The equlibrium constant for the reaction 2NO2(g) leftharpoons 2NO(g)+O2(g) is 2× 10-6 at 185° C . Then the equilibrium constant for the reaction, 4NO(g)+2O2(g) leftharpoons 2NO2(g) at the same temperature would be

Question

The equlibrium constant for the reaction 2NO2(g) leftharpoons 2NO(g)+O2(g) is 2× 10-6 at 185° C . Then the equilibrium constant for the reaction, 4NO(g)+2O2(g) leftharpoons 2NO2(g) at the same temperature would be (A)  2.5× 10-5  (B)  4× 10-12  (C)  2.5× 1111  (D)  2× 106

Tardigrade Admin · Accepted Answer

2NO2(g) leftharpoons 2NO(g)+O2(g) K=([NO]2[O2]/[NO2]2)=2× 10-6 4NO2(g)+2O2(g)4NO2(g) K=([NO2]4/[NO]4[O2]2) =(1/(K)2)=(1/(2× 10-6)2) Equilibrium constant K=0.25× 1012 =2.5× 1011

Q. The equlibrium constant for the reaction $2 N O_{2} (g) ⇌ 2 NO (g) + O_{2} (g)$ is $2 \times 10^{- 6}$ at $185^{\circ} C$ . Then the equilibrium constant for the reaction, $4 NO (g) + 2 O_{2} (g) ⇌$ $2 N O_{2} (g)$ at the same temperature would be

$2.5 \times 10^{- 5}$

$4 \times 10^{- 12}$

$2.5 \times 11^{11}$

$2 \times 10^{6}$

$5 \times 10^{5}$

Q. The equlibrium constant for the reaction 2NO2​(g)⇌2NO(g)+O2​(g) is 2×10−6 at 185∘C . Then the equilibrium constant for the reaction, 4NO(g)+2O2​(g)⇌ 2NO2​(g) at the same temperature would be

2.5×10−5

4×10−12

2.5×1111

2×106

5×105