Question

For a reversible reaction, the rate constants for the forward and backward reactions are 2.38 × ${10^{ - 4}}$ and 8.15 × ${10^{ - 5}}$ respectively. The equilibrium constant for the reaction is:
(A) 0.342
(B) 2.92
(C) 0.292
(D) 3.42

Answer 1

Hint: It is important to understand the terms rate constant and equilibrium constant. After that, try to understand the relation between the two terms. You can substitute the values given in question in the formula mentioned below:
${{K}_{eq}}=\dfrac{{{K}_{f}}}{{{K}_{b}}}$
Where,
${{K}_{eq}}$ denotes equilibrium constant,
${{K}_{f}}$ denotes rate constant for forward reaction,
${{K}_{b}}$ denotes rate constant for forward reaction.

Complete Solution :
Let us first learn what is rate constant and equilibrium constant.
- The rate constant is the proportionality constant which explains the relationship between the molar concentration of the reactants and the rate of a chemical reaction. It is denoted by k. Rate constant is also called as reaction rate constant for reaction rate coefficient. It is temperature dependent.

The two possible ways to calculate rate constant are:
- Using the Arrhenius equation.
- Using the molar concentrations of the reactants and the order of the reaction.
We know that at equilibrium, Rate of the forward reaction = Rate of the backward reaction.
Now let us see the data given in the question to find the equilibrium constant,
Rate constants of the forward reaction (${K_f}$) = 2.38×${10^{ - 4}}$
Rate constants of the backward reaction (${K_b}$) = 8.15×${10^{ - 5}}$
We know that ${K_{eq}}$= $\dfrac{{{K_f}}}{{{K_b}}}$, on substituting the values
${K_{eq}}$= $\dfrac{{{K_f}}}{{{K_b}}}$= $\dfrac{{2.38 \times {{10}^{ - 4}}}}{{8.15 \times {{10}^{ - 5}}}}$ = 2.92
So, the correct answer is “Option B”.

Note: Do remember that there is no unit for equilibrium constant. The unit of rate constant varies based on the order of reaction.