A) \[{{K}_{2}}=\frac{1}{{{K}_{1}}}\]
B) \[{{K}_{2}}=K_{1}^{2}\]
C) \[{{K}_{2}}=\frac{{{K}_{1}}}{2}\]
D) \[{{K}_{2}}=\frac{1}{K_{1}^{2}}\]
Correct Answer: D
Solution :
(i) \[NO(g)+\frac{1}{2}{{O}_{2}}(g)N{{O}_{2}}(g)\] \[So\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{K}_{1}}=\frac{[N{{O}_{2}}]}{[NO]\,{{[{{O}_{2}}]}^{1/2}}}.....(i)\] (ii) \[2N{{O}_{2}}(g)2NO+{{O}_{2}}(g)\] \[So{{K}_{2}}=\frac{{{[NO]}^{2}}\,[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}....(ii)\] From eq. (i) \[K_{1}^{2}=\frac{{{[N{{O}_{2}}]}^{2}}}{{{[NO]}^{2}}\,[{{O}_{2}}]}\] \[or\frac{1}{K_{1}^{2}}=\frac{{{[NO]}^{2}}\,[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}....(iii)\] From Eqs. (ii) and (iii) \[{{K}_{2}}=\frac{1}{K_{1}^{2}}\]You need to login to perform this action.
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