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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
At 700K, equilibrium constant for the reaction:
$\text{H}_2\text{ (g) + I}_2\text{ (g)}\rightleftharpoons2\text{HI (g)}$ is 54.8. If $0.5 ~mol L^{–1}$ of $HI(g)$ is present at equilibrium at 700K, what are the concentration of $H^2(g)$ and $I^2(g)$ assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
✓

Answer

It is given that equilibrium constant $K_c$ for the reaction
$\text{H}_{2\text{(g)}}\text{ + I}_{2\text{(g)}}\leftrightarrow2\text{HI}_{\text{(g)}} \text{ is }54.8.$
Therefore, at equilibrium, the equilibrium constant $K_c$ for the reaction
$2\text{HI}_{\text{(g)}}\leftrightarrow\text{H}_{2\text{(g)}}\text{ + I}_{2\text{(g)}}\text{ Will be }\frac{1}{54.8}$
$[\text{HI]}=0.5\text{mol L}^{-1}$
Let the concentrations of hydrogen and iodine at equilibrium be x mol $L^{–1}$
$[\text{H}_2]=[\text{I}_2]=\text{xmol L}^{-1}$
Therefore, $\frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}=\text{K}'_{\text{c}}$
$\Rightarrow\frac{\text{x}\times\text{x}}{(0.5)^2}=\frac{1}{54.8}$
$\Rightarrow\text{x}^2=\frac{0.25}{54.8}$
$\Rightarrow\text{x}=0.06754$
$\text{x}=0.068\text{mol L}^{-1}\text{(approximately)}$
Hence, at equilibrium, $[\text{H}_2]=[\text{I}_2]=\text{0.068mol L}^{-1}$

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