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 .

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.

Calculate the standard entropy change for the following reaction: H 2 g + Cl 2 g → 2 HCl g at 298 K . Given, S ° H 2 = 131 J K - 1 mol - 1 , S ° Cl 2 = 223 J K - 1 mol - 1 , S ° HCl = 187 J K - 1 mol - 1 .

Given, H 2 g + Cl 2 g → 2 HCl g . The standard entropy of the reactants and the products are given as, S ° H 2 = 131 J K - 1 mol - 1 , S ° Cl 2 = 223 J K - 1 mol - 1 , S ° HCl = 187 J K - 1 mol - 1 . The formula used to find the entropy change is, ∆ S ° = 2 S ° HCl - S ° H 2 + S ° Cl 2 . ∆ S ° = 2 × 187 - 131 + 223 . ∆ S ° = 374 - 354 = 20 J K - 1 mol - 1 .

Calculate the standard Gibbs' energy change for the formation of propane at 298 K . 3 C graphite + 4 H 2 g → C 3 H 8 g . ∆ f H ° for propane, C 3 H 8 ( g ) , is - 103 . 8 kJ mol - 1 . Given, S m ° C 3 H 8 g = 270 . 2 J K - 1 mol - 1 , S m ° C graphite = 5 . 70 J K - 1 mol - 1 and S m ° H 2 = 130 . 7 J K - 1 mol - 1 .

Given, 3 C graphite + 4 H 2 g → C 3 H 8 g , ∆ f H ° = - 103 . 8 kJ mol - 1 . S ° C 3 H 8 = 270 . 2 J K - 1 mol - 1 , S ° C = 5 . 70 J K - 1 mol - 1 , S ° H 2 = 130 . 7 J K - 1 mol - 1 . Formula Used, ∆ S ° = S ° C 3 H 8 - { 3 S ° C + 4 S ° H 2 } . ∆ S ° = 270.2 - 3 × 5.70 + 4 × 130.7 . ∆ S ° = 270.2 - 17.1 + 522.8 = 270.2 - 539.9 = - 269.7 J K - 1 mol - 1 . Using formula, ∆ G ° = ∆ H ° - T ∆ S ° ∆ G ° = - 103 .8 - 298 × - 269 .7 . ∆ G ° = - 103800 + 80370 .6 = - 23429 .4 . ∆ G ° = - 23 . 429 kJ mol - 1 .

HARD

11th CBSE

IMPORTANT

Earn 100

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

Important Questions on Thermodynamics

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}

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 $- 13 \cdot 6 kJ {mol}^{- 1}$ .

MEDIUM

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,

$H_{2} (g) + l_{2} (g) \to 2 HI (g), dH ° = + 51 \cdot 9 kJ$ .

Given that the standard entropies ( $S °$ ) of $H_{2}, I_{2}$ and $HI$ are $130.6, 116.7$ and $206.3 J K^{- 1} {mol}^{- 1}$ respectively. Predict whether the reaction is feasible at the standard state or not.

MEDIUM

11th CBSE

IMPORTANT

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

Calculate the standard entropy change for the following reaction:

$H_{2} (g) + {Cl}_{2} (g) \to 2 HCl (g) at 298 K$ .

Given, $S °_{H_{2}} = 131 J K^{- 1} {mol}^{- 1}, S °_{{Cl}_{2}} = 223 J K^{- 1} {mol}^{- 1}, S °_{HCl} = 187 J K^{- 1} {mol}^{- 1} .$

MEDIUM

11th CBSE

IMPORTANT

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

Calculate the standard Gibbs' energy change for the formation of propane at $298 K .$

$3 C (graphite) + 4 H_{2} (g) \to C_{3} H_{8} (g)$ .

$∆_{f} H °$ for propane, $C_{3} H_{8} (g)$ , is $- 103.8 kJ {mol}^{- 1}$ .

Given, ${S_{m}}^{°}_{C_{3} H_{8} (g)} = 270.2 J K^{- 1} {mol}^{- 1}$ , ${S_{m}}^{°}_{C (graphite)} = 5.70 J K^{- 1} {mol}^{- 1}$ and ${S_{m}}^{°}_{(H_{2})} = 130.7 J K^{- 1} {mol}^{- 1} .$

MEDIUM

11th CBSE

IMPORTANT

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

For oxidation of iron,

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

, entropy change is

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

298 K

. Inspite of the negative entropy change of this reaction, why is the reaction spontaneous? (

∆ H °

for this reaction is

- 1648 \times 10^{3} J {mol}^{- 1}

EASY

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>TEST YOUR GRIP>Q 1

Which one of the following pairs does not represent example for intensive property?

Calculate the equilibrium constant for the following reaction at 298 K and 1 atm pressure: NOg+12 O2g⇌NO2g. Given:∆fH° at 298 K. For NO(g)=90·4 kJ mol-1; For NO2(g)=33·8 kJ mol-1. ∆S° at 298 K for the reaction =-70.8 J K-1mol-1. Gas constant, R=8.314 J K-1mol-1.

Important Questions on Thermodynamics

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