = o, then will the process be adiabatic?
Answer:

In an adiabatic process, heat is not exchanged between a system and its surroundings at any stage of the process. In the given process, heat is being exchanged between the system and the surroundings. Thus, the process is not adiabatic though the sum of the amounts of heat absorbed and released is zero for the process.

Question 25. Why is infinite time required for the completion of an ideal reversible process?
Answer:

In an ideal reversible process, the system maintains equilibrium at every intermediate step and the process is extremely slow. Thus, from a theoretical point of view, an ideal reversible process should require an infinite time for Its completion.

Question 26. At 25°C, is the standard reaction enthalpy for the reaction 2H(g) + O(g) → H₂O(I) the same as the standard enthalpy of formation of H₂O(f)
Answer:

1. The given reaction does not indicate the formation reaction of H₂O(f) because the standard states of hydrogen and oxygen at 25°C are H₂(g) and O₂(g), respectively.
2. Hence, the standard reaction enthalpy of the given reaction is not the same as the standard enthalpy of the formation of H₂O(l).

Question 27. Which condition does not satisfy the spontaneity criteria of a reaction at constant temperature and pressure: ΔH<0, ΔS<0, ΔH>0, ΔS<0, ΔH>0, ΔH<0, ΔS>0?
Answer:

1. At constant temperature and pressure, a reaction will be spontaneous if ΔG = -vc for the reaction at that temperature and pressure.
2. If ΔG = +vc, the reaction will be nonspontaneous.
3. At a constant temperature and pressure, ΔG = ΔH- TΔS. If ΔH > 0 and ΔS<0, then ΔG = +i/e.
4. Thus, a reaction will be noil-spontaneous if ΔH > 0 and ΔS < 0.

Question 28. According to the definition of a thermodynamic system, which system do living beings belong to, and why?
Answer:

According to the definition of a thermodynamic system, every living being in nature belongs to an open system.

Explanation: All living beings (systems) take food (matter) from the surroundings and excrete waste materials (matter) to the surroundings. They also exchange heat (energy) with the surroundings.

Question 29. Comment on (lie thermodynamic stability of NO(g).
Given:

⇒ \(\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}(g); \Delta_r H^0=90 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒  \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g); \Delta_r H^0=-74 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)
Answer:

1. For the first reaction, refer to the standard enthalpy of formation (ΔrH°) for NO because 1 mol of NO forms from its constituent elements.
2. The positive value of Δr-Hº of a compound implies that the compound has enthalpy (or energy) than its constituent elements. Hence, the compound will be unstable.
3. Therefore, the positive value of ΔH° for the first reaction indicates that NO is unstable.

Question 30. Calculate the entropy change in the surroundings when 1.00 mol of ΔfHº(J) is formed under standard conditions. AH° = -286 kj.mol^-1
Answer:

For the given process

⇒ \(\Delta S_{\text {surr }}=-\frac{\Delta_f H^0}{T}=-\frac{-286 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{298 \mathrm{~K}}\)

= 959.73 J-K^-1– mol^-1

Question 31. For the reaction, 2A + B→C; ΔH = 400 kj. mol 1 & ΔS = 0.2 kj.K^-1 .mol^-1 at 298 K. At what temperature will the reaction become spontaneous considering ΔH, ΔS to be constant over the temperature range?
Answer:

We know, ΔG = ΔH- TΔS. For a spontaneous reaction at a given temperature and pressure ΔG < 0.

Given:

ΔH = 400 kj.mol^-1 and

ΔS = 0.2 kj. K^-1 mol^-1

So, ΔG = (400- T ×  0.2) kj. mol^-1

According to this relation, ΔG will be <0 when T × 0.2 > 400 i.e., T> 2000K.

Question 32. For the reaction 2Cl(g)→Cl₂(g), what are the signs of ΔH andΔS?
Answer:

The process involves the formation of a bond, which is always exothermic. Hence, AH < 0 for this process. The no. of gaseous particles decreases in the process. Consequently, the randomness of the system decreases. Hence, ΔS < 0 for this process.

Question 33. State the second law of thermodynamics based on entropy. The boiling point of ethanol is 78.4°C. The change in enthalpy during the vaporization of ethanol is 96 J- mol^-1. Calculate the change in entropy of vaporization of ethanol.
Answer:

We know,

⇒ \(\Delta S_{\text {vap }}\)

= \(\frac{\Delta H_{\text {vap }}}{T_b}\)

= \(\frac{96}{(273,+78.4)} \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

= \(0.2732 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

Question 34. For the following reaction at 298 K 2X + Y → Z , ΔH = 300 kj.mol^-1 and ΔS = 0.2 kj K^-1.mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
Answer:

For spontaneous process, ΔG < 0

∴ \(\Delta H-T \Delta S<0 \text { or, } \Delta H<T \Delta S_1 \text { or, } \frac{\Delta H}{\Delta S}<T\)

⇒ \(\text { or, } \frac{300}{0.2}<T \text { or, } 1500<T\)

Therefore, the given reaction becomes spontaneous above 1500 K temperature.

