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If the equilibrium constant for the reaction,
$\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \text { is } K$
what is the equilibrium constant of
$\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) ?$
Options:
$\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \text { is } K$
what is the equilibrium constant of
$\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) ?$
Solution:
1278 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\sqrt{K}}$
$\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) ; K$
$\therefore \quad K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}$ ...(i)
For the reaction,
$\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g)$
Equilibrium constant,
$K^{\prime}=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]}$
On squaring both sides, we get
$\left(K^{\prime}\right)^2=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}$ ...(ii)
On multiplying Eqs. (i) and (ii), we get $K \cdot\left(K^{\prime}\right)^2=1$
$K^{\prime}=\frac{1}{\sqrt{K}}$
$\therefore \quad K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}$ ...(i)
For the reaction,
$\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g)$
Equilibrium constant,
$K^{\prime}=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]}$
On squaring both sides, we get
$\left(K^{\prime}\right)^2=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}$ ...(ii)
On multiplying Eqs. (i) and (ii), we get $K \cdot\left(K^{\prime}\right)^2=1$
$K^{\prime}=\frac{1}{\sqrt{K}}$
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