Chapter 19, Problem 19.46P

Interpretation Introduction

Interpretation:

The total pressure for the given reaction after equilibrium is reestablished has to be calculated.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{p}}\right)$:

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration.  Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$ is defined as ratio of partial pressure of products to partial pressure of reactants.  Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

    $\text{aA}\rightleftharpoons \text{bB}$

    $\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\text{P}}_{\text{A}}{}^{\text{a}}{\text{=k}}_{\text{r}}{\text{P}}_{\text{B}}{}^{\text{a}}\end{array}$

On rearranging,

    $\frac{{\text{P}}_{\text{B}}{}^{\text{b}}}{{\text{P}}_{\text{A}}{}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{p}}$

Where,

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

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

    ${\text{K}}_{\text{p}}$ is the equilibrium constant.

Explanation of Solution

Given:

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ gas is introduced in an empty reaction container at an initial pressure of $\text{0}\text{.554 bar}$.  The total pressure after equilibrium is $0.770\text{ bar}$.  On addition of some ${\text{NO}}_{\text{2}}$, the total pressure of the system jumps to $0.906\text{ bar}$.

The equilibrium chemical equation of the reaction is,

  ${\text{N}}_{\text{2}}{\text{O}}_{\text{4(g)}}\stackrel{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2(g)}}$

Calculation of new partial pressures:

The equilibrium constant expression of the given reaction is,

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\text{P}}_{{\text{NO}}_{\text{2}}}^2}{{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}}$

From total pressure, calculate the initial partial pressure of ${\text{NO}}_{\text{2}}$.

  $\begin{array}{l}{\text{P}}_{\text{tot}}{\text{= P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{ + P}}_{{\text{NO}}_{\text{2}}}\\ 0.770=0.554\text{}{\text{+P}}_{{\text{NO}}_{\text{2}}}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}=0.770\text{ bar - }0.554\text{ bar}\\ =0.216\text{ bar}\end{array}$

Calculate the equilibrium constant.

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{= }\frac{{\text{P}}_{{\text{NO}}_{\text{2}}}^2}{{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}}\\ =\frac{{(0.216)}^2}{0.554}\\ =0.0842\end{array}$

Determine the change in pressure and the new partial pressure for ${\text{NO}}_{\text{2}}$.

Initial total pressure of the reaction is $\text{0}\text{.770 bar}$.

Final total pressure after addition of ${\text{NO}}_{\text{2}}$ is $\text{0}\text{.906 bar}$.

The change in pressure is,

  $\begin{array}{l}\text{Change in P }\text{= Final total P - Initial total P}\\ \text{= 0}\text{.906 bar - 0}\text{.770 bar}\\ \text{=0}\text{.136 bar}\end{array}$

The new partial pressure of ${\text{NO}}_{\text{2}}$ is $\text{0}\text{.136 bar + 0}\text{.216 bar = 0}\text{.352 bar}$

Construct ICE table for the reaction.

	${\text{N}}_{\text{2}}{\text{O}}_{\text{4(g)}}\stackrel{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2(g)}}$
Initial $\text{(bar)}$	0.554	0.352
Change $\text{(bar)}$	-x	+2x
Equilibrium $\text{(bar)}$	(0.554 – x)	(0.352 + 2x)

Plug the equilibrium pressures from ICE table in equilibrium constant expression and solve for x.

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{= }\frac{{\text{P}}_{{\text{NO}}_{\text{2}}}^2}{{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}}\\ 0.0842=\frac{{(0.352+2\text{x)}}^2}{\text{(0}\text{.554 - x)}}\\ 0.0842\text{(0}\text{.554 - x)=}{(0.352+2\text{x)}}^2\\ {\text{-4x}}^2\text{ - 1}\text{.4922x - 0}\text{.077257 =0}\\ \text{solving x using quadratic equation, we get}\\ \text{x = 0}\text{.311 or 0}\text{.062}\\ \text{Neglect x = 0}\text{.311 since it gives negative values}\text{. Finally,}\\ \text{x = 0}\text{.062 bar}\end{array}$

Substitute ‘x’ in the equilibrium partial pressure of ICE table and solve for new partial pressures.

  $\begin{array}{l}{\text{P}}_{{\text{NO}}_{\text{2}}}=\text{ 0}\text{.352+2x}\\ \text{=0}\text{.352+2(0}\text{.062)}\\ \text{=0}\text{.352+0}\text{.124}\\ \text{=0}\text{.476 bar}\\ {\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}=0.554-\text{0}\text{.062}\\ =0.492\text{ bar}\end{array}$

Therefore, the new equilibrium partial pressures of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{NO}}_{\text{2}}$ are,

  $\begin{array}{l}{\text{P}}_{{\text{NO}}_{\text{2}}}=\text{0}\text{.476 bar}\\ {\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}=0.492\text{ bar}\end{array}$

The total pressure after equilibrium is re-established is,

  $\begin{array}{l}{\text{P}}_{\text{tot}}{\text{= P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{ + P}}_{{\text{NO}}_{\text{2}}}\\ =0.492\text{ bar}\text{}\text{+0}\text{.476 bar}\\ =0.968\text{ bar}\end{array}$

Therefore, the total pressure after equilibrium is re-established is found as $0.968\text{ bar}$.