The equilibrium constant for the reaction- 12H2g - 12I2g - HIg is Kc Equilibrium constant for the reaction-2HIg - H2g - I2g will be

The given reaction is, 12H_2(g) + 12I_2(g) ⇌ HI(g) ....eqn(i) nbsp; Hence, K_c = [HI][H_2]^1/2[I_2]^1/2 .....eqn(ii) Now, reverse the eqn(i) and multiply b ...

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Standard XII

Chemistry

Introduction

The equilibri...

Question

The equilibrium constant for the reaction,
$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ H I (g) is K_{c}$
Equilibrium constant for the reaction,
$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ will be

\frac{1}{K_{c}}

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\frac{1}{(K_{c})^{2}}

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\frac{2}{K_{c}}

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\frac{2}{(K_{c})^{2}}

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Solution

The correct option is B $\frac{1}{(K_{c})^{2}}$
The given reaction is,
$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ H I (g) . . . . e q n (i)$
Hence, $K_{c} = \frac{[H I]}{[H_{2}]^{1 / 2} [I_{2}]^{1 / 2}} . . . . . e q n (i i)$

Now, reverse the eqn(i) and multiply by 2, we get
$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$
Hence, $K_{c}^{'} = \frac{[H_{2}] [I_{2}]}{[H I]^{2}} . . . . . e q n (i i i)$
Equating equation (ii) and (iii), we get
$K_{c}^{'} = \frac{1}{(K_{c})^{2}}$

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The equilibrium constant for the reaction, 12H2(g)+12I2(g)⇌HI(g) is Kc Equilibrium constant for the reaction, 2HI(g)⇌H2(g)+I2(g) will be

The correct option is B 1(Kc)2The given reaction is, 12H2(g)+12I2(g)⇌HI(g) ....eqn(i) Hence, Kc=[HI][H2]1/2[I2]1/2 .....eqn(ii) Now, reverse the eqn(i) and multiply by 2, we get 2HI(g)⇌H2(g)+I2(g) Hence, K′c=[H2][I2][HI]2 .....eqn(iii) Equating equation (ii) and (iii), we get K′c=1(Kc)2

The equilibrium constant for the reaction,
$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ H I (g) is K_{c}$
Equilibrium constant for the reaction,
$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ will be