Question

Equilibrium concentration of $ HI,\,\,{{I}_{2}} $ and $ {{H}_{2}} $ are 0.7, 0.1 and 0.1 M respectively. The value of equilibrium constant for the reaction, $ {{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI $ is:

• A. 36
• B. 49
• C. 0.49
• D. 0.36

Answer 1

Answer: (B)

$ \begin{align} & \begin{matrix} {} & {{H}_{2}}+ & {{I}_{2}}\rightleftharpoons & 2HI \\ \text{initial}\,\text{conc}\text{.} & a & a & 0 \\ \text{conc}\text{.}\,\text{at}\,\text{equil}\text{.} & (a-x) & (a-x) & x=0.7 \\ \end{matrix} \\ & \begin{matrix} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.1 & \,\,\,\,=0.1 \\ \end{matrix} \\ \end{align} $ $ \therefore \,\,\,\,{{K}_{c}}\,=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,[{{I}_{2}}]}\,=\frac{{{(0.7)}^{2}}}{(0.1)\,(0.1)} $ $ =\frac{0.49}{0.01}\,=49 $