Chapter 14, Problem 14.129QP

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 14.129QP

The equilibrium mixture contains $0.429\text{mol}{\text{I}}_{\text{2}}$, $0.429\text{mol}{\text{Br}}_{\text{2}}$ and $\text{1}\text{.429mol}\text{HBr}$

Explanation of Solution

Given,

The initial amount of ${\text{I}}_{\text{2}}\text{=}\text{0}\text{.500}\text{mol}$

The initial amount of ${\text{Br}}_{\text{2}}\text{=}\text{0}\text{.500}\text{mol}$

The final amount of $\text{IBr=}\text{0}\text{.4221}\text{mol}$

To find the equilibrium constant.

$\begin{array}{l}\text{Amount}\text{(mol)}{\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}}{}_{\text{(g)}}\rightleftharpoons {\text{2HI}}_{\text{(g)}}\\ \end{array}$

$\begin{array}{l}\text{Initial}0.500+0.500\rightleftharpoons \text{0}\text{}\\ \text{Change}\text{-x}+-x\rightleftharpoons +2x(=0.4221)\\ \text{Equillibrium}\left(0.500-0.21105\right)=0.2889+\left(0.500-0.21105\right)=0.2889\rightleftharpoons 0.4211\\ \end{array}$

The equilibrium concentration values are then substituted into the equilibrium expression to get the value of equilibrium constant.

${\text{K}}_{\text{c}}\text{=}\frac{{\left[\text{IBr}\right]}^{\text{2}}}{\left[{\text{I}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}=\frac{{(0.4221)}^2}{(0.2889)(0.2889)}=2.134$

To find the equilibrium composition.

The initial  amount of ${\text{I}}_{\text{2}}\text{=}2.00\text{mol}$

The initial  amount of ${\text{Br}}_{\text{2}}\text{=}2.00\text{mol}$

The equilibrium constant ${\text{K}}_{\text{c}}\text{=}\text{2}\text{.314}$

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

$\text{Amount}\text{(mol)}{\text{I}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{Br}}_{\text{2}}{}_{\text{(g)}}\rightleftharpoons {\text{2IBr}}_{\text{(g)}}$

$\begin{array}{l}\text{Initial}1.00+1.00\rightleftharpoons \text{0}\text{}\\ \text{Change}\text{-x}+-x\rightleftharpoons +2x\\ \text{Equillibrium}\left(1.00-x\right)+\left(1.00-x\right)\rightleftharpoons 2x\end{array}$

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

${\text{K}}_{\text{c}}\text{=2}\text{.134=}\frac{{\left[\text{IBr}\right]}^{\text{2}}}{\left[{\text{I}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]}=\frac{{(2x)}^2}{(1.00-x)(1.00-x)}$

On rearranging we get a quadratic equation.

$1.866x^2+6.402x-4.268=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{-(6.402)\pm \sqrt{{(6.40)}^2-4(1.866)(-4.268)}}{2(1.866)}$

On solving we get two values for x, the positive value for x is taken.

$\text{x}=0.515\text{mol}$

Hence,

$\begin{array}{l}{\text{The equilibrium amout of Br}}_2\text{=}1.00-x\\ =1.00-(0.5715)\\ =0.429\text{mol}\end{array}$

$\begin{array}{l}{\text{The equilibrium amout of I}}_2\text{=}1.00-x\\ =1.00-(0.5715)\\ =0.429\text{mol}\end{array}$

$\begin{array}{l}\text{The equilibrium amout of HBr}\text{=}1.00-2x\\ =1.00-(2\times 0.5715)\\ =1.143\text{mol}\end{array}$

Conclusion

The equilibrium composition of the given mixture was found by using the value of equilibrium constant and initial amount of the reactants.