At a given temperature, the equilibrium constants for the reactions, NO (g)+(1/2) O 2(g) leftharpoons NO 2(g) and 2 NO 2(g)leftharpoons 2 NO (g)+ O 2(g) are K1 and K2 respectively. If K1 is 4 × 10-3 then will be

Question

At a given temperature, the equilibrium constants for the reactions, NO (g)+(1/2) O 2(g) leftharpoons NO 2(g) and 2 NO 2(g)leftharpoons 2 NO (g)+ O 2(g) are K1 and K2 respectively. If K1 is 4 × 10-3 then will be (A)  8× 10-3  (B)  16× 10-3  (C)  6.25× 104  (D)  6.25× 106

Tardigrade Admin · Accepted Answer

NO(g) + (1/2) O2(g) leftharpoons NO2(g) K1 = 4 × 10-3 4 × 10-3 = ([NO2]/[NO][O2]1/2) On squaring, we get 16 × 10-6 = ([NO2]2/[NO]2[O2]) .... (i) 2NO2(g) leftharpoons 2NO(g) + O2(g) K2 = ? K2=([N O]2[O2]/[N O2]2)....(ii) From Eqs. (i) and (ii), K2=(1/K1)=(1/16 × 10-6)=6.25 × 104

Q. At a given temperature, the equilibrium constants for the reactions,
$NO (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)$
and $2 N O_{2} (g) ⇌ 2 NO (g) + O_{2} (g)$
are $K_{1}$ and $K_{2}$ respectively. If $K_{1}$ is $4 \times 1 0^{- 3}$ then will be

$8 \times 1 0^{- 3}$

$16 \times 1 0^{- 3}$

$6.25 \times 1 0^{4}$

$6.25 \times 1 0^{6}$

Q. At a given temperature, the equilibrium constants for the reactions, NO(g)+21​O2​(g)⇌NO2​(g) and 2NO2​(g)⇌2NO(g)+O2​(g) are K1​ and K2​ respectively. If K1​ is 4×10−3 then will be

8×10−3

16×10−3

6.25×104

6.25×106

NO(g)+21​O2​(g)⇌NO2​(g) K1​=4×10−3 4×10−3=[NO][O2​]1/2[NO2​]​ On squaring, we get 16×10−6=[NO]2[O2​][NO2​]2​ .... (i) 2NO2​(g)⇌2NO(g)+O2​(g) K2​=? K2​=[NO2​]2[NO]2[O2​]​....(ii) From Eqs. (i) and (ii), K2​=K1​1​=16×10−61​=6.25×104

Q. At a given temperature, the equilibrium constants for the reactions,
$NO (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)$
and $2 N O_{2} (g) ⇌ 2 NO (g) + O_{2} (g)$
are $K_{1}$ and $K_{2}$ respectively. If $K_{1}$ is $4 \times 1 0^{- 3}$ then will be

$8 \times 1 0^{- 3}$

$16 \times 1 0^{- 3}$

$6.25 \times 1 0^{4}$

$6.25 \times 1 0^{6}$