Question 35. What is meant by an isolated system?
Answer:

We know, ΔG = ΔH- TΔS

or, AG = (29.3- 298 ×104.1 ×10^-3) kj.mol^-1

= -1.7218 kj-mol^-1

At a particular pressure and 298 K temperature, the free energy change of the given reaction is negative which indicates the spontaneity of the reaction.

Question 36. The initial pressure, temperature & volume of 1 mol of gas are P1, T₁ and V₁ respectively. The state of the gas is changed in the following two ways. Will the internal energy change be the same in both cases?
Answer:

The internal energy of a system is a state function. The change in internal energy in a process depends only on the initial and final states of the system. It does not depend on the path used to arrive at the state. Since the initial and final states are the same in 1 and 2, the internal energy change will also be the same.

Question 37. If one mole of an ideal gas is expanded in the following two ways, then will the value of P₂ and P₂ be greater than, less than or equal to P₁?
Answer:

The temperature of the gas remains the same during isothermal expansion. Therefore, P₂ < P₁. On the other hand, the temperature of the gas decreases during an adiabatic expansion. Therefore, in the process, T₂ is less than T₁. In this process. P₁ < P₂ since T₂<T₁

Question 38. Why does the value of All for a chemical reaction depend on the physical states of the reactant (s) and produces)?
Answer:

In the case of solids and liquids, ΔH = ΔH for a chemical reaction (as AV is negligible here). If the participating substances are gases, then ΔH = ΔH + ΔnRT. Hence, AH for a chemical reaction depends on the physical states of the reactant(s) and product(s)

Question 39. Will the transformation ofice into water be spontaneous -2°C and latm pressure? Will the reverse process be spontaneous at this pressure and temperature?
Answer:

No. The transformation of ice into water at -2°C and 1 atm pressure is not spontaneous. This is because the sum ofthe increase in entropy of the system and the decrease in entropy of the surrounding is less than zero.

The reverse process, i.e., the transformation of water into ice is spontaneous. This is because at -2°C and 1 atm pressure the sum of die decreases in the entropy ofthe system and the increase in entropy of the surroundings is greater than zero.

Question 40. Heat is not exchanged between the system and its surroundings during the free expansion of an ideal gas. Therefore, in this process, q = 0. Will the change in entropy in this process be zero?
Answer:

1. In the free expansion of an ideal gas, no exchange of heat takes place between the system and its surroundings.
2. Because of the expansion, the volume of the gas increases, and the larger space is now available to the gas molecules for their movement.
3. This results in an increase in randomness in the system, and hence the entropy ofthe system increases.

Question 41. Why does the entropy of the gaseous system Increase with the temperature rise?
Answer:

1. Due to large intramolecular distance and weak intermolecular forces, the molecules in a gas can move about freely.
2. The motion ofthe molecules becomes more random and disordered with the rise in temperature as the average speed of the molecules increases.
3. Now, the entropy ofa system is a measure of the disorderliness of the constituent molecules. Therefore, the entropy ofa gas increases with the temperature rise.

Question 42. Give an example of a process for each of the given cases: ΔG = 0, ΔS <0, ΔG= 0, ΔS > 0 ΔG < 0, ΔS > 0 ΔG<0, ΔS<0 in a system.
Answer:

Fusion of ice at 0°C and 1 atm pressure.

Condensation of water vapour at 100°C and 1 atm.

⇒ \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \text { at } 25^{\circ} \mathrm{C} \text { and } 1 \mathrm{~atm}\)

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \text { at } 25^{\circ} \mathrm{C} \text { and } 1 \mathrm{~atm} .\)

Question 43. A particular amount of an ideal gas participates in a reversible process as given in the figure. What type of process is this? Explain the changes in each step.
Answer:

1. This process is cyclic because the system returns to its initial state after undergoing consecutive processes AB, BC and CA.
2. In step AB, the gas expands reversibly at constant pressure. p A Hence, step AB indicates an isobaric change. In step BC, the temperature of the gas decreases at constant volume.
3. Thus BC indicates an isochoric change.
4. In step CA, the gas is compressed reversibly at a constant temperature. Thus CA indicates a reversible isothermal change.

Question 44. The transformation of A to B can be carried out in the following two ways in which the initial and final states are identical.

What will be the value of ΔH during the transformation of C to B?

Answer:

According to Hess’s law, the change in enthalpy in the process

1 = total change in enthalpy in process  2

∴ -xkJ =-yKJ + ΔH = (y-x)kJ.

