A) 64
B) 12
C) 8
D) 0.8
Correct Answer: A
Solution :
\[{{H}_{2}}+{{I}_{2}}\]⇌ 2HI; [HI] = 0.80, \[[{{H}_{2}}]=0.10\], \[[{{I}_{2}}]=0.10\] \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,\,[{{I}_{2}}]}=\frac{0.80\times 0.80}{0.10\times 0.10}=64\]You need to login to perform this action.
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