The value of the equilibrium constant for the reaction
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
at $720$ K is $48$ . What is the value of the equilibrium constant for the following reaction?
$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ H I (g)$

Open in App

Solution

Solution:-
${H_{2}}_{(g)} + {I_{2}}_{(g)} ⇌ 2 {H I}_{(g)}$
$K = \frac{{[H I]}^{2}}{[H_{2}] [I_{2}]} = 48 . . . . . (i)$
$\frac{1}{2} {H_{2}}_{(g)} + \frac{1}{2} {I_{2}}_{(g)} ⇌ {H I}_{(g)}$
$K^{'} = \frac{[H I]}{{[H_{2}]}^{\frac{1}{2}} {[I_{2}]}^{\frac{1}{2}}}$
Squring both sides we have,
$K^{2} = \frac{{[H I]}^{2}}{[H_{2}] [I_{2}]}$
$\Rightarrow {K^{'}}^{2} = 48 [From (i)]$
$\Rightarrow K^{'} = \sqrt{48} = 4 \sqrt{3}$
Hence, for the reaction $\frac{1}{2} {H_{2}}_{(g)} + \frac{1}{2} {I_{2}}_{(g)} ⇌ {H I}_{(g)}$ , the value of equillibrium constant is $4 \sqrt{3}$ .

Suggest Corrections

Similar questions

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

8.0

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

500

K_{c}

5

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

K_{c}

2 H I_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}

H_{2_{(g)}} + I_{2_{(g)}} ⇌ {2 H I}_{(g)}

{2 H I}_{(g)} ⇌ H_{2_{(g)}} + I_{2_{(g)}}

\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ H I (g) is K_{c}

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

H I (g) ⇌ 1 / 2 H_{2} (g) + 1 / 2 I_{2} (g)

8.0

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

Join BYJU'S Learning Program