Question

At $500$K, the equilibrium constant for the reaction $H_{2\left(g\right)}+I_{2\left(g\right)}\rightleftharpoons 2H{I}_{(}g)is24.8$. If $\frac{1}{2}mol/L$ of HI is present at equilibrium. What are the concentrations of $H_2$ and $I_2$, assuming that we started by taking HI and reached the equilibrium at $500$K?

• A. $0.068mol{L}^{-1}$
• B. $1.020mol{L}^{-1}$
• C. $0.10mol{L}^{-1}$
• D. $1.20mol{L}^{-1}$

Answer 1

Answer: (C)

$0.10mol{L}^{-1}$

The given reaction is :-

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right);K_C=24.8$

Now, for the reaction :-

$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

${K^{\prime }}_C=\frac{1}{K_C}=\frac{1}{24.8}$

${K^{\prime }}_C=\frac{1}{K_C}=\frac{1}{24.8}$

Now, $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

Initial : $2X$ $0$ $0$ (let)

At eqm : $(2X-2x)$ $x$ $x$ (let)

Now, ${K^{\prime }}_C=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}$

given that, At eqm concentration of $HI=0.5mol/L$

$\Rightarrow 2X-2x=0.5$

$\Rightarrow X-x=\frac{0.5}{2}=0.25$

Now, $\frac{1}{24.8}=\frac{x.x}{{(2X-2x)}^2}$

$\Rightarrow \frac{1}{24.8}=\frac{x^2}{{4(X-x)}^2}$

$\Rightarrow x^2=\frac{4{(X-x)}^2}{24.8}=\frac{0.5\times 0.5}{24.8}$

$\Rightarrow x=\frac{0.5}{4.98}=0.10mol{L}^{-1}$