Question 45. Write three differences between reversible and irreversible processes. Melting office at 0°C and 1 atm pressure is a reversible process— explain.
Answer:

1. It is a reversible process. Ice melts at 0°C under normal atmospheric pressure. Latent heat for the fusion of Ice Is lit) cal .g-1, i.e., 80 cal of heat is required to melt logfile. If 80 cal heat IB is extracted from the surroundings, 1 g of ice gets converted Into water.
2. Therefore, at normal pressure and temperature, ice and water remain in an equilibrium state.
3. By Increasing or decreasing the value of the driving force (by the supply or extraction of heat) the process can be made to move In the forward or backward direction.
4. So, the melting of ice at normal atmospheric pressure and temperature is an example of a reversible process.

Question 46. The boiling point of benzene is 80.1 °C. At ordinary pressure and 70°C, the benzene vapour spontaneously transforms into liquid benzene. In this process, what will the signs of ΔH, ΔS and ΔG be?
Answer:

1. The entropy of the system decreases when a vapour transforms into a liquid.
2. So ΔS < 0. Again, the condensation is an exothermic process. So, in this process, ΔH < 0.
3. Under the given conditions, the benzene vapour spontaneously condenses into liquid. So, in this process ΔG < 0.

Question 47. What is meant by the terms change of entropy (ΔS) and change in free energy (ΔG) of a system? Write down the mathematical relation between them. At 0°C, liquid water and ice remain in equilibrium. If lg of liquid water under equilibrium conditions is converted to ice, explain with reason whether the process is endothermic or exothermic.
Answer:

1. In the conversion of water into ice, the entropy of the system decreases, and Hence AS_sys < 0
2. We know, ΔG = ΔH- TΔS
3. For the given process, ΔG< 0. As the process occurs spontaneously, ΔG < 0 for the process.
4. According to the relation (1), if ΔS<0, then AG will be negative only when ΔH < 0. So, the process is exothermic.

Question 48. Given: C(s) + O₂(g)→CO₂(g); ΔH = -393.5 kj 2H₂(g) + O₂(g)→2H₂O(g) ; ΔH = -571.6 kj. Calculate ΔH of the reaction:

⇒ \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g})\)

Answer:

C(s) +O₂(g)→CO₂(g); ΔH = -393.5 kj……………… (1)

2H₂(g) + O₂(g)→2H₂O(g) ; ΔH = -571.6 kJ……………… (2)

Subtracting equation (2) from equation (1), we have C(s) + 2H₂O(g)→CO₂(g) + 2H₂(g);

ΔH = [-393.5- (-571.6)] k] = 178.1J.

Question 49. Calculate the enthalpy of the formation of liquid ethyl alcohol from the following data.
Answer:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-1368 \mathrm{~kJ}\)

⇒ \(\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H=-393 \mathrm{~kJ}\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \( \Delta H=-287 \mathrm{~kJ}\)

Question 50. N₂(g) + 3H₂(g)Δ₂NH₃(g) ; ΔrH°=-92.4 kj.mol-1. What is the standard enthalpy of the formation of NH₃?
Answer:

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) ; \Delta_r H^0=-92.4 \mathrm{~kJ}\)

⇒ \(\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightarrow \mathrm{NH}_3(g) ; \Delta_r H^0=-46.2 \mathrm{~kJ}\)

Equation (1) represents the formation of NH₃(g) from the constituent elements. So, the standard enthalpy change for the reaction represented by equation (1) = the standard enthalpy of formation for NH₃(g) = -46.2 kj.mol^-1

Question 51. For the reaction, 2A(g) + B(g)→2D(g) ; ΔUº=-l0.5kJ and ASº = -44.11.K^-1 Calculate ΔG° for the reaction and predict whether it may occur spontaneously.
Answer:

The temperature has not been mentioned in the problem. Here, the calculation has been done by considering temperature as the normal temperature (298 K). For the reaction, An = 2- (2 + 1) = -1 . So, for this reaction,

ΔH° =ΔU° + ΔnRT =- 10.5-1 × 8.314 × 10^-3×298 kj.

=-12.98kj

We know, ΔG° = ΔHº -TΔSº

∴ AGº =(- 12.98 + 298 × 44.1 ×10^-3) kl =0.16kl

The positive value. of AG° indicates; that the cannot occur spontaneously.

Question 52. The equilibrium constant for a mission is 10. Find the value of AG0 t R a 0.814 MC ^–1mol^–1, T = 300K.
Answer:

We know, ΔG° w -2.303/log K

Given K=10 and T = 300k

∴ A (1° a 11.303 × 11.3 1 4 × 300 log10).mol^-1

=-5.74kJ.mol^-1