Question

If the equilibrium concentrations for ${X_2}$ , ${Y_2}$ and $XY$ are $1.2 \times {10^3}M$ , $5.83 \times {10^{12}}M$ and $8.9 \times {10^2}M$ respectively, what was the initial concentration ${X_2}$ before the reaction occurred to form $XY$?
A. $8.83 \times {10^{ - 1}}M$
B. $1.65 \times {10^3}M$
C. $1.13 \times {10^3}M$
D. $2.09 \times {10^3}M$

Answer 1

A reaction is in equilibrium when the rate of forward reaction becomes equal to the rate of backward reaction. The concentration of the reactants and products are fixed at equilibrium.

Complete step by step answer:
When the concentration of the reactants and the products does not change any more during a chemical reaction, then the reaction is said to be in the state of chemical equilibrium. When the equilibrium is reached both the rate of forward reaction and backward reaction are the same.
If the initial concentrations of the reactants and products are known the equilibrium concentration is calculated from the values. Likewise if the equilibrium concentration of the reactants and the products are known the initial concentration is determined.
The equilibrium concentration is calculated using the following steps:
i. A chemical equation needs to be balanced
ii. Determine the change in concentration of the reactants and product with time.
iii. Determine the degree of dissociation of the reactants using the equilibrium constant Keq.
iv. Subtract the degree of dissociation from the initial concentration of the reactants.
The given starting materials for the reaction are ${X_2}$ and ${Y_2}$. The product of the reaction is $XY$. Thus the corresponding balanced chemical equation for the reaction is written as
${X_2} + {Y_2} \rightleftharpoons 2XY$
Let the concentrations of the reactants and products in the beginning and at the equilibrium are written as

| ${X_2}$| ${Y_2}$| $XY$
---|---|---|---
In the beginning| $a$ | $b$ | $0$
At equilibrium | $a - x$ | $b - x$ | $2x$
| $1.2 \times {10^3}M$ | $5.83 \times {10^{12}}M$ | $8.9 \times {10^2}M$

So from the table the initial concentration of ${X_2}$ is $a$ and that of ${Y_2}$ is $b$. Let the amount of dissociation is $x$.
Given$a - x = 1.2 \times {10^3}M$ , $b - x = 5.83 \times {10^{12}}M$ and$2x = 8.9 \times {10^2}M$.
So, $x = 4.45 \times {10^2}M$
Or $x = 0.445 \times {10^3}M$
Using the value of x,
$a - x = 1.2 \times {10^3}M$
or $a = \left( {1.2 \times {{10}^3} + x} \right)M$
$a = 1.2 \times {10^3} + 0.445 \times {10^3} = 1.645 \times {10^3}M$.
Hence the initial concentration ${X_2}$ before the reaction occurred to form $XY$ is $1.645 \times {10^3}M$, i.e. option B is the correct answer.

Note:
The concentration of the substances whose value changes with time is only considered in the equilibrium constant. Equilibrium constant of a reactant depends on temperature. It is not dependent on concentrations, pressures and volumes of reactants and products.