Answer:
The given
reaction is:
\[\underset{At\,equilibrium}{\overset{{}}{\mathop{{}}}}\,\underset{\frac{0.3}{1}M}{\mathop{CO(g)}}\,+\underset{\frac{0.1}{1}M}{\mathop{3{{H}_{2}}(g)}}\,\rightleftharpoons
\underset{\frac{x}{1}M}{\mathop{C{{H}_{4}}(g)}}\,+\underset{\frac{0.02}{1}M}{\mathop{{{H}_{2}}O(g)}}\,\] \[{{K}_{c}}=\frac{[C{{H}_{4}}][{{H}_{2}}O]}{[CO]{{[{{H}_{2}}]}^{3}}}\]
\[3.90=\frac{x\times 0.02}{0.3\times
{{(0.1)}^{2}}}\]
\[x=0.0585M\]
\[\therefore \]Concentration of
CH4 will be 0.0585 M.
You need to login to perform this action.
You will be redirected in
3 sec