Chapter 14, Problem 14.120QP

The K_P for the reaction

${\text{SO}}_2{\text{Cl}}_2(g)\rightleftharpoons {\text{SO}}_2(g)+{\text{Cl}}_2(g)$

is 2.05 at 648 K. A sample of SO₂Cl₂ is placed in a container and heated to 648 K while the total pressure is kept constant at 9.00 atm. Calculate the partial pressures of the gases at equilibrium.

Interpretation Introduction

Interpretation:

Partial pressures of the gases at equilibrium has to be calculated

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

$\begin{array}{l}\text{aA}\text{+}\text{bB}\to \text{cC}\text{+}\text{dD}\\ \\ {\text{K}}_{eq}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}\text{for}\text{aqueous}\\ \\ {\text{K}}_{\text{eq}}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}\text{for}\text{gases}\\ \end{array}$

Multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{A}\text{+ B }\rightleftarrows {\text{ C + D K}}_{\text{c}}^{\text{'}}\\ \text{C}\text{+ D }\rightleftarrows {\text{ E + F K}}_{\text{c}}^{\text{''}}\end{array}}\\ \text{overall}\text{reaction:}\underset{\_}{\text{A}\text{+ B }\rightleftarrows \text{ E + F}{\text{K}}_{\text{c}}}\end{array}$

The equilibrium constant for two separate equilibrium constants are,

${\text{K}}_{\text{c}}^{\text{'}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\text{and}{\text{K}}_{\text{c}}^{\text{''}}\text{=}\frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{C}\right]\left[\text{D}\right]}$

For overall reaction, the equilibrium constant ${\text{K}}_{\text{c}}$ is,

${\text{K}}_{\text{c}}^{\text{'}}{\text{K}}_{\text{c}}^{\text{''}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\times \frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{C}\right]\left[\text{D}\right]}=\frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Therefore, $\boxed{{\text{K}}_c\text{=}{\text{K}}_{\text{c}}^{\text{'}}\times {\text{K}}_{\text{c}}^{\text{''}}}$

Calculating equilibrium concentration:

• Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
• Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
• Having solved for x, calculate the equilibrium concentrations of all species.

Explanation of Solution

The given reaction is,

${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\rightleftharpoons }{\text{SO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

Given:  Total pressure is $\text{9 atm}$; ${\text{K}}_P=2.05$; $T=648K$.

Construct ICE table as follows,

$\begin{array}{l}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\rightleftharpoons }{\text{SO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\\ Initial\left(atm\right):9.0000\\ Change\left(atm\right):-2x+x+x\\ Equilibrium\left(atm\right):9.0-2xxx\end{array}$

Initially, the pressure of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is $\text{9 atm}$. The pressure is held constant, so after the reaction reaches equilibrium $P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}+P_{{\text{SO}}_{\text{2}}}+P_{{\text{Cl}}_{\text{2}}}=9.\text{00}\text{atm}$. The amount (pressure) of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ reacted must equal the pressure of ${\text{SO}}_{\text{2}}$ and ${\text{Cl}}_{\text{2}}$ produced for the pressure to remain constant. If we let $P_{{\text{SO}}_{\text{2}}}+P_{{\text{Cl}}_{\text{2}}}=x$, then the pressure of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ reacted must be $\text{2x}$.

Again the change in pressure for ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$$\text{-2x}$ does not match the stoichiometry of the reaction, because change in pressure is expressed. The total pressure is kept at 9.00atm throughout.

$\begin{array}{l}{K}_P=\frac{P_{{\text{SO}}_{\text{2}}}{P}_{{\text{Cl}}_{\text{2}}}}{P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}}\\ \\ 2.05=\frac{\left(x\right)\left(x\right)}{9.00-2x}\\ \\ x^2+4.10x-18.45=0\end{array}$

Solving quadratic equation as follows,

$\begin{array}{l}{x}^2+4.10x-18.45=0\\ where,\\ a=1,b=4.10,andc=-18.45\\ \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \\ =\frac{-4.10\pm \sqrt{{\left(4.10\right)}^2+73.8}}{2}=\frac{-4.10\pm \sqrt{90.61}}{2}\\ \\ =\frac{-4.10+9.5}{2};\frac{-4.10-9.5}{2}\\ \\ x=2.7;6.8\end{array}$

Out of the two values, the value of ‘x’ has to be less than the original concentration of reactants. So, the value of $6.8$ is physically possible.

At equilibrium,

$\begin{array}{l}{P}_{S{O}_2}=P_{C{l}_2}=x=2.71atm\\ \\ P_{S{O}_{2}C{l}_2}=9.00-2\left(2.71\right)=3.58atm\end{array}$

Therefore, Partial pressures of the gases at equilibrium were calculated.