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