A) 0.0625
B) 16
C) 0.0156
D) 64
Correct Answer: C
Solution :
\[\begin{matrix} {} & 2HI(g)\rightleftharpoons & {{H}_{2}}(g)+ & {{I}_{2}}(g) \\ {} & 1 & 0 & 0 \\ \text{at}\,\text{Equil}\text{.} & 1-0.20 & \frac{0.20}{2} & \frac{0.20}{2} \\ {} & =0.80 & =0.10 & =0.10 \\ \end{matrix}\] According to law of mass-action \[{{K}_{c}}=\frac{[{{H}_{2}}]\,\,[{{I}_{2}}]}{{{[HI]}^{2}}}\] \[=\frac{(0.10)\,(0.10)}{{{(0.80)}^{2}}}=\frac{1}{64}\] \[=0.0156\]You need to login to perform this action.
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