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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

The equlibrium constant for the reaction 2NO2(g)⇌2NO(g)+O2(g)2N{{O}_{2}}(g) \rightleftharpoons 2NO(g)+{{O}_{2}}(g) is 2×10−62\times {{10}^{-6}} at 185∘C185{}^\circ C . Then the equilibrium constant for the reaction, 4NO(g)+2O2(g)⇌4NO(g)+2{{O}_{2}}(g) \rightleftharpoons 2NO2(g)2N{{O}_{2}}(g) at the same temperature would be

A

2.5×10−52.5\times {{10}^{-5}}

B

4×10−124\times {{10}^{-12}}

C

2.5×11112.5\times {{11}^{11}}

D

2×1062\times {{10}^{6}}

Answer

2.5×11112.5\times {{11}^{11}}

Explanation

Solution

2NO2(g)⇌2NO(g)+O2(g)2N{{O}_{2}}(g) \rightleftharpoons 2NO(g)+{{O}_{2}}(g) K=[NO]2[O2][NO2]2=2×10−6K=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}=2\times {{10}^{-6}} 4NO2(g)+2O2(g)4NO2(g)4N{{O}_{2}}(g)+2{{O}_{2}}(g)4N{{O}_{2}}(g) K=[NO2]4[NO]4[O2]2K=\frac{{{[N{{O}_{2}}]}^{4}}}{{{[NO]}^{4}}{{[{{O}_{2}}]}^{2}}} =1(K)2=1(2×10−6)2=\frac{1}{{{(K)}^{2}}}=\frac{1}{{{(2\times {{10}^{-6}})}^{2}}} Equilibrium constant K=0.25×1012K=0.25\times {{10}^{12}} =2.5×1011=2.5\times {{10}^{11}}