Question

Equimolar concentration of ${{H}_{2}}$ and ${{I}_{2}}$ are heated at equilibrium in a flask of $1L$. The forward and backward rate constants are found to be equal. What percentage of the initial concentration of ${{H}_{2}}$ has reached equilibrium?
A. $33%$
B. $66%$
C. $50%$
D. $40%$

Answer 1

Hint: Think about the definitions and formulae of the equilibrium constant in terms of the concentration of the reactants and products. Also consider how the equilibrium constant is defined in terms of the rate constants of the forward and backward reactions.

Complete step by step answer:
First, we will consider how the equilibrium constant is defined in terms of the concentrations of the reactants and the products and then we will determine the concentrations. The reaction defined in this question is:
\[{{H}_{2}}+{{I}_{2}}2HI\]
So, the reactants here are ${{H}_{2}}$ and ${{I}_{2}}$ , and the product is $HI$. We can see that one mole each of the reactants gives us two moles of the product. According to this reaction. The formula for the equilibrium constant will be defined as:
\[{{K}_{c}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}\]
Here, ${{K}_{c}}$ is the equilibrium constant. Now, we will determine the concentrations of the reactants and the products.
It is given that both the reactants are present in equimolar concentrations, so we will assume that one mole of each reactant is present. We know that the reaction is reversible, so at equilibrium only a fraction of the total concentration of the reactants will have been converted into the products. We will assume that this concentration of the reactants that has been converted into the products is $x$.
So, initially, the concentration of reactants present will be one mole each. And, at equilibrium, the concentration of the reactants present will be $(1-x)$. We can represent this data in a tabular form as:

Reaction	${{H}_{2}}$	${{I}_{2}}$	$HI$
Initially	1	1	2
At equilibrium	$(1-x)$	$(1-x)$	$2x$

The amount of reactant that has reacted will be converted into the product. Now before we put these values into the equation given above, we need to find the value of ${{K}_{c}}$ so that finding the value of $x$ becomes possible.
It has been given that the rate constant of the forward and the backward reaction are the same. We know that the equilibrium constant is the ratio of the forward reaction to the backward reaction. So, this can be expressed as:
\[{{K}_{c}}=\dfrac{{{K}_{f}}}{{{K}_{b}}}\]
Where, ${{K}_{f}}$ is the rate of the forward reaction and ${{K}_{b}}$ is the rate of the backward reaction. So, from this we can say that ${{K}_{c}}=1$. Now, putting all these values in the formula given above, we get:
\[1=\dfrac{{{(2x)}^{2}}}{(1-x)(1-x)}\]
Now, we can solve for $x$
\[\begin{align}
 & {{\left( \dfrac{2x}{1-x} \right)}^{2}}=1 \\
 & \dfrac{2x}{1-x}=1 \\
 & 2x=1-x \\
 & x=\dfrac{1}{3} \\
\end{align}\]
So, we can see that the value of $x$ is 1/3. This means that one-thirds of the initial concentration of the reactants will be converted into the products. If we convert this value to percentage, we get $x=33%$.
So, the correct answer is “Option A”.

Note: While calculating the value of $x$, we took the square root on both the sides. While doing this, we did not consider the value $\sqrt{1}=-1$. We did this because if we consider this value, $x$ will turn out to be 1. This implies that the reaction has been completed completely, which is not true. So, we will not consider this value. Also, pay attention to what is asked, the percentage of reactants that have reached equilibrium is $33%$, and the percentage of reactants that are unreacted is $66%$.