Question

The rate constants of forward and backward reaction are $8.5 \times {10^{ - 5}}$ and $2.38 \times {10^{ - 4}}$ respectively. The equilibrium constant is
(A) $0.35$
(B) $0.42$
(C) $12.92$
(D) $0.292$

Answer 1

Hint: Here in this question of equilibrium firstly we understand what is equilibrium so, equilibrium is the state which balances the forces which are applied in opposite directions or ways. Equilibrium is a part of physical chemistry so this question is a formula based question. Before starting this question we must have to know which formula is going to be used in this question.

Complete Step by Step Solution:
Firstly, we write the given values which are used to solve the question,
In this question we have to find the equilibrium constant which is denoted by ${K_{eq}}$ ,
From which now we can write suitable formula which is used in this question,
${K_{eq}} = \dfrac{{{K_f}}}{{{K_b}}}$
Where, in above formula there are given the values which are as follows,
Here ${K_f}$ is stands for rate constant for the forward reaction,
And ${K_b}$ stands for rate constant for backward reaction.
Where the given values of both are,
${K_f} = 8.5 \times {10^{ - 5}}$
And ${K_b} = 2.38 \times {10^{ - 4}}$
Now, by rearranging the values in the formula, we get,
${K_{eq}} = \dfrac{{8.5 \times {{10}^{ - 5}}}}{{2.38 \times {{10}^{ - 4}}}}$
As per doing further solution we get,
${K_{eq}} = 0.35$
As by doing the calculation we get the answer as $0.35$ .
Hence, the correct option is (A).

Note: As per doing the questions of equilibrium we have to know that there is no unit for equilibrium constant. The unit of the solution is based on the given units in the question of given rate constants or on the order of the reaction.