The equilibrium constant for the reaction N2( g)+O2( g)⇌2NO(g) at temperature (T) is 4×10-4 The value of K. for the reaction NO(g)⇌12 N2(g)+12O2(g) at the same temperature is

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4×10-4

2.5×102

0.02

answer is D.

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Detailed Solution

N2+O2⇄2NO,Δng=0 ⇒kc=kp=kx=4×10-4 So for NO⇌12 N2+12O2 ,kx1 =1kx =14×10-4 kx1=4×10-4-1/2 =2×10-2 =0.02

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