Question

In a chemical equilibrium, the rate constant for the backward reaction is $2\times {{10}^{-4}}$ and the equilibrium constant is 1.5. The rate constant for the forward reaction is:
(A) $2\times {{10}^{-3}}$
(B) $5\times {{10}^{-4}}$
(C) $3\times {{10}^{-4}}$
(D) $9.0\times {{10}^{-4}}$

Answer 1

Hint: The equilibrium constant ${{K}_{C}}$ can be defined as the ratio of concentrations of products over the concentration of reactants each raised to the power to their stoichiometric coefficients in equilibrium conditions. The value of ${{K}_{C}}$ is equal to the rate constant for the forward reaction divided by the rate constant for the backward reaction.

Complete step by step solution:
- Let’s start with the concepts of equilibrium constant and rate constant. As we know, at equilibrium condition, the forward rate equals the reverse rate of the reaction. The equilibrium constant ${{K}_{C}}$ can be defined as the ratio of concentrations of products over the concentration of reactants each raised to the power to their stoichiometric coefficients in equilibrium conditions.
- Let’s look at a reaction and derive the relationship between equilibrium constant and rate constant
\[aA+bB\to cC+dD\]
As we mentioned at equilibrium condition, the forward rate equals the reverse rate of the reaction. Hence, by the equilibrium constant can be written as follows
\[{{K}_{C}}=\dfrac{{{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}}{{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}}=\dfrac{{{K}_{f}}}{{{K}_{b}}}\]
Where ${{K}_{f}}$ and ${{K}_{b}}$ are the rate constants of forward and backward reactions respectively. ${{K}_{C}}$ is the equilibrium constant.
- The value of ${{K}_{b}}$ is given as $2\times {{10}^{-4}}$ and ${{K}_{C}}$ is given as 1.5. We are asked to find the value of the rate constant for the forward reaction ${{K}_{f}}$ . Let’s substitute the given values in the above equation
\[{{K}_{c}}=\dfrac{{{K}_{f}}}{{{K}_{b}}}\]
\[{{K}_{f}}={{K}_{c}}\times {{K}_{b}}\]
\[{{K}_{f}}=1.5\times 2\times {{10}^{-4}}=3\times {{10}^{-4}}\]
Thus the rate constant for the forward reaction is $3\times {{10}^{-4}}$.

Therefore the answer is option (C) $3\times {{10}^{-4}}$.

Note: The equilibrium constant ${{K}_{C}}$ gives us an idea about how far the reaction can proceed at a given temperature. The reaction almost goes to completion when ${{K}_{C}}$ much greater than one is and when ${{K}_{C}}$ is much less than one, the reaction will hardly take place.