For the following reaction: NO(g) +O3(g) leftharpoons NO2(g) +O2(g) The value of Kc is 8.2 × 104. What will be the value of Kc for the reverse reaction?

Question

For the following reaction: NO(g) +O3(g) leftharpoons NO2(g) +O2(g) The value of Kc is 8.2 × 104. What will be the value of Kc for the reverse reaction? (A) 8.2 × 104 (B) (1/8.2×104) (C) (8.2× 104)2 (D) √8.2×104

Tardigrade Admin · Accepted Answer

The value of Kc for the reverse reaction will be (1/8.2× 104).

Q. For the following reaction :
$N O_{(g)} + O_{3 (g)} ⇌ N O_{2 (g)} + O_{2 (g)}$
The value of $K_{c}$ is $8.2 \times 1 0^{4}$ . What will be the value of $K_{c}$ for the reverse reaction?

$8.2 \times 1 0^{4}$

$\frac{1}{8.2 \times 1 0 ^{4}}$

$(8.2 \times 1 0^{4})^{2}$

$8.2 \times 1 0^{4}$

The value of $K_{c}$ for the reverse reaction will be $\frac{1}{8.2 \times 1 0 ^{4}}$ .

Q. For the following reaction : NO(g)​+O3(g)​⇌NO2(g)​+O2(g)​ The value of Kc​ is 8.2×104. What will be the value of Kc​ for the reverse reaction?

8.2×104

8.2×1041​

(8.2×104)2

8.2×104​

The value of Kc​ for the reverse reaction will be 8.2×1041​.

Q. For the following reaction :
$N O_{(g)} + O_{3 (g)} ⇌ N O_{2 (g)} + O_{2 (g)}$
The value of $K_{c}$ is $8.2 \times 1 0^{4}$ . What will be the value of $K_{c}$ for the reverse reaction?

$8.2 \times 1 0^{4}$

$\frac{1}{8.2 \times 1 0 ^{4}}$

$(8.2 \times 1 0^{4})^{2}$

$8.2 \times 1 0^{4}$

The value of $K_{c}$ for the reverse reaction will be $\frac{1}{8.2 \times 1 0 ^{4}}$ .