At 500 K, the equilibrium constant, Kc, for the following reaction is 5.

\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) \rightleftharpoons HI (g)
What would be the equilibrium constant K_{c} for the reaction
2HI (g)\rightleftharpoons H_{2} (g) + I_{2} (g)

(i) 0.04

(ii) 0.4

(iii) 25

(iv) 2.5

Answers (1)

The answer is the option (i) 0.04

Explanation: If the given equation is multiplied by 2, the equilibrium constant for the new equation will be squared, and on reversing the reaction the value of the equilibrium constant is reciprocated. Thus

 K=5_{2}=25

For the required reaction equation K=\frac{1}{25}=0.04

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions