A) 0.15 mole
B) 0.06 mole
C) 0.03 mole
D) 0.2 mole
Correct Answer: B
Solution :
\[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,\,[{{I}_{2}}]}\]; \[64=\frac{{{x}^{2}}}{0.03\times 0.03}\] \[{{x}^{2}}=64\times 9\times {{10}^{-4}}\] \[x=8\times 3\times {{10}^{-2}}=0.24\] x is the amount of HI at equilibrium amount of \[{{I}_{2}}\]at equilibrium will be \[0.30-0.24=0.06\]You need to login to perform this action.
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