\[{{N}_{2}}+{{O}_{2}}2NO\] ......I |
\[\frac{1}{2}{{N}_{2}}+\frac{1}{2}{{O}_{2}}NO\] ......II |
A) \[{{K}_{1}}={{K}_{2}}\]
B) \[{{K}_{2}}=\sqrt{{{K}_{1}}}\]
C) \[{{K}_{1}}=2{{K}_{2}}\]
D) \[{{K}_{1}}=\frac{1}{2}{{K}_{2}}\]
Correct Answer: B
Solution :
\[{{N}_{2}}+{{O}_{2}}2NO\] \[{{K}_{1}}=\frac{{{\left[ NO \right]}^{2}}}{\left[ {{N}_{2}} \right]\,\left[ {{O}_{2}} \right]}\] ?..I \[\frac{1}{2}{{N}_{2}}+\frac{1}{2}{{O}_{2}}NO\] \[{{K}_{2}}=\frac{\left[ NO \right]}{{{\left[ {{N}_{2}} \right]}^{{\scriptstyle{}^{1}/{}_{2}}}}{{\left[ {{O}_{2}} \right]}^{{\scriptstyle{}^{1}/{}_{2}}}}}\] ?.II \[K_{2}^{2}=\frac{{{\left[ NO \right]}^{2}}}{\left[ {{N}_{2}} \right]\left[ {{O}_{2}} \right]}\] ?..III Hence, from equation I and III \[{{K}_{1}}=K_{2}^{2}\] \[{{K}_{2}}=\sqrt{{{K}_{1}}}\]You need to login to perform this action.
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