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**Q.6.1. ****Choose the correct answer. A thermodynamic state function is a quantity****(i) used to determine heat changes****(ii) whose value is independent of path****(iii) used to determine pressure volume work****(iv) whose value depends on temperature only.****Ans. **A thermodynamic state function is a quantity whose value is independent of a path.

Functions like P, V, T etc. depend only on the state of a system and not on the path. Hence, alternative (ii) is correct.**Q.6.2. ****For the process to occur under adiabatic conditions, the correct condition is:****(i)**** ΔT = 0(ii) Δp = 0(iii) q = 0(iv) w = 0**

Therefore, alternative (iii) is correct.

Therefore, alternative (ii) is correct.

Δ

⇒ Δ

Therefore, alternative (iii) is correct.

ΔH = -890.3 kJ mol

Thus, the desired equation is the one that represents the formation of CH

= [-393.5 + 2(-285.8) - (-890.3)]kJ mol^{-1}

=-74.8 kJ mol^{-1}

∴ Enthalpy of formation of CH_{4(g) }= -74.8 kJ mol^{-1}

Hence, alternative (i) is correct.

**Q.6.6. ****A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be**

Δ

According to the question, for the given reaction,

Δ

Δ

⇒ Δ

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Δ

Where,

Δ

Given,

Substituting the values in expression (i), we get

Δ

Δ

Hence, the change in internal energy for the given process is 307 J.

Δ

Where,

Δ

Δ

For the given reaction,

Δ

= (2 – 1.5) moles

Δ

And,

Δ

R = 8.314 × 10

Substituting the values in the expression of Δ

Δ

= –742.7 + 1.2

Δ

Where,

c = molar heat capacity

Δ

Substituting the values in the expression of

q = (60/27 mol)(24 mol

Total ΔH = Cp [H

= (75.3 J mol^{–1} K^{–1}) (0 – 10)K + (–6.03 × 10^{3} J mol^{–1}) + (36.8 J mol^{–1} K^{–1}) (–10 – 0)K

= –753 J mol^{–1} – 6030 J mol^{–1} – 368 J mol^{–1}

= –7151 J mol^{–1}

= –7.151 kJ mol^{–1}

Hence, the enthalpy change involved in the transformation is –7.151 kJ mol^{–1}.**Q.6.11. ****Enthalpy of combustion of carbon to CO _{2} is –393.5 kJ mol^{–1}. Calculate the heat released upon formation of 35.2 g of CO_{2} from carbon and dioxygen gas.**

C

(1 mole = 44 g)

Heat released on formation of 44 g CO_{2} = –393.5 kJ mol^{–1}

∴ Heat released on formation of 35.2 g CO_{2}

= –314.8 kJ mol^{–1}**Q. 6.12. ****Enthalpies of formation of CO _{(}_{g}_{), }CO_{2(}_{g}_{), }N_{2}O_{(}_{g}_{) }and N_{2}O_{4(}_{g}_{) }are –110 kJ mol^{–1}, – 393 kJ mol^{–1}, 81 kJ mol^{–1}and 9.7 kJ mol^{–1} respectively. Find the value of Δ_{r}H for the reaction:**

**N _{2}O_{4(}_{g}_{)} + 3CO_{(}_{g}_{)} N_{2}O_{(}_{g}_{)} + 3CO_{2(}_{g}_{)}**

Δ

For the given reaction,

N_{2}O_{4(}_{g}_{)} + 3CO_{(}_{g}_{)} → N_{2}O_{(}_{g}_{)} + 3CO_{2(}_{g}_{)}

Δ_{r}H = [{Δ_{f}H (N_{2}O) + 3Δ_{f}H (CO_{2})} - {Δ_{f}H (N_{2}O_{4}) + 3Δ_{f}H (CO)}]

Substituting the values of Δ_{f}*H* for N_{2}O, CO_{2}, N_{2}O_{4,} and CO from the question, we get:

Δ_{r}H = [{81 kJ mol^{-1} + 3(-393) kJ mol^{-1}} - {9.7 kJmol^{-1} + 3(-110)kJ mol^{-1}}]

