For the reaction
$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
and $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {N H}_{3}$
write down the expression for equilibrium constant $K_{c}$ and $K_{c}^{'}$ . How is $K_{c}$ related to $K_{c}^{'}$ ?

Open in App

Solution

For the equation $N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
$K_{c} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} . . . (1)$
For equation $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {N H}_{3}$
$K_{c}^{'} = \frac{{[{N H}_{3}]}}{{[N_{2}]}^{1 / 2} {[H_{2}]}^{3 / 2}} . . . (2)$
Taking square of the expression for $K_{c}^{'}$ , we get
${(K_{c}^{'})}^{2} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
This is equal to the expression for $K_{c}$ .
Thus, $K_{c} = {(K_{c}^{'})}^{2}$
or $\sqrt{K_{c}} = K_{c}^{'}$

Suggest Corrections

Similar questions

K_{c}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

7

2 N_{2} + 6 H_{2} ⇌ 4 N H_{3}

K_{c}

2 N H_{3} ⇌ N_{2} + 3 H_{2}

The equilibrium constant $K_{c}$ for the reaction $N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ 2 $N H_{3}$ (g) is:

K_{1}

2 N H_{3} ⇌ N_{2} + 3 H_{2}

\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3}

K_{c}

N_{2} + 3 H_{2} ⇋ 2 N H_{3}

Join BYJU'S Learning Program