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.

The value of K p for the water gas reaction, CO + H 2 O → CO 2 + H 2 , is 1 . 06 × 10 5 at 25 ° C . Calculate the standard free energy change for the reaction at 25 ° C .

Given, CO + H 2 O → CO 2 + H 2 . K p = 10 . 6 × 10 5 . T = 25 ° C = 298 K . ∆ r G = ? Using the following formula, the standard free energy change can be calculated. ∆ r G = - 2 .303 RT log K p . ∆ r G = - 2 .303 × 8 .314 × 298 × log 10 . 6 × 10 5 . ∆ r G = - 2 .303 × 8 .314 × 298 × 6 .02 = - 34 . 35 kJ mol - 1 . The standard free energy change of the given reaction is - 34 . 35 kJ mol - 1 .

Calculate the equilibrium constant for the following reaction at 298 K and 1 atm pressure: NO g + 1 2 O 2 g ⇌ NO 2 g . Given: ∆ f H ° at 298 K . For NO ( g ) = 90 · 4 kJ mol - 1 ; For NO 2 ( g ) = 33 · 8 kJ mol - 1 . ∆ S ° at 298 K for the reaction = - 70 . 8 J K - 1 mol - 1 . Gas constant, R = 8 . 314 J K - 1 mol - 1 .

Given: NO g + 1 2 O 2 g ⇌ NO 2 g . T = 298 K . ∆ f H NO = 90 . 4 kJ mol - 1 ; ∆ f H NO 2 = 33 . 8 kJ mol - 1 ; ∆ S = - 70 . 8 J K - 1 mol - 1 ; R = 8 . 314 J K - 1 mol - 1 . Using the formula for enthalpy change, ∆ r H = ∆ f H NO 2 - ∆ f H NO + 1 2 ∆ f H O 2 . ∆ r H = 33 . 8 - 90 . 4 + 0 . ∆ r H = 33 . 8 - 90 . 4 = - 56 . 6 kJ mol - 1 . ∆ r H = - 56600 kJ mol - 1 . Using the formula for free energy change, ∆ G = ∆ H - T ∆ S . ∆ G = - 56600 - 298 × - 70 . 8 . ∆ G = - 56600 + 20979 . 2 = - 35501 . 6 J mol - 1 . Using the formula for equilibrium constant, ∆ r G = - 2 .303 × RT × log K . - 35501 . 6 = - 2 . 303 × 8 . 314 × 298 × log K ⇒ - 5703 . 1 × log K . log K = - 35501 . 6 - 5703 . 1 = 6 . 224 . K = 1 . 679 × 10 6 .

Using the data given below, calculate the value of equilibrium constant for the reaction, 3 HC ≡ CH g ⇌ C 6 H 6 g Acetylene benzene at 298 K , assuming ideal gas behaviour ∆ f G ° for HC = CH ( g ) = 2 · 09 × 10 5 J mol - 1 ; ∆ G ° for C 6 H 6 ( g ) = 1 . 24 × 10 5 J mol - 1 ; R = 8 · 314 J K - 1 mol - 1 . Based on your calculated value, comment whether this process can be recommended as a practical method for making benzene.

Given: 3 HC ≡ CH ( g ) ⇌ C 6 H 6 ( g ) . ∆ f G ° HC ≡ CH = 2 .09 × 10 5 J mol - 1 . ∆ f G ° C 6 H 6 = 1 .24 × 10 5 J mol - 1 . R = 8 . 314 J K - 1 mol - 1 . Using the formula for standard free energy change, ∆ r G ° = ∆ f G ° C 6 H 6 - 3 × ∆ f G ° HC ≡ CH . ∆ r G ° = 1 . 24 × 10 5 - 3 × 2 . 09 × 10 5 . ∆ r G ° = 1 . 24 × 10 5 - 6 . 27 × 10 5 . ∆ r G ° = - 5 . 03 × 10 5 J mol - 1 . ∆ r G ° = - 2 .303 × RT × log K . - 5 . 03 × 10 5 = - 2 . 303 × 8 . 314 × 298 × log K . log K = - 5 . 03 × 10 5 - 2 . 303 × 8 . 314 × 298 = 503000 5705 . 8 = 88 . 15 . Because value of K is very high and ∆ r G ° is negative. Therefore, the process can occur easily. Based on the calculated value, we can recommend this process as a practical method for making benzene.

S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE

Author:S C Kheterpal, S N Dhawan & P N Kapil

S C Kheterpal Chemistry Solutions for Exercise - S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE

Attempt the practice questions on Chapter 7: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE with hints and solutions to strengthen your understanding. Pradeep's Chemistry Vol 1 solutions are prepared by Experienced Embibe Experts.

Questions from S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE with Hints & Solutions

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.

EASY

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 3

The value of $K_{p}$ for the water gas reaction, $CO + H_{2} O \to {CO}_{2} + H_{2}$ , is $1.06 \times 10^{5}$ at $25 ° C$ . Calculate the standard free energy change for the reaction at $25 ° C$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 4

Calculate the equilibrium constant for the following reaction at $298 K$ and $1 atm$ pressure:

$NO (g) + \frac{1}{2} O_{2} (g) ⇌ {NO}_{2} (g)$ .

Given: $∆_{f} H °$ at $298 K$ .

For $NO (g) = 90 \cdot 4 kJ {mol}^{- 1}$ ; For ${NO}_{2} (g) = 33 \cdot 8 kJ {mol}^{- 1}$ .

$∆ S °$ at $298 K$ for the reaction $= - 70.8 J K^{- 1} {mol}^{- 1}$ .

Gas constant, $R = 8.314 J K^{- 1} {mol}^{- 1}$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 5

Calculate the standard entropy change for the reaction $X ⇌ Y$ , if the value of $∆ H ° = 28.40 kJ$ and equilibrium constant is $1 \cdot 8 \times 10^{- 7}$ at $298 K$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 6

Using the data given below, calculate the value of equilibrium constant for the reaction,

$3 HC \equiv CH (g) ⇌ C_{6} H_{6} (g) Acetylene benzene$

at $298 K$ , assuming ideal gas behaviour $∆_{f} G °$ for $HC = CH (g) = 2 \cdot 09 \times 10^{5} J {mol}^{- 1}$ ; $∆ G °$ for $C_{6} H_{6} (g) = 1.24 \times 10^{5} J {mol}^{- 1}$ ; $R = 8 \cdot 314 J K^{- 1} {mol}^{- 1}$ .

Based on your calculated value, comment whether this process can be recommended as a practical method for making benzene.

MEDIUM

11th CBSE

IMPORTANT

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

Find out the value of equilibrium constant for the following reaction at $298 K$ :

$2 {NH}_{3} (g) + {CO}_{2} (g) ⇌ {NH}_{2} {CONH}_{2} (aq) + H_{2} O (l)$ .

Standard Gibbs energy change $∆_{f} G °$ at the given temperature is $- 13 \cdot 6 kJ {mol}^{- 1}$ .

S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE

S C Kheterpal Chemistry Solutions for Exercise - S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE

Questions from S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Thermodynamics, Exercise 13: PROBLEMS FOR PRACTICE with Hints & Solutions

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