Δ_{r}H = -777.7 kJ mol^{-1}

Hence, the value of Δ_{r}*H *for the reaction is -777.7 kJ mol^{-1}.**Q.6.13. ****Given****N _{2(g)} + 3H_{2(g)} → 2NH_{3(g)}**

Re-writing the given equation for 1 mole of NH

Standard enthalpy of formation of NH_{3(}_{g}_{)}

= ½ Δ_{r}*H*^{θ}

= ½ (–92.4 kJ mol^{–1})

= –46.2 kJ mol^{–1}**Q.6.14. ****Calculate the standard enthalpy of formation of CH _{3}OH_{(}_{l}_{)} from the following data:**

**C _{(}_{g}_{)} + O_{2(}_{g}_{)} CO_{2(}_{g}_{)} ; Δ_{c}H^{θ} = –393 kJ mol^{–1}**

**Ans. **The reaction that takes place during the formation of CH_{3}OH_{(}_{l}_{)} can be written as:

The reaction (**1**) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

Δ_{f}*H*^{θ} [CH_{3}OH_{(}_{l}_{)}] = Δ_{c}*H*^{θ }+ 2Δ_{f}*H*^{θ} [H_{2}O_{(}_{l}_{)}] – Δ_{r}*H*^{θ}

= (–393 kJ mol^{–1}) + 2(–286 kJ mol^{–1}) – (–726 kJ mol^{–1})

= (–393 – 572 + 726) kJ mol^{–1}

∴Δ_{f}*H*^{θ} [CH_{3}OH_{(}_{l}_{)}] = –239 kJ mol^{–1}**Q.6.15. ****Calculate the enthalpy change for the process ****CCl _{4(}_{g}_{)} → C_{(}_{g}_{)} + 4Cl_{(}_{g}_{) }**

(iv) C

Enthalpy change for the given process CCl

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

Δ

= (715.0 kJ mol

∴Δ

Bond enthalpy of C–Cl bond in CCl

= 326 kJ mol

Since Δ

Δ

Assuming the reaction at equilibrium, Δ

= ΔH/ΔS (ΔG = 0 at equilibrium)

*T* = 2000 K

For the reaction to be spontaneous, Δ*G* must be negative. Hence, for the given reaction to be spontaneous, *T* should be greater than 2000 K.**Q.6.18. ****For the reaction,****2Cl _{(}_{g)} → Cl_{2(}_{g}_{),} what are the signs of ΔH and ΔS ?**

The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, Δ

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, Δ

2 A

Δ

= –1 mole

Substituting the value of Δ

Δ

= (–10.5 kJ) – (–1) (8.314 × 10

= –10.5 kJ – 2.48 kJ

Δ

Substituting the values of Δ

Δ

= –12.98 kJ – (298 K) (–44.1 J K

= –12.98 kJ + 13.14 kJ

Δ

Since Δ

Δ

Δ

= (2.303) (8.314 JK

= –5744.14 Jmol

= –5.744 kJ mol

**Ans. **The positive value of Δ_{r}*H* indicates that heat is absorbed during the formation of NO_{(}_{g}_{)}. This means that NO_{(}_{g}_{)} has higher energy than the reactants (N_{2} and O_{2}). Hence, NO_{(}_{g}_{)} is unstable.

The negative value of Δ_{r}*H* indicates that heat is evolved during the formation of NO_{2(}_{g}_{)} from NO_{(}_{g}_{)} and O_{2(}_{g}_{)}. The product, NO_{2(}_{g}_{)} is stabilized with minimum energy.

Hence, unstable NO_{(}_{g}_{)} changes to stable NO_{2(}_{g}_{)}.

**Q.6.22. Calculate the entropy change in surroundings when 1.00 mol of H _{2}O_{(}_{l}_{)} is formed under standard conditions. Δ_{f}H^{θ} = –286 kJ mol^{–1}.**

Entropy change (Δ*S*_{surr}) for the surroundings = q_{surr}/7

Δ*S*_{surr} = 959.73 J mol^{–1} K^{–1 }

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