Question

Mention the relation between equilibrium constants of forward and reverse reactions.

Answer 1

equilibrium constant Kc = kf / kr

Answer 2

Answer:

Equilibrium constant (k) =$\frac{k_{f}}{k_{r}}$

Explanation:

• When the rate of  the forward reaction is equal to the rate of the reverse reaction, then the reaction is said to be in the state of equilibrium.
• Example: the decomposition reaction of $N_{2}O_{2}$ to $NO_{2}$

• forward rate = $k_{f}[N_{2}O_{4}]$

• Reverse rate =$k_{r}[NO_{2}]^2$

At equilibrium both the rates of reaction are equal.

So,

$k_{f}[N_{2}O_{4}]$ = $k_{r}[NO_{2}]^2$

• $\frac{k_{f}}{k_{y}}$ is the ratio of rate of forward and reverse reaction which is denoted by a new constant known as equilibrium constant (k).
• Hence, (k) =$\frac{k_{f}}{k_{r}}$