A) \[{{K}_{c}}=\frac{4{{x}^{2}}}{(a-x)\,(b-x)}.\frac{1}{V}\]
B) \[{{K}_{c}}=\frac{4{{x}^{2}}}{(a-x)\,(b-x)}.V\]
C) \[{{K}_{c}}=\frac{{{x}^{2}}}{(a-x)\,(b-x)}.V\]
D) \[{{K}_{c}}=\frac{4{{x}^{2}}}{(a-x)\,(b-x)}\]
Correct Answer: D
Solution :
For this reaction \[\underset{(a-x)}{\mathop{\underset{a\,mol}{\mathop{{{N}_{2}}}}\,}}\,+\underset{(b-x)}{\mathop{\underset{b\,mol}{\mathop{{{O}_{2}}}}\,}}\,\underset{(2x)}{\mathop{\underset{0}{\mathop{2NO}}\,}}\,\] Thus- \[{{K}_{c}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}\] \[=\frac{\frac{{{(2x)}^{2}}}{{{V}^{2}}}}{\frac{(a-x)}{V}.\frac{(b-x)}{V}}\] \[=\frac{4{{x}^{2}}}{(a-x)(b-x)}\]You need to login to perform this action.
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