The equilibrium constant $K_{c}$ for, $A (g) ⇋ B (g)$ is $2.5 \times 10^{- 2}$ . The rate constant of forward reaction is $0.05 s e c^{- 1}$ . Therefore, the rate constant of the reverse reaction is:

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Solution

The correct option is A 2 seconds
We have, $K_{e q} = \frac{k_{f}}{k_{b}}$ , $k_{f} \to$ Rate constant of forward reaction
$↙$ $k_{b} \to$ rate constant of backward reaction
Equilibrium
constant

$\Rightarrow k_{b} = \frac{k_{f}}{K_{e q}} = \frac{0.05 \times 10^{2}}{2.5} = \frac{50}{25} = 2 s e c$ .

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The equilibrium constant Kc for, A(g)⇋B(g) is 2.5×10−2. The rate constant of forward reaction is 0.05sec−1. Therefore, the rate constant of the reverse reaction is:

The correct option is A 2 secondsWe have, Keq=kfkb, kf→ Rate constant of forward reaction ↙ kb→ rate constant of backward reactionEquilibriumconstant⇒kb=kfKeq=0.05×1022.5=5025=2sec.

The equilibrium constant $K_{c}$ for, $A (g) ⇋ B (g)$ is $2.5 \times 10^{- 2}$ . The rate constant of forward reaction is $0.05 s e c^{- 1}$ . Therefore, the rate constant of the reverse reaction is: