Question

In the reaction : $ {{H}_{2}}+{{I}_{2}}=2HI $ In a 2 L 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

Answer 1

Answer: (D)

$ \underset{\text{At}\,\text{equilibrium}\,=0.15}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4-0.25}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{H}_{2}}}}\,}}\,}}\,\,\,\,\,\,\,+\underset{=\,0.15}{\mathop{\underset{0.4-0.25}{\mathop{\underset{0.4}{\mathop{{{I}_{2}}}}\,}}\,}}\,=\,\,\,\,\,\,\,\,\,\,\underset{0.05}{\mathop{\underset{0}{\mathop{2HI}}\,}}\, $ $ {{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]} $ $ =\frac{{{\left( \frac{0.50}{2} \right)}^{2}}}{\left( \frac{0.50}{2} \right)\left( \frac{0.50}{2} \right)}=\frac{0.5\times 0.5}{0.15\times 0.15}=11.11 $