Chapter 14, Problem 14.96AP

Interpretation Introduction

Interpretation: The composition of the mixture is to be calculated when equilibrium was reestablished of the given reaction.

Concept introduction: The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

$K_{\text{c}}=\frac{{\left[\text{Product}\right]}^y}{{\left[\text{Reactant}\right]}^x}$

To determine: The composition of the mixture when equilibrium was reestablished.

Answer to Problem 14.96AP

Solution

The composition mixture when equilibrium reestablished is the concentration of $\text{NO}$ is $\underset{\_}{2.055M}$, the concentration of ${\text{SO}}_3$ is $\underset{\_}{0.655M}$, the concentration of ${\text{NO}}_2$ is $\underset{\_}{0.045M}$ and the concentration of ${\text{SO}}_2$ is $\underset{\_}{0.800M}$.

Explanation of Solution

Explanation

Given

The balanced chemical equation is,

${\text{NO}}_2\left(g\right)+{\text{SO}}_2\left(g\right)\rightleftharpoons \text{NO}\left(g\right)+{\text{SO}}_3\left(g\right)$

The equilibrium concentration of ${\text{NO}}_2$ is $0.100\text{M}$.

The equilibrium concentration of ${\text{SO}}_2$ is $0.300\text{M}$.

The equilibrium concentration of $\text{NO}$ is $2.00\text{M}$.

The equilibrium concentration of ${\text{SO}}_3$ is $0.600\text{M}$.

The concentration of ${\text{SO}}_2$ after extra addition is $0.800\text{M}$.

The equilibrium constant is calculated by the formula,

$K_{\text{c}}=\frac{\left[\text{NO}\right]\left[{\text{SO}}_3\right]}{\left[{\text{NO}}_2\right]\left[{\text{SO}}_2\right]}$ (1)

Where,

• $\left[\text{NO}\right]$ is the equilibrium concentration of $\text{NO}$.
• $\left[{\text{SO}}_3\right]$ is the equilibrium concentration of ${\text{SO}}_3$.
• $\left[{\text{NO}}_2\right]$ is the equilibrium concentration of ${\text{NO}}_2$.
• $\left[{\text{SO}}_2\right]$ is the equilibrium concentration of ${\text{SO}}_2$.

Substitute the equilibrium concentrations of $\text{NO}$, ${\text{SO}}_3$, ${\text{NO}}_2$, ${\text{SO}}_2$ in equation (1).

$\begin{array}{c}{K}_{\text{c}}=\frac{\left[\text{2}\text{.00}\right]\left[0.600\right]}{\left[0.100\right]\left[0.300\right]}\\ =\frac{1.2}{0.03}\\ =40\end{array}$

Consider the following table showing the values of concentration at initial and at equilibrium stage after addition of extra ${\text{SO}}_2$,

$\begin{array}{cccccccc}& {\text{NO}}_2\left(g\right)& +& {\text{SO}}_2\left(g\right)& \rightleftharpoons & \text{NO}\left(g\right)& +& {\text{SO}}_3\left(g\right)\\ \text{Initial}& 0.100& & 0.300& & 2.00& & 0.600\\ \text{Equilibrium}& 0.100-x& & 0.800& & 2.00+x& & 0.600+x\end{array}$ (2)

Where,

• $x$ is the change in concentration occurred during the reaction.

Substitute the values of concentration of $\text{NO}$, ${\text{SO}}_3$, ${\text{NO}}_2$ ${\text{SO}}_2$ at equilibrium and equilibrium constant  in equation (1).

$\begin{array}{c}40=\frac{\left(2.00+x\right)\left(0.600+x\right)}{\left(0.100-x\right)\left(0.800\right)}\\ x^2+34.6x-2=0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.
• $b$ is the coefficient of $x$.
• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{c}x=\frac{-\left(34.6\right)\pm \sqrt{{\left(34.6\right)}^2-4\times 1\times \left(-2\right)}}{2\times 1}\\ =\frac{-34.6\pm 34.71}{2}\\ =\frac{0.11}{2}\\ =0.055\end{array}$

The value if $x$ is $0.055$.

Substitute the value of $x$  in equation (2).

The equilibrium concentration of $\text{NO}$ is $\begin{array}{l}=2.00+x\\ =2.00-0.055\\ =\underset{\_}{2.055M}\end{array}$

The equilibrium concentration of ${\text{SO}}_3$ is $\begin{array}{l}=0.600-x\\ =0.600-0.055\\ =\underset{\_}{0.655M}\end{array}$

The equilibrium concentration of ${\text{NO}}_2$ is $\begin{array}{l}=0.100-x\\ =0.100-0.055\\ =\underset{\_}{0.045M}\end{array}$

Therefore, the equilibrium concentration of $\text{NO}$ is $\underset{\_}{2.055M}$, the equilibrium concentration of ${\text{SO}}_3$ is $\underset{\_}{0.655M}$, the equilibrium concentration of ${\text{NO}}_2$ is $\underset{\_}{0.045M}$ and the equilibrium concentration of ${\text{SO}}_2$ is $\underset{\_}{0.800M}$.

Conclusion

The composition mixture when equilibrium reestablished is the concentration of $\text{NO}$ is $\underset{\_}{2.055M}$, the concentration of ${\text{SO}}_3$ is $\underset{\_}{0.655M}$, the concentration of ${\text{NO}}_2$ is $\underset{\_}{0.045M}$ and the concentration of ${\text{SO}}_2$ is $\underset{\_}{0.800M}$.