Chapter 14, Problem 14.96QP

Interpretation Introduction

Interpretation:

The equilibrium composition of the given 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 14.96QP

The equilibrium mixture contains $0.87\text{mol}{\text{N}}_2$, $3.62\text{mol}{\text{H}}_{\text{2}}$ and $0.125\text{mol}{\text{NH}}_3$

Explanation of Solution

 Explanation:

Given,

The equilibrium constant ${\text{K}}_{\text{c}}=0.153$

The initial amount of ${\text{N}}_{\text{2}}\text{=}\text{1}\text{.00 mol}$

The initial amount of ${\text{H}}_{\text{2}}\text{=}4.\text{00 mol}$

The volume of the vessel $\text{=}10.0\text{L}$

To find initial concentration of reactants

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

$\begin{array}{l}\text{Initial concentration of}{N}_2\text{=}\frac{\text{Num of moles}}{\text{Volume}}\\ =\frac{\text{1}\text{.00mol}}{\text{10}\text{.00L}}\\ =0.100\text{M}\end{array}$

$\begin{array}{l}\text{Initial concentration of}{\text{H}}_{\text{2}}\text{=}\frac{\text{Num of moles}}{\text{Volume}}\\ =\frac{\text{4}\text{.00mol}}{\text{10}\text{.00L}}\\ =0.400\text{M}\end{array}$

To find the equilibrium composition.

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

$\text{Amount}\text{(M)}{\text{N}}_{\text{2(g)}}+{\text{3H}}_{\text{2(g)}}\rightleftharpoons {\text{2NH}}_3{}_{\text{(g)}}$

$\begin{array}{l}\text{Initial}0.1000.400\rightleftharpoons \text{0}\\ \text{Change}\text{-x}-3x\rightleftharpoons +2x\\ \text{Equillibrium}\left(0.100-x\right)\left(0.400-3x\right)\rightleftharpoons x\end{array}$

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

${\text{K}}_{\text{c}}\text{=}\frac{{\left[{\text{NH}}_{\text{3}}\right]}^{\text{2}}}{\left[{\text{N}}_{\text{2}}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{3}}}$

$0.153=\frac{{\left(2x\right)}^2}{(0.100-x){(0.400-3x)}^2}$

Given,

$f(x)=0.153$

Or

$f(x)=\frac{{\left(2x\right)}^2}{(0.100-x){(0.400-3x)}^2}=0.153$

The unknown value x is then calculated by guessing various values.

Since, ${\text{K}}_{\text{c}}<1$, we can assume that below 50% of the reaction is complete.  The value of x is assumed to be $0.010$ and that is used as the first value and entered in the table given below.

x	$\text{f(x)}$	Interpretation
$0.010$	$0.0877$	x too small
$0.020$	$0.5089$	x too big
$0.015$	$0.237$	x still too big
$0.012$	$0.136$	$x<0.153\text{(but}\text{close)}$
$0.013$	$0.1651$	$\text{f(x)}\text{of0}\text{.1651}\cong 0.153$
$0.0125$	$0.150$	Best value for $\text{f(x)}$

Hence, the value of x is can be taken as $0.0125$

$\begin{array}{l}\text{The equilibrium concentration of}{N}_2\text{=0}\text{.100-x}\\ \text{=0}\text{.100-0}\text{.0125}\\ =\text{0}\text{.087M}\end{array}$

$\begin{array}{l}{\text{Number of moles of N}}_2\text{ at equilibrium}\text{=Equilibrium}\text{concentration×Volume}\\ =0.087\times 10.0\\ =0.87\text{mol}\end{array}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{H}}_{\text{2}}\text{=0}\text{.400-3x}\\ \text{=0}\text{.400-(3}\times \text{0}\text{.0125)}\\ =\text{0}\text{.362M}\end{array}$

$\begin{array}{l}{\text{Number of moles of H}}_2\text{at equilibrium}\text{=Equilibrium}\text{concentration×Volume}\\ =0.362\times 10.0\\ =3.62\text{mol}\end{array}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{NH}}_{\text{3}}\text{=x}\\ \text{=0}\text{.0125M}\end{array}$

$\begin{array}{l}{\text{Number of moles of NH}}_3\text{ at equilibrium}\text{=Equilibrium}\text{concentration×Volume}\\ =0.0125\times 10.0\\ =0.125\text{mol}\end{array}$

Conclusion

The equilibrium composition of the given reaction mixture at $\text{450°C}$ was found by using the value of equilibrium constant.