The rate constants of the forward reaction and reverse reaction are 0.05 $sec^{-1}$ and 2 $sec^{-1}$ respectively, then $K_c$ is : (a) 0.035 (b) 0.025 (c) 0.045 (d) 0.02

The rate constants of the forward reaction and reverse reaction are 0.05 sec⁻¹ and 2 sec⁻¹ respectively, then

Question 61 of 80

sec^{-1}

(a) 0.035

(b) 0.025

(d) 0.02

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At equilibrium the forward and reverse rates are equal, so the equilibrium constant equals the ratio of the forward to the reverse rate constant: $K_c = k_f/k_r$ .

For a reversible reaction $A \rightleftharpoons B$ , at equilibrium the rate of the forward reaction equals the rate of the reverse reaction:

$k_f[A] = k_r[B]$

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Q.The rate constants of the forward reaction and reverse reaction are 0.05 $sec^{-1}$ and 2 $sec^{-1}$ respectively, then $K_c$ is :

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Q.The rate constants of the forward reaction and reverse reaction are 0.05 $sec^{-1}$ and 2 $sec^{-1}$ respectively, then $K_c$ is :

Unlock everything free for 14 days