The free energy changes for the two reactions given below are: SO 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 2270 cal . S ( rhom ) + O 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 74060 cal . Find ∆ G for the reaction: S ( rhom ) + O 2 ( g ) → SO 2 ( g ) .

Given, SO 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 2270 cal ----- i S ( rhom ) + O 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 74060 cal ----- ii The free energy changes for two reactions are given above. To calculate Gibbs free energy, Eq ii - Eq i , = ∆ G ii - ∆ G i . = - 74060 - - 2270 cal . = - 71790 cal . Here, the free energy change value is negative. i.e. it is also a spontaneous process.

The free energy changes for the two reactions given below are: SO 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 2270 cal . S ( rhom ) + O 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 74060 cal . Find ∆ G for the reaction: S ( rhom ) + O 2 ( g ) → SO 2 ( g ) .

Given, SO 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 2270 cal ----- i S ( rhom ) + O 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 74060 cal ----- ii The free energy changes for two reactions are given above. To calculate Gibbs free energy, Eq ii - Eq i , = ∆ G ii - ∆ G i . = - 74060 - - 2270 cal . = - 71790 cal . Here, the free energy change value is negative. i.e. it is also a spontaneous process.

At what temperature, reduction of lead oxide to lead by carbon, i.e., PbO ( s ) + C ( s ) → Pb ( s ) + CO ( g ) becomes spontaneous. For the reaction, ∆ H and ∆ S are 108 . 4 kJ mol - 1 and 190 . 0 J K - 1 mol - 1 respectively.

Given, PbO ( s ) + C ( s ) → Pb ( s ) + CO ( g ) . ∆ H = 108 . 4 kJ mol - 1 , ∆ S = 190 J K - 1 mol - 1 . At equilibrium, the free energy change, ∆ G = 0 . Formula Used, ∆ G = ∆ H - T ∆ S . 0 = ∆ H - T ∆ S . ∆ H = T ∆ S . T = ∆ H ∆ S . T = 108400 J mol - 1 190 J K - 1 mol - 1 = 570 . 5 K . (Above 570 .5 K , ∆ G , the free energy will be negative and reaction will be spontaneous).

For the reaction: 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) , the enthalpy and entropy changes are - 113 · 0 kJ mol - 1 and - 145 J K - 1 mol - 1 respectively. Find the temperature below which this reaction is spontaneous.

Given, 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) . ∆ H = - 113 . 0 kJ mol - 1 , ∆ S = - 145 J K - 1 mol - 1 . At equilibrium, ∆ G = 0 . Formula Used, ∆ G = ∆ H - T ∆ S . T = ∆ H ∆ S . T = - 113 × 1000 - 145 T = 779 .3 K . (Below 779 .3 K , ∆ G will be negative and reaction will be spontaneous).

A chemist claims that the following reaction is feasible at 298 K . SF 6 ( g ) + 8 Hl ( g ) → H 2 S g + 6 HF ( g ) + 4 I 2 ( s ) . Verify his claim. Given that ∆ G ° f for SF 6 ( g ) , Hl ( g ) , H 2 S ( g ) and HF ( g ) are - 991 . 61 , 1 . 30 , - 33 . 01 and - 270 . 73 kJ mo 1 - 1 respectively.

Given reaction is, SF 6 ( g ) + 8 Hl ( g ) → H 2 S g + 6 HF ( g ) + 4 I 2 ( s ) . Given that ∆ G for, SF 6 = - 991 . 61 kJ mol - 1 . HI = 1 . 30 kJ mol - 1 . H 2 S = - 33 . 01 kJ mol - 1 . HF = - 270 . 73 kJ mol - 1 . For the reaction to be feasible, depends on the sign of the free energy change. If ∆ G = positive, the reaction is not feasible. If ∆ G = negative, the reaction is feasible. For finding the free energy change, ∆ G = ∆ H - T ∆ S . ∆ G = ∆ G H 2 S + 6 ∆ G ( HF ) - ∆ G SF 6 + 8 ∆ G HI ∆ G = - 33 . 01 + 6 ( - 270 . 73 ) - - 991 . 61 + 8 1 . 30 ∆ G = - 1657 . 39 - - 981 . 21 ∆ G = - 676 . 18 kJ mol 1 . As the free energy, ∆ G is negative, the reaction is feasible.

A chemist claims that the following reaction is feasible at 298 K . SF 6 ( g ) + 8 Hl ( g ) → H 2 S g + 6 HF ( g ) + 4 I 2 ( s ) . Verify his claim. Given that ∆ f G ° for SF 6 ( g ) , Hl ( g ) , H 2 S ( g ) and HF ( g ) are - 991 . 61 , 1 . 30 , - 33 . 01 and - 270 . 73 kJ mo 1 - 1 respectively.

