Chapter 15, Problem 15.82QP

Interpretation Introduction

Interpretation:

The concentrations of the gases have to be calculated using the data.

Concept Introduction:

• Equilibrium constant: at equilibrium the ratio of products to reactants (each raised to the power corresponding to its stoichiometric coefficient) has a constant value, K

  For a general reaction, $\text{aA+bB}\rightleftarrows \text{cC+dD}$

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

• For a general reaction $\text{aA+bB}\rightleftarrows \text{cC+dD}$, the ratio of product to reactant concentrations at any point in the reaction is the reaction quotient, $Q$

    $\text{Reaction}\text{quotient}\text{=}\text{Q}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

Explanation of Solution

Given data:

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\text{54}\text{.3}\\ \\ {\text{n}}_{{\text{H}}_{\text{2}}}\text{=}\text{0}\text{.714}\text{mol}\\ \\ {\text{n}}_{{\text{I}}_{\text{2}}}\text{=}\text{0}\text{.984}\text{mol}\\ \\ {\text{n}}_{\text{HI}}\text{=}\text{0}\text{.886}\text{mol}\\ \\ \text{V}\text{=}\text{2}\text{.40}\text{L}\end{array}$

The reaction is given below:

  $H_{\text{2}}\left(\text{g}\right)+I_2\left(g\right)\stackrel{}{\rightleftharpoons }\text{2}H\text{I}\left(\text{g}\right)$

The initial concentration of the gases can be calculated by the equation given below:

  $\text{Concentration}\text{=}\frac{\text{Number}\text{.}\text{of}\text{moles}}{\text{Volume}}$

$\text{Concentration}\text{of}{\left[{\text{H}}_{\text{2}}\right]}_{\text{0}}\text{=}\frac{\text{0}\text{.714}\text{mol}}{\text{2}\text{.40}}\text{=}\text{0}\text{.298}\text{M}$

$\text{Concentration}\text{of}{\left[I_{\text{2}}\right]}_{\text{0}}\text{=}\frac{\text{0}\text{.984}\text{mol}}{\text{2}\text{.40}}\text{=}\text{0}\text{.410}\text{M}$

$\text{Concentration}\text{of}{\left[\text{HI}\right]}_{\text{0}}\text{=}\frac{\text{0}\text{.886}\text{mol}}{\text{2}\text{.40}}\text{=}\text{0}\text{.369}\text{M}$

The reaction quotient can be calculated using the initial concentration as follows:

$Q_c=\frac{{\left[HI\right]}^2{}_0}{{\left[H_2\right]}_0{\left[I_2\right]}_0}=\frac{{\left(0.369\right)}^2}{\left(0.298\right)\left(0.410\right)}=1.11$

The $Q_c$ is less than ${\text{K}}_{\text{c}}$. Therefore, the equilibrium will shift to the right.

The decrease in concentration of ${\text{H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}$ is assumed to be x.

The ICE table for the reaction is given below

     $H_{\text{2}}\left(\text{g}\right)+I_2\left(g\right)\stackrel{}{\rightleftharpoons }\text{2}H\text{I}\left(\text{g}\right)$

  $\begin{array}{l}\begin{array}{cccccc}& H_2\left(g\right)& +I_2\left(g\right)& & \rightleftarrows & 2HI\left(g\right)\\ Initial& 0.298& 0.410& & & 0.369\\ Change& -x& -x& & & +2x\end{array}\\ \underset{\_}{}\\ \begin{array}{cccccc}Equilibirium& 0.298-x& 0.410-x& & & 0.369+2x\end{array}\end{array}$

Substituting the values in the equilibrium constant equation ,

  $\begin{array}{c}{\text{K}}_{\text{c}}\text{=}\frac{{\left[\text{HI}\right]}^{\text{2}}}{\left[H_2\right]\left[{\text{I}}_2\right]}\text{=}\frac{{\left(\text{0}\text{.369+2x}\right)}^{\text{2}}}{\left(\text{0}\text{.298-x}\right)\left(\text{0}\text{.410-x}\right)}\text{=}54.3\\ \\ \text{50}{\text{.3x}}^{\text{2}}\text{+}39.9\text{x+6}\text{.48}\text{=}\text{0}\end{array}$

By solving the above quadratic equation, the value of x can be determined.

$\text{x=}\text{0}\text{.228}\text{M}$

Therefore, the equilibrium concentrations of gases are given below:

$\begin{array}{l}\left[H_{\text{2}}\right]\text{=}\text{0}\text{.298-x}\text{=}\left(\text{0}\text{.298-0}\text{.228}\right)\text{M}\text{=}\text{0}\text{.070}\text{M}\\ \\ \left[{\text{I}}_{\text{2}}\right]\text{=}\text{0}\text{.410-x}\text{=}\left(\text{0}\text{.410-0}\text{.228}\right)\text{M}\text{=}\text{0}\text{.182}\text{M}\\ \\ \left[\text{HI}\right]\text{=}\text{0}\text{.369+2x}\text{=}\left(\text{0}\text{.369+2×0}\text{.228}\right)\text{M}\text{=}\text{0}\text{.825}\text{M}\end{array}$