Question

At ${120}^{\circ }\mathrm{C},K_c=0.090$ for the reaction $${\mathrm{SO}}_2{\mathrm{Cl}}_2(g)\rightleftharpoons {\mathrm{SO}}_2(g)+{\mathrm{Cl}}_2(g)$$ In an equilibrium mixture of the three gases, the concentrations of ${\mathrm{SO}}_2,{\mathrm{Cl}}_2,$ and ${\mathrm{SO}}_2$ are $0.100 \mathrm{M}$and$0.075 \mathrm{M}$, respectively. What is the partial pressure of ${\mathrm{Cl}}_2$ in the equilibrium mixture?

Short answer

The partial pressure of Cl2 in the equilibrium mixture is 0.0434 atm.

Step-by-step solution

Write the Balanced Chemical Equation

The balanced chemical equation is already given: ${\mathrm{SO}}_2{\mathrm{Cl}}_2(g)\rightleftharpoons {\mathrm{SO}}_2(g)+{\mathrm{Cl}}_2(g)$

Write the Expression for the Equilibrium Constant, Kc

For the reaction, the expression for Kc is: $K_c=\frac{[{\mathrm{SO}}_2][{\mathrm{Cl}}_2]}{[{\mathrm{SO}}_2{\mathrm{Cl}}_2]}$ We are given that $K_c=0.090$ at 120°C.

Find the Change in Concentrations during the Reaction

Let the change in the concentration of reactants and products during the reaction be x. Since the balanced equation has a 1:1:1 stoichiometric ratio, we can find the change in concentration for each species. At equilibrium: $[S{O}_{2}C{l}_2]=0.100-x$ $[S{O}_2]=0.075+x$ $[C{l}_2]=x$

Substitute the Concentrations back into the Kc Expression

Substitute the concentrations into the Kc expression: $0.090=\frac{(0.075+x)(x)}{(0.100-x)}$

Solve for x

Solve the above equation for x: $0.090=\frac{0.075x+x^2}{0.100-x}$ Multiply both sides by $(0.100-x)$: $0.009-0.090x=0.075x+x^2$ Rearrange the equation: $x^2+0.165x-0.009=0$ Now, solve the quadratic equation for x. We can use the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ In this case, $a=1$, $b=0.165$, and $c=-0.009$. Upon solving for x, we find that there are two possible values: $x=-0.996$ and $x=0.00906$. Since the negative value of x doesn't make sense in the context of concentrations, we choose the positive value for x: $x=0.00906\mathrm{M}$ This means that the equilibrium concentration of Cl2 is 0.00906 M.

Calculate the Partial Pressure of Cl2

To calculate the partial pressure of Cl2, first, we can assume that the total pressure (P) is 1 atm since it is not given. Then, we can use the mole fraction of Cl2 in the gas mixture to find its partial pressure. Mole fraction of Cl2, $X_{Cl2}=\frac{[C{l}_2]}{([S{O}_2]+[S{O}_{2}C{l}_2]+[C{l}_2])}=\frac{0.00906}{(0.075+0.100-0.00906+0.00906)}$ Calculate $X_{Cl2}$: $X_{C{l}_2}=0.0434$ Now, calculate the partial pressure of Cl2 using the total pressure and the mole fraction: $P_{C{l}_2}=X_{C{l}_2}\times P$ $P_{C{l}_2}=0.0434\times 1\mathrm{atm}$ $P_{C{l}_2}=0.0434\mathrm{atm}$ So, the partial pressure of Cl2 in the equilibrium mixture is 0.0434 atm.