| \[NO(g)+\frac{1}{2}{{O}_{2}}(g)\rightleftharpoons N{{O}_{2}}(g);{{K}_{1}}\] |
| \[2N{{O}_{2}}(g)\rightleftharpoons 2NO(g)+{{O}_{2}}(g);{{K}_{2}}\] |
A) \[{{K}_{2}}=\frac{1}{K_{1}^{2}}\]
B) \[{{K}_{2}}=\frac{1}{{{K}_{1}}}\]
C) \[{{K}_{2}}=K_{1}^{2}\]
D) \[{{K}_{2}}=\frac{{{K}_{1}}}{2}\]
Correct Answer: A
Solution :
\[{{K}_{1}}=\frac{\left[ N{{O}_{2}} \right]}{\left[ NO \right]\left[ {{O}_{2}} \right]}\]and \[{{K}_{2}}=\frac{{{\left[ NO \right]}^{2}}\left[ {{O}_{2}} \right]}{{{\left[ N{{O}_{2}} \right]}^{2}}}\] If eqn. is multiplied by n, then new equilibrium constant is \[k_{1}^{n}\]and if eqn. is reversed, then ne equilibrium constant is reciprocal of pervious \[\therefore \] \[{{K}_{2}}=\frac{1}{K_{1}^{2}}\]
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