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

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

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infoexpert22

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