A) 20.2
B) 25.4
C) 0.284
D) 11.1
Correct Answer: D
Solution :
\[\underset{\text{At}\,\text{equilibrium}\,=0.15}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4-0.25}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{H}_{2}}}}\,}}\,}}\,\,\,\,\,\,\,+\underset{=\,0.15}{\mathop{\underset{0.4-0.25}{\mathop{\underset{0.4}{\mathop{{{I}_{2}}}}\,}}\,}}\,=\,\,\,\,\,\,\,\,\,\,\underset{0.05}{\mathop{\underset{0}{\mathop{2HI}}\,}}\,\] \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}\] \[=\frac{{{\left( \frac{0.50}{2} \right)}^{2}}}{\left( \frac{0.50}{2} \right)\left( \frac{0.50}{2} \right)}=\frac{0.5\times 0.5}{0.15\times 0.15}=11.11\]You need to login to perform this action.
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