Here given reaction is, SF 6 ( g ) + 8 HI ( g ) → H 2 S g + 6 HF ( g ) + 4 I 2 ( s ) . Given that ∆ G for SF 6 = - 991 . 61 kJ mol - 1 . HI = 1 . 30 kJ mol - 1 . H 2 S = - 33 . 01 kJ mol - 1 . HF = - 270 . 73 kJ mol - 1 . For the reaction to be feasible, depends on the sign of free energy. If ∆ G = positive, the reaction is not feasible. If ∆ G = negative, the reaction is feasible. For finding the free energy change of the chemical reaction is given below, ∆ G = ∆ H - T ∆ S . ∆ G = ∆ G H 2 S + 6 ∆ G ( HF ) - ∆ G SF 6 + 8 ∆ G HI . ∆ G = - 33 . 01 + 6 ( - 270 . 73 ) - - 991 . 61 + 8 1 . 30 . ∆ G = - 1657 . 39 - - 981 . 21 . ∆ G = - 676 . 18 kJ mol - 1 . The ∆ G is negative. So, the reaction is feasible in nature.

HARD

11th CBSE

IMPORTANT

Earn 100

The free energy changes for the two reactions given below are:

${SO}_{2} (g) + {Cl}_{2} (g) \to {SO}_{2} {Cl}_{2} (g), ∆ G = - 2270 cal$ .

$S (rhom) + O_{2} (g) + {Cl}_{2} (g) \to {SO}_{2} {Cl}_{2} (g), ∆ G = - 74060 cal$ .

Find $∆ G$ for the reaction: $S (rhom) + O_{2} (g) \to {SO}_{2} (g)$ .

Important Questions on Thermodynamics

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 8

At what temperature, reduction of lead oxide to lead by carbon, i.e., $PbO (s) + C (s) \to Pb (s) + CO (g)$ becomes spontaneous. For the reaction, $∆ H$ and $∆ S$ are $108.4 kJ {mol}^{- 1}$ and $190.0 J K^{- 1} {mol}^{- 1}$ respectively.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 9

For the reaction:

2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)

, the enthalpy and entropy changes are

- 113 \cdot 0 kJ {mol}^{- 1}

and

- 145 J K^{- 1} {mol}^{- 1}

respectively. Find the temperature below which this reaction is spontaneous.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 10

For a hypothetical reaction,

X \to Y

, the enthalpy and entropy changes are

46.3 kJ {mol}^{- 1}

and

108.80 J K^{- 1} {mol}^{- 1}

respectively. Find the temperature above which this reaction is spontaneous.

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 11

The standard free energy change for a reaction is

- 212.3 kJ {mol}^{- 1}

. If the enthalpy change of the reaction is

- 216.7 kJ {mol}^{- 1}

, calculate the entropy change for the reaction.

(T = 25 ° C = 298 K)

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 12

For the reaction

N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)

ΔH = - 95.4 kJ

and

ΔS = - 198.300 J K^{- 1}

. Calculate the temperature in centigrade at which it attains equilibrium.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 13

A chemist claims that the following reaction is feasible at $298 K .$

${SF}_{6} (g) + 8 Hl (g) \to H_{2} S (g) + 6 HF (g) + 4 I_{2} (s)$ .

Verify his claim. Given that $∆ G °_{f}$ for ${SF}_{6} (g), Hl (g), H_{2} S (g)$ and $HF (g)$ are $- 991.61, 1.30, - 33.01$ and $- 270.73 kJ mo 1^{- 1}$ respectively.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 1

Calculate the standard free energy change for the reaction:

{Fe}_{2} O_{3} (s) + 4 H_{2} (g) \to 2 Fe (s) + 3 H_{2} O (l)

. Given that the standard free energies of formation of

{Fe}_{2} O_{3}

and

H_{2} O

are

- 741 \cdot 0

and

- 237 \cdot 2 kJ {mol}^{- 1}

respectively. Can the above reaction occur at the standard state?

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 2

A chemist claims that the following reaction is feasible at $298 K$ .

${SF}_{6} (g) + 8 Hl (g) \to H_{2} S (g) + 6 HF (g) + 4 I_{2} (s)$ .
Verify his claim. Given that $∆_{f} G °$ for ${SF}_{6} (g), Hl (g), H_{2} S (g)$ and $HF (g)$ are $- 991.61, 1.30, - 33.01$ and $- 270.73 kJ mo 1^{- 1}$ respectively.

The free energy changes for the two reactions given below are: SO2(g)+Cl2(g)→SO2Cl2(g), ∆G=-2270 cal. S(rhom)+O2(g)+Cl2(g)→SO2Cl2(g), ∆G=-74060 cal. Find ∆G for the reaction: S(rhom)+O2(g)→SO2(g).

Important Questions on Thermodynamics

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