Ncert Exercises & Solutions

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

$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ HI (g)$

What would be the equilibrium constant K_c for the reaction?

$2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$

1. 0.04

2. 0.4

3. 25

4. 2.5

Hint:

K_{c} = \frac{[HI]}{{[H_{2}]}^{1 / 2} {[I_{2}]}^{1 / 2}}

Step 1:

The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward

direction

For the reaction,

\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ HI (g)

K_{c} = \frac{[HI]}{{[H_{2}]}^{1 / 2} {[I_{2}]}^{1 / 2}} = 5

Step 2:

Thus, for the reaction,

2 HI (g) ⇌ H_{2} (g) + I_{2} (g)

K_{c_{1}} = \frac{[H_{2}] [I_{2}]}{{[HI]}^{2}} = {(\frac{1}{K_{c}})}^{2} = {(\frac{1}{5})}^{2} = \frac{1}{25} = 0.04