• # question_answer In the reaction, ${{H}_{2}}+{{I}_{2}}$$\rightleftharpoons$$2HI$. In a 2 litre flask 0.4 moles of each ${{H}_{2}}$ and ${{I}_{2}}$ are taken. At equilibrium 0.5 moles of $HI$ are formed. What will be the value of equilibrium constant, ${{K}_{c}}$    A) 20.2      B) 25.4      C) 0.284 D) 11.1

 [d]     $\underset{0.4-0.25\ =\ 0.15}{\mathop{\underset{0.4}{\mathop{{{H}_{2}}}}\,}}\,$$+$ $\underset{0.4-0.25\ =\ 0.15/2}{\mathop{\underset{0.4}{\mathop{{{I}_{2}}}}\,}}\,$$\rightleftharpoons$ $\underset{0.50/2}{\mathop{\underset{0.50\,\,\,\,}{\mathop{2HI\,\,\,\,}}\,}}\,$ ${{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{\left[ \frac{0.5}{2} \right]}^{2}}}{\left[ \frac{0.15}{2} \right]\,\left[ \frac{0.15}{2} \right]}$ $=\frac{0.5\times 0.5}{0.15\times 0.15}=11.11$