Chapter 19, Problem 19.23P

Interpretation Introduction

Interpretation:

The total gas pressure in the vessel for the given reaction, ${\text{NH}}_{\text{4}}{\text{HS}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ NH}}_{\text{3(g)}}{\text{ + H}}_{\text{2}}{\text{S}}_{\text{(g)}}$ 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.

Relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$:

The relation between and ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ is given by the following equation.

  ${\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta \text{ngas}}$

Where,

${\text{K}}_{\text{p}}$ is the equilibrium constant in terms of partial pressure

${\text{K}}_{\text{c}}$ is the equilibrium constant in terms of concentration

$\Delta \text{ngas = moles of gaseous product - moles of gaseous reactant}$

Only moles of gaseous products and reactants is used to calculate $\Delta \text{ngas}$

Explanation of Solution

Given:

Ammonium hydrogen sulphide decomposes according to the below chemical equation.

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

The equilibrium constant, ${\text{K}}_{\text{c}}$ for the reaction is $1.81\times {10}^{-4}{\text{ M}}^2$ at $25°\text{C}$.

Determination of total gas pressure:

First determine, ${\text{K}}_{\text{p}}$ for the reaction using given ${\text{K}}_{\text{c}}$ value.

$\begin{array}{l}\Delta \text{ngas }\text{= moles of gaseous product - moles of gaseous reactant}\\ \text{=2 - 0}\\ \text{=2}\end{array}$

Then,

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{= K}}_{\text{c}}{\text{(RT)}}^{\Delta \text{ngas}}\\ \text{= (}1.81\times {10}^{-4}){[(0.0821\text{ atm}·\text{L/mol}·\text{K)(298 K)]}}^2\\ =(1.81\times {10}^{-4}){(24.4658)}^2\\ =(1.81\times {10}^{-4})(598.58)\\ =0.108\text{ atm}\end{array}$

Convert the obtained ${\text{K}}_{\text{p}}$ value in units of $\text{bar}$.

  ${\text{K}}_{\text{p}}\text{ = 0}\text{.108 atm = 0}\text{.109 bar}$

The equilibrium constant expression for the given reaction is,

  ${\text{K}}_{\text{p}}{\text{ = P}}_{{\text{NH}}_3}{\text{ P}}_{{\text{H}}_{\text{2}}\text{S}}$

According to the stoichiometry of the equation, it is known that both ${\text{NH}}_{\text{3(g)}}$ and $\text{H}{\text{2}}_{}{\text{S}}_{\text{(g)}}$ are present with same number of moles.

Thus, their equilibrium pressures would also be same.  Now, consider equilibrium partial pressure of each gas as ‘x’.

Using equilibrium constant expression, solve for ‘x’.

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{= P}}_{{\text{NH}}_3}{\text{ P}}_{{\text{H}}_{\text{2}}\text{S}}\\ \text{ 0}{\text{.109 bar}}^2\text{ = (x)(x)}\\ \text{ 0}{\text{.109 bar}}^2{\text{ = (x)}}^2\\ \text{x}\text{=}\sqrt{\text{ 0}{\text{.109 bar}}^2}\\ \text{=}\text{0}\text{.330 bar}\\ \approx 0.33\text{ bar}\end{array}$

Therefore, the equilibrium partial pressures of ${\text{NH}}_{\text{3(g)}}$ and $\text{H}{\text{2}}_{}{\text{S}}_{\text{(g)}}$ are,

  $\begin{array}{l}{\text{ P}}_{{\text{NH}}_3}\text{ = 0}\text{.33 bar}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{S}}\text{ = 0}\text{.33 bar}\end{array}$

Now, sum up the obtained partial pressures and find the total gas pressure as follows,

  $\begin{array}{l}\text{Total Pressure }{\text{= P}}_{{\text{NH}}_3}{\text{ + P}}_{{\text{H}}_{\text{2}}\text{S}}\\ \text{= 0}\text{.33 bar + 0}\text{.33 bar}\\ \text{=0}\text{.66 bar}\end{array}$

Therefore, the total pressure in the reaction vessel at equilibrium is $\text{0}\text{.66 bar}$.