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**Ques 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.**Ques 6.2: ****For the process to occur under adiabatic conditions, the correct condition is:****(i)**** Î”T = 0(ii) Î”p = 0(iii) q = 0(iv) w = 0**

Therefore, alternative (iii) is correct.

The enthalpy of all elements in their standard state is zero.

Therefore, alternative (ii) is correct.

Since Î”

Î”

â‡’ Î”

Therefore, alternative (iii) is correct.

According to the question,

Thus, the desired equation is the one that represents the formation of CH_{4} _{(}_{g}_{)} i.e.,

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

=-74.8 kJ mol^{-1}

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

Hence, alternative (i) is correct.

**Ques 6.6: ****A reaction, A + B â†’ C + D + q is found to have a positive entropy change. The reaction will be**

For a reaction to be spontaneous, Î”

Î”

According to the question, for the given reaction,

Î”

Î”

â‡’ Î”

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

According to the first law of thermodynamics,

Î”

Where,

Î”

Given,

Substituting the values in expression (i), we get

Î”

Î”

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

Enthalpy change for a reaction (Î”

Î”

Where,

Î”

Î”

For the given reaction,

Î”

= (2 â€“ 1.5) moles

Î”

And,

Î”

R = 8.314 Ã— 10

Substituting the values in the expression of Î”

Î”

= â€“742.7 + 1.2

Î”

From the expression of heat (

Where,

c = molar heat capacity

Î”

Substituting the values in the expression of

*q* = 1066.7 J*q* = 1.07 kJ**Ques 6.10: ****Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0Â°C to ice at â€“10.0Â°C. Î” _{fus}H = 6.03 kJ mol^{â€“1} at 0Â°C.**

Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10Â°C to 1 mol of water at 0Â°C.

(b) Energy change involved in the transformation of 1 mol of water at 0Â° to 1 mol of ice at 0Â°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0Â°C to 1 mol of ice at â€“10Â°C.

= (75.3 J mol^{â€“1} K^{â€“1}) (0 â€“ 10)K + (â€“6.03 Ã— 10^{3} J mol^{â€“1}) + (36.8 J mol^{â€“1} K^{â€“1}) (â€“10 â€“ 0)K

= â€“753 J mol^{â€“1} â€“ 6030 J mol^{â€“1} â€“ 368 J mol^{â€“1}

= â€“7151 J mol^{â€“1}

= â€“7.151 kJ mol^{â€“1}

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

Formation of CO

(1 mole = 44 g)

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

Heat released on formation of 35.2 g CO_{2}

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

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

Î”

For the given reaction,

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

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

Hence, the value of Î”_{r}*H *for the reaction is.

**Ques 6.13:****Given****; Î” _{r}H^{Î¸} = â€“92.4 kJ mol^{â€“1}**

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.

Re-writing the given equation for 1 mole of NH

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

= Â½ Î”_{r}*H*^{Î¸}

= Â½ (â€“92.4 kJ mol^{â€“1})

= â€“46.2 kJ mol^{â€“1}**Ques 6.14:****Calculate the standard enthalpy of formation of CH _{3}OH_{(}_{l}_{)} from the following data:**

**Ans **

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

C_{(}_{s}_{)} + 2H_{2}O_{(}_{g}_{)} + O_{2(}_{g}_{)} CH_{3}OH_{(}_{l}_{)} (**1**)

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

Equation (ii) + 2 Ã— equation (iii) â€“ equation (i)

Î”_{f}*H*^{Î¸} [CH_{3}OH_{(}_{l}_{)}] = Î”_{c}*H*^{Î¸ }+ 2Î”_{f}*H*^{Î¸} [H_{2}O_{(}_{l}_{)}] â€“ Î”_{r}*H*^{Î¸}

= (â€“393 kJ mol^{â€“1}) + 2(â€“286 kJ mol^{â€“1}) â€“ (â€“726 kJ mol^{â€“1})

= (â€“393 â€“ 572 + 726) kJ mol^{â€“1}

Î”_{f}*H*^{Î¸} [CH_{3}OH_{(}_{l}_{)}] = â€“239 kJ mol^{â€“1}**Ques 6.15: ****Calculate the enthalpy change for the process ****CCl _{4(}_{g}_{)} â†’ C_{(}_{g}_{)} + 4Cl_{(}_{g}_{) }**

The chemical equations implying to the given values of enthalpies are:

Î”

Î”

Î”

Î”

Enthalpy change for the given process can be calculated using the following algebraic calculations as:

Equation (ii) + 2 Ã— Equation (iii) â€“ Equation (i) â€“ Equation (iv)

Î”

= (715.0 kJ mol

Î”

Bond enthalpy of Câ€“Cl bond in CCl

= 326 kJ mol

**Ques 6.16: For an isolated system, Î” U = 0, what will be Î”S?**

Î”S will be positive i.e., greater than zero

Since Î”

**Ques 6.17: For the reaction at 298 K,****2A + B â†’ C****Î” H = 400 kJ mol^{â€“1} and Î”S = 0.2 kJ K^{â€“1} mol^{â€“1}**

**Ans:**

From the expression,

Î”*G* = Î”*H* â€“ *T*Î”*S*

Assuming the reaction at equilibrium, Î”*T* for the reaction would be:

(Î”*G* = 0 at equilibrium)

*T* = 2000 K

For the reaction to be spontaneous, Î”*G* must be negative. Hence, for the given reaction to be spontaneous, *T* should be greater than 2000 K.**Ques 6.18:****For the reaction,****2Cl _{(}_{g)} â†’ Cl_{2(}_{g}_{),} what are the signs of Î”H and Î”S ?**

Î”

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

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

For the given reaction,

2 A

Î”

= â€“1 mole

Substituting the value of Î”

Î”

= (â€“10.5 kJ) â€“ (â€“1) (8.314 Ã— 10

= â€“10.5 kJ â€“ 2.48 kJ

Î”

Substituting the values of Î”

Î”

= â€“12.98 kJ â€“ (298 K) (â€“44.1 J K

= â€“12.98 kJ + 13.14 kJ

Î”

Since Î”

From the expression,

Î”

Î”

= (2.303) (8.314 JK

= â€“5744.14 Jmol

= â€“5.744 kJ mol

**Ans:**

The positive value of Î”_{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 Î”_{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}_{)}.

**Ques 6.22:Calculate the entropy change in surroundings when 1.00 mol of H _{2}O_{(}_{l}_{)} is formed under standard conditions. Î”_{f}H^{Î¸} = â€“286 kJ mol^{â€“1}.**

It is given that 286 kJ mol

Entropy change (Î”*S*_{surr}) for the surroundings

Î”*S*_{surr} = 959.73 J mol^{â€“1} K^{â€“1 }

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