Question

Equimolar concentration of ${{H}_{2}}$ and ${{I}_{2}}$ are heated to equilibrium in a 2L flask. At equilibrium the forward and backward rate constants are found to be equal. What percent of initial concentration of ${{H}_{2}}$ has reacted at equilibrium?
(A) 40%
(B) 66%
(C) 33%
(D) 20%

Answer 1

Attempt this question by using the definitions and formulas of the equilibrium constant in terms of concentration of the reactants and products. Also consider the definition of equilibrium constant in terms of rate constants of the forward and backward reactions.

Complete answer:
First of all let's begin with equilibrium constant in terms of concentration of the reactants and products and then we will determine their concentration. The reaction is given below:
${{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI$
Since one mole of each of the reactants gives 2 moles of product. Therefore equilibrium constant (${{K}_{C}}$ ) will be:
${{K}_{C}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$
-Determination of concentration of the products:
As it is given that both the reactants are present in equimolar concentration, so we will consider that 1 mole of each reactant is present. Since the reaction is reversible therefore only a few amount of reactants is converted into product and this amount is supposed to be ‘x’. The initial and equilibrium concentration of reactants and products are shown below:-

Reaction	${{H}_{2}}$	${{I}_{2}}$	HI
Initial	1	1	0
At equilibrium	(1-x)	(1-x)	2x

Also the value of equilibrium constants in terms of rate constant of backward and forward reaction is given below:-
${{K}_{C}}=\dfrac{{{K}_{f}}}{{{K}_{b}}}$
Since it is given in the question that rate constant of forward and backward reaction are same therefore ${{K}_{C}}$=1 Now we can use this value to the concentration formula of ${{K}_{C}}$.
$\begin{aligned} & {{K}_{C}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]} \\\ & 1=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]} \\\ \end{aligned}$
Concentration is the number of moles in per liter solution.
$\begin{aligned} & [HI]=2x/2L \\\ & [{{H}_{2}}]=(1-x)/2L \\\ & [{{I}_{2}}]=(1-x)/2L \\\ \end{aligned}$
On substituting all these values we get:-
$\begin{aligned} & \Rightarrow 1=\dfrac{{{[2x/2L]}^{2}}}{[(1-x)/2L][(1-x)/2L]} \\\ & \Rightarrow 1=\dfrac{{{[2x]}^{2}}}{{{[(1-x)]}^{2}}} \\\ & \text{On square root of both sides we get:-} \\\ & \Rightarrow \pm 1=\dfrac{[2x]}{[(1-x)]}\text{ }\\!\\!\\{\\!\\!\text{ neglecting (-1) }\\!\\!\\}\\!\\!\text{ } \\\ & \Rightarrow 1=\dfrac{[2x]}{[(1-x)]}\text{ } \\\ & \Rightarrow 2x=1-x \\\ & \Rightarrow x=\dfrac{1}{3}=0.333 \\\ \end{aligned}$
On converting this value to percentage we get x= 33.3%

Hence, initial concentration of ${{H}_{2}}$ has reacted at equilibrium is (C) 33%

Note:
We neglected -1 value while calculating the value of x because if we consider this value, then x would be 1. This means that the reaction has completed which ought to be not true. So we don’t consider this value.
-Also while solving such kind of questions, always look for what value is asked as it can be the amount of reactant left or the amount of reactant already reacted in the reaction at equilibrium.