Chapter 14, Problem 14.71QP

Suppose 1.000 mol CO and 3.000 mol H₂ are put in a 10.00-L vessel at 1200 K. The equilibrium constant K_c for

$\text{CO}(g)+3{\text{H}}_2(g)\stackrel{}{\rightleftharpoons }{\text{CH}}_4(g)+{\text{H}}_2\text{O}(g)$

equals 3.92. Find the equilibrium composition of the reaction mixture.

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found.

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Answer to Problem 14.71QP

The equilibrium mixture contains, $0.0613\text{M}\text{CO}$, $0.1893\text{M}{\text{H}}_{\text{2}}$, $0.0387\text{M}{\text{H}}_{\text{2}}\text{O}$ and $0.0387\text{M}{\text{CH}}_4$

Explanation of Solution

Given,

The equilibrium constant ${\text{K}}_c=3.92$

The initial amount of $\text{CO}\text{=}1.000\text{mol}$

The initial amount of ${\text{H}}_{\text{2}}\text{ =}3.000\text{mol}$

The volume of the vessel $\text{=}10.00\text{L}$

To find initial concentration of reactants

The initial concentrations of the gaseous reactant are found as given below.

$\begin{array}{l}\text{Initial concentration of}CO\text{=}\frac{\text{Num of moles}}{\text{Volume}}\\ =\frac{\text{1}\text{.000mol}}{10.0\text{L}}\\ =0.1000\text{M}\end{array}$

$\begin{array}{l}\text{Initial concentration of}{\text{H}}_2\text{=}\frac{\text{Num of moles}}{\text{Volume}}\\ =\frac{\text{3}\text{.000mol}}{10.0\text{L}}\\ =0.3000\text{M}\end{array}$

To find the equilibrium composition.

Using the table approach, the equilibrium concentrations of the reactants and the products can be found.

$\begin{array}{l}\text{Amount}\text{(M)}{\text{CO}}_{\text{(g)}}+{\text{3H}}_{\text{2}}{}_{\text{(g)}}\rightleftharpoons {\text{CH}}_4{}_{\text{(g)}}+{\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \end{array}$

$\begin{array}{l}\text{Initial}0.10000.3000\rightleftharpoons \text{0}\text{}0\\ \text{Change}\text{-x}-3x\rightleftharpoons +x+x\\ \text{Equillibrium}\left(0.1000-x\right)\left(0.3000-3x\right)\rightleftharpoons xx\end{array}$

The equilibrium concentration values are then substituted into the equilibrium expression to get the change in concentration x.

${\text{K}}_{\text{c}}\text{=}\frac{\left[{\text{CH}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{3}}}$

$\begin{array}{l}3.92=\frac{x^2}{\left[0.1000-x\right]{\left[3\left(0.1000-x\right)\right]}^3}\\ 3.92=\frac{x^2}{27{\left(0.1000-x\right)}^4}\end{array}$

Both the sides are multiplied by 27 and taking the square root, we get

$10.29=\frac{x}{{\left(0.1000-x\right)}^2}$

On rearranging we get a quadratic equation.

$10.29x^2-3.058x+0.1029=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{3.058\pm \sqrt{{(-3.058)}^2-4(10.29)(0.1029)}}{2(10.29)}$

On solving we get two values for x

$\text{x}=0.2585\text{M}\text{or}\text{0}\text{.0387M}$

It is not possible to lose $0.2585$ from $0.1000$.  Hence, the logical value for x is $\text{0}\text{.0387}$

$\begin{array}{l}\text{The equilibrium concentration of CO}\text{=}0.1000-x\\ =0.1000-0.0387\\ =0.0613\text{M}\end{array}$

$\begin{array}{l}{\text{The equilibrium concentration of H}}_2\text{=}0.3000-3x\\ =0.3000-\left(0.0387\times 3\right)\\ =0.1839\text{M}\end{array}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{CH}}_{\text{4}}\text{=}\text{x}\\ \text{=0}\text{.0387M}\end{array}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{H}}_2\text{O}\text{=}\text{x}\\ \text{=0}\text{.0387M}\end{array}$

Conclusion

The equilibrium composition of the given reaction mixture at ${957}^{°}\text{C}$ was found by using the value of equilibrium constant.