Chapter 19, Problem 19.45P

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:

The decomposition of Ammonium hydrogen sulfide takes place as follows,

  ${\text{NH}}_{\text{4}}{\text{HS}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ NH}}_{\text{3(g)}}{\text{ + H}}_{\text{2}}{\text{S}}_{\text{(g)}}$

${\text{NH}}_{\text{4}}{\text{HS}}_{\text{(s)}}$ is placed in a sealed container.  The total pressure after equilibrium is $0.664\text{ bar}$.  On addition of some ${\text{NH}}_3$, the total pressure of the system jumps to $0.906\text{ bar}$.

Calculation of new partial pressures:

The equilibrium constant expression of the given reaction is,

  ${\text{K}}_{\text{p}}{\text{ = P}}_{{\text{NH}}_{\text{3}}}{\text{ P}}_{{\text{H}}_{\text{2}}\text{S}}$

Since ${\text{NH}}_{\text{4}}{\text{HS}}_{\text{(s)}}$ is a solid, it is not involved in the equilibrium constant expression.  Form equilibrium reaction equation, it is known that one mole each of both ${\text{NH}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{S}$ are formed.  Since same mole of both gases are formed, consider their partial pressures as ‘x’.

The partial pressure of ${\text{NH}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{S}$ are calculated as follows,

  $\begin{array}{l}{\text{P}}_{\text{tot}}{\text{= P}}_{{\text{NH}}_3}{\text{ + P}}_{{\text{H}}_{\text{2}}\text{S}}\\ 0.664=\text{x + x}\\ 0.664=2\text{x}\\ \text{ x}\text{=}\frac{0.664}{2}\\ =0.332\text{ bar}\end{array}$

Calculate the equilibrium constant.

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{= P}}_{{\text{NH}}_{\text{3}}}{\text{ P}}_{{\text{H}}_{\text{2}}\text{S}}\\ =0.332\times 0.332\\ =0.110\end{array}$

Determine the change in pressure and the new partial pressure for ${\text{NH}}_3$.

Initial total pressure of the reaction is $\text{0}\text{.664 bar}$.

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

The change in pressure is,

  $\begin{array}{l}\text{Change in P }\text{= Final total P - Initial total P}\\ \text{= 0}\text{.906 bar - 0}\text{.664 bar}\\ \text{=0}\text{.242 bar}\end{array}$

The new partial pressure of ${\text{NH}}_3$ is $\text{0}\text{.242 bar + 0}\text{.332 bar = 0}\text{.574 bar}$

Construct ICE table for the reaction.

	${\text{ NH}}_{\text{4}}{\text{HS}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ NH}}_{\text{3(g)}}{\text{ + H}}_{\text{2}}{\text{S}}_{\text{(g)}}$
Initial $\text{(bar)}$		0.574	0.332
Change $\text{(bar)}$		+x	+x
Equilibrium $\text{(bar)}$		(0.574 + x)	(0.332 + x)

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

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{= P}}_{{\text{NH}}_{\text{3}}}{\text{ P}}_{{\text{H}}_{\text{2}}\text{S}}\\ 0.110=\text{ (0}\text{.574+x)(0}\text{.332+x)}\\ 0.0842\text{(0}\text{.554 - x)=}{(0.352+2\text{x)}}^2\\ {\text{-4x}}^2\text{ - 1}\text{.4922x - 0}\text{.077257 =0}\\ \text{solving x using quadratic equation, we get}\\ \text{x = 0}\text{.806 or 0}\text{.1}\\ \text{Neglect x = 0}\text{.806 since it gives negative values}\text{. Finally,}\\ \text{x = 0}\text{.1 bar}\end{array}$

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

  $\begin{array}{l}{\text{P}}_{{\text{NH}}_{\text{3}}}=0.574+0.1\\ \text{=0}\text{.674 bar}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{S}}=0.574+0.1\\ =\text{0}\text{.674 bar}\end{array}$

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

  $\begin{array}{l}{\text{P}}_{{\text{NH}}_{\text{3}}}\text{=0}\text{.674 bar}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{S}}=\text{0}\text{.674 bar}\end{array}$

The total pressure after equilibrium is re-established is,

  $\begin{array}{l}{\text{P}}_{\text{tot}}{\text{= P}}_{{\text{NH}}_{\text{3}}}{\text{ + P}}_{{\text{H}}_{\text{2}}\text{S}}\\ =\text{0}\text{.674 bar }\text{}\text{+ 0}\text{.674 bar}\\ =1.348\text{ bar}\end{array}$

Therefore, the total pressure after equilibrium is re-established is found as $1.348\text{ bar}$.