Chapter 19, Problem 19.44P

Interpretation Introduction

Interpretation:

The new partial pressure of ${\text{PCl}}_{\text{5}}$, ${\text{PCl}}_{\text{3}}$ and ${\text{Cl}}_{\text{2}}$ 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:

The partial pressures of ${\text{PCl}}_{\text{5}}$, ${\text{PCl}}_{\text{3}}$ and ${\text{Cl}}_{\text{2}}$ in an equilibrium mixture is $\text{217 Torr}$, $\text{13}\text{.2 Torr}$ and $13.2\text{ Torr}$ respectively.  On addition of some ${\text{Cl}}_{\text{2}}$, the total pressure of the system jumps to $263\text{ Torr}$.

The equilibrium chemical equation of the reaction is,

  ${\text{PCl}}_{\text{3(g)}}{\text{ + Cl}}_{\text{2(g)}}\stackrel{}{\rightleftharpoons }{\text{ PCl}}_{\text{5(g)}}$

Calculation of new partial pressures:

The equilibrium constant expression of the given reaction is,

  ${\text{K}}_{\text{p}}\text{ = }\frac{{\text{P}}_{{\text{PCl}}_{\text{5}}}}{{\text{P}}_{{\text{PCl}}_3}{\text{ P}}_{{\text{Cl}}_{\text{2}}}}$

Calculate the equilibrium constant.

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{= }\frac{{\text{P}}_{{\text{PCl}}_{\text{5}}}}{{\text{P}}_{{\text{PCl}}_3}{\text{ P}}_{{\text{Cl}}_{\text{2}}}}\\ =\frac{217}{{(13.2)}^2}\\ =1.25\end{array}$

Determine the change in pressure and the new partial pressure for ${\text{Cl}}_{\text{2}}$.

Initial total pressure of the reaction is $\text{243}\text{.4 Torr}$.

Final total pressure after addition of ${\text{Cl}}_{\text{2}}$ is $263\text{ Torr}$.

The change in pressure is,

  $\begin{array}{l}\text{Change in P }\text{= Final total P - Initial total P}\\ \text{= 263 Torr - 243}\text{.4 Torr}\\ \text{=19}\text{.6 Torr}\end{array}$

The new partial pressure of ${\text{Cl}}_{\text{2}}$ is $\text{13}\text{.2 Torr + 19}\text{.6 Torr = 32}\text{.8 Torr}$

Construct ICE table for the reaction.

	${\text{PCl}}_{\text{3(g)}}{\text{ + Cl}}_{\text{2(g)}}\stackrel{}{\rightleftharpoons }{\text{ PCl}}_{\text{5(g)}}$
Initial $\text{(Torr)}$	13.2	32.8	217
Change $\text{(Torr)}$	-x	-x	+x
Equilibrium $\text{(Torr)}$	13.2 - x	32.8 - x	217+x

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

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{= }\frac{{\text{P}}_{{\text{PCl}}_{\text{5}}}}{{\text{P}}_{{\text{PCl}}_3}{\text{ P}}_{{\text{Cl}}_{\text{2}}}}\\ 1.25=\frac{(217\text{ + x)}}{[\text{(13}\text{.2 - x)(32}\text{.8 - x)}}\\ 1.25({\text{x}}^{\text{2}}\text{ - 46x+432}\text{.96) = 217 + x}\\ \text{1}{\text{.25x}}^2\text{ - 58}\text{.5x+324}\text{.2}\text{=0}\\ \text{solving x using quadratic equation, we get}\\ \text{x = 40}\text{.4 or 6}\text{.42}\\ \text{Neglect x = 40}\text{.4 since it gives negative values}\text{. Finally,}\\ \text{x = 6}\text{.42}\end{array}$

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

  $\begin{array}{l}{\text{P}}_{{\text{PCl}}_{\text{5}}}\text{= 217 + x}\\ \text{=217+6}\text{.42}\\ \text{=223}\text{.42 Torr}\\ {\text{P}}_{{\text{PCl}}_3}=13.2\text{ -x}\\ \text{=13}\text{.2 - 6}\text{.42}\\ \text{=6}\text{.78 Torr}\\ {\text{P}}_{{\text{Cl}}_{\text{2}}}=32.8\text{ -x}\\ \text{=32}\text{.8 - 6}\text{.42}\\ \text{=26}\text{.38 Torr}\end{array}$

Therefore, the new equilibrium partial pressures of ${\text{PCl}}_{\text{5}}$, ${\text{PCl}}_{\text{3}}$ and ${\text{Cl}}_{\text{2}}$ are,

  $\begin{array}{l}{\text{P}}_{{\text{PCl}}_{\text{5}}}\text{=223}\text{.42 Torr}\\ {\text{P}}_{{\text{PCl}}_3}\text{=6}\text{.78 Torr}\\ {\text{P}}_{{\text{Cl}}_{\text{2}}}\text{=26}\text{.38 Torr}\end{array}$