Chapter 15, Problem 75AP

Consider the following reaction at a certain temperature:

${\text{A}}_{\text{2}}\text{+ }{\text{B}}_{\text{2}}\rightleftarrows \text{ 2AB}$

The mixing of 1 mole of ${\text{A}}_{\text{2}}$ with 3 moles of ${\text{B}}_{\text{2}}$ gives rise to x mole of AB at equilibrium. The addition of 2 more moles of ${\text{A}}_{\text{2}}$ produces another x mole of AB. What is the equilibrium constant for the reaction?

Interpretation Introduction

Interpretation:

The equilibrium constant of the reaction is to be calculated.

Concept introduction:

For a general reaction: $a\text{A}\text{+}b\text{B}\rightleftarrows c\text{C}\text{+}d\text{D}$

Equilibrium constant is defined as the ratio of concentration of products and the concentration of reactants, which are in equilibrium, and its value is constant at a particular temperature. It is denoted by $K$.

$\text{Equilibrium constant, }K=\frac{{\left[\text{C}\right]}^{\text{c}}{}_{\text{eqm}}{\left[\text{D}\right]}^{\text{d}}{}_{\text{eqm}}}{{\left[\text{A}\right]}^{\text{a}}{}_{\text{eqm}}{\left[\text{B}\right]}^{\text{b}}{}_{\text{eqm}}}$.

Answer to Problem 75AP

Solution: $K=4.0$.

Explanation of Solution

Given information: The given reaction is as follows:

$A_2+B_2\rightleftarrows 2AB$

In the given reaction, the mixture of 1 mole of $A_2$ with 3 moles of $B_2$ gives rise to $x$ moles of AB, which is given in the ICE table is as follows:

$\begin{array}{cccc}& {\text{A}}_2& {\text{B}}_2& 2\text{AB}\\ \text{Initial}\left(\text{mol}\right):& 1& 3& 0\\ \text{Change}\left(\text{mol}\right):& -\frac{x}{2}& -\frac{x}{2}& +x\\ \text{Equilibrium}\left(\text{mol}\right):& 1-\frac{x}{2}& 3-\frac{x}{2}& x\end{array}$

After the addition of 2 moles of A, the ICE table is given as follows:

$\begin{array}{cccc}& {\text{A}}_2& {\text{B}}_2& 2\text{AB}\\ \text{Initial}\left(\text{mol}\right):& 3-\frac{x}{2}& 3-\frac{x}{2}& x\\ \text{Change}\left(\text{mol}\right):& -\frac{x}{2}& -\frac{x}{2}& +x\\ \text{Equilibrium}\left(\text{mol}\right):& 3-x& 3-x& 2x\end{array}$

The equilibrium constant is given by the expression as follows:

$\text{Equilibrium constant, }K=\frac{{\left[\text{C}\right]}^{\text{c}}{}_{\text{eqm}}{\left[\text{D}\right]}^{\text{d}}{}_{\text{eqm}}}{{\left[\text{A}\right]}^{\text{a}}{}_{\text{eqm}}{\left[\text{B}\right]}^{\text{b}}{}_{\text{eqm}}}$

$K=\frac{{\left[\text{AB}\right]}^2}{\left[{\text{A}}_2\right]\left[{\text{B}}_2\right]}$

Here, $K$ is the equilibrium constant, $\text{AB}$ is the concentration of product, and ${\text{A}}_2$ and ${\text{B}}_2$ are the concentrations of reactants.

Substitute the values of $\text{AB}$, ${\text{A}}_2$, and ${\text{B}}_2$ in the equation given above,

$K=\frac{x^2}{\left[1-\frac{x}{2}\right]\left[3-\frac{x}{2}\right]}$ $\begin{array}{c}\\ \end{array}$

After addition of two moles, the equilibrium constant expression is given as follows:

$K=\frac{{\left(2x\right)}^2}{\left[3-x\right]\left[3-x\right]}$

Solve for $x$,

$\begin{array}{c}\frac{x^2}{\left[1-\frac{x}{2}\right]\left[3-\frac{x}{2}\right]}=\frac{{\left(2x\right)}^2}{\left[3-x\right]\left[3-x\right]}\\ \frac{\cancel{x^2}}{\left[\frac{2-x}{2}\right]\left[\frac{6-x}{2}\right]}=\frac{4\cancel{x^2}}{x^2-6x+9}\\ \frac{1}{\frac{1}{\cancel{4}}\left[2-x\right]\left[6-x\right]}=\frac{\cancel{4}}{x^2-6x+9}\\ \frac{1}{\left[2-x\right]\left[6-x\right]}=\frac{\cancel{4}}{x^2-6x+9}\end{array}$

Again, on solving further, we get

$\begin{array}{c}\frac{1}{x^2-8x+12}=\frac{1}{x^2-6x+9}\\ \cancel{x^2}-6x+9=\cancel{x^2}-8x+12\\ -6x+9=-8x+12\\ -6x+8x=12-9\end{array}$

The value of $x$ is given as follows:

$\begin{array}{c}2x=3\\ x=\frac{3}{2}\\ x=1.5\end{array}$

Substitute the value of $x$ in one of the equilibrium constant’s expressions,

$\begin{array}{c}K=\frac{{\left(2x\right)}^2}{\left[3-x\right]\left[3-x\right]}\\ =\frac{{\left(2\left(1.5\right)\right)}^2}{\left[3-1.5\right]\left[3-1.5\right]}\\ =\frac{{\left(3\right)}^2}{\left[1.5\right]\left[1.5\right]}\\ =4\end{array}$

Conclusion

The value of the equilibrium constant is $4.0$.