You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction to Le Chatelier's Principle

Increase in t...

Question

Increase in temperature favors the forward reaction in:

2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 O 3 ⇋ 3 O 2 (g)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C (g r a p h i t e) ⇋ C (d i a m o n d)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2} (g) + \frac{1}{2} O_{2} (g) ⇋ H_{2} O (g)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $C (g r a p h i t e) ⇋ C (d i a m o n d)$
The equilibrium reaction $C (g r a p h i t e) ⇋ C (d i a m o n d)$ is an endothermic reaction. When the temperature is increased, the equilibrium shifts in the forward direction so that more heat is absorbed. This nullifies the effect of an increase in temperature.

Suggest Corrections

Similar questions

Δ S

H_{2} (g) ⟶ 2 H (g)

N_{2} (g, 1 a t m) ⟶ N_{2} (g, 5 a t m)

2 S O_{3} (g) ⟶ 2 S O_{2} (g) + O_{2} (g)

C (d i a m o n d) ⟶ C (g r a p h i t e)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)

Δ H

k J

C (g) + O_{2} (g) \to C O_{2} (g)

H_{2} O (g) + C (g) \to C O (g) + H_{2} (g)

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H = - 242 k J

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.2 k J

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l); Δ H = - 1410.8 k J

S (s) + O_{2} (g) ⇌ {S O}_{2} (g); K_{c 1} = 5 \times 10^{52}

2 S (s) + 3 O_{2} (g) ⇌ 2 S O_{3} (g); K_{c 2} = 10^{29}

K_{c}

2 {S O}_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

Join BYJU'S Learning Program