Question

At 1100 K, for the following reaction:

${\mathbf{\text{2SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)+O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\stackrel{}{\mathbf{\rightleftharpoons }}{\mathbf{\text{2SO}}}_{\mathbf{\text{3}}}\mathbf{\text{(g)}}$

Calculate the equilibrium partial pressures of SO₂, O₂, and SO₃ produced from an initial mixture in which P_SO2 = P_O2 = 0.50 atm and P_SO3 = 0.

Short answer

The equilibrium partial pressures of SO₂, O₂, and SO₃ at 1100 K are 0.38 atm, 0.44 atm and 0.12 atm, respectively.

Step-by-step solution

The expression for equilibrium constant

The expression for the equilibrium constant, K_p of the given reaction in terms of partial pressures is written below:

${\text{K}}_{\text{P}}\text{=}\frac{{{\text{P}}_{{\text{SO}}_{\text{3}}}}^{\text{2}}}{\left({{\text{P}}_{{\text{SO}}_{\text{2}}}}^{\text{2}}\right)\left({\text{P}}_{{\text{O}}_{\text{2}}}\right)}$ …(1)

The given initial partial pressures of SO₂, O₂, and SO₃ are 0.50 atm, 0.50 atm and 0 atm, respectively.

ICE table

Let x be the change in partial pressure.

Setup the ICE table as:

$\begin{array}{cccccc}& {\text{2SO}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& {\text{2SO}}_{\text{3}}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{atm}\right)& \text{0.50}& & \text{0.50}& & \text{0}\\ \text{Change}\left(\text{atm}\right)& \text{\hspace{0.17em}–\hspace{0.17em}2x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+2x}\\ \text{Equilibrium}\left(\text{atm}\right)& \text{0.50\hspace{0.17em}–\hspace{0.17em}2x}& & \text{0.50\hspace{0.17em}–\hspace{0.17em}x}& & \text{2x}\end{array}$

The value of “x”

Now, plug the values of K_p and the terms of the equilibrium partial pressures of the gases from ICE table into equation (1) to obtain the value of x.

$\text{0.25=}\frac{{\left(\text{2x}\right)}^{\text{2}}}{{\left(\text{0.50\hspace{0.17em}–\hspace{0.17em}2x}\right)}^{\text{2}}\left(\text{0.50\hspace{0.17em}–\hspace{0.17em}x}\right)}$

Solve for x.

$\text{x=0.062}$

Equilibrium partial pressures

The equilibrium partial pressures of SO₂, O₂, and SO₃ can be calculated by using the x value as follows:

$\begin{array}{c}{P}_{S{O}_2}=0.50-2x\\ =0.50-2\left(0.062\right)\\ =0.376\\ \cong 0.38atm\end{array}$

$\begin{array}{c}{P}_{O_2}=0.50-x\\ =0.50-\left(0.062\right)\\ =0.438\\ \cong 0.44atm\end{array}$

$\begin{array}{c}{P}_{S{O}_3}=2x\\ =2\left(0.062\right)\\ =0.124\\ \cong 0.12atm\end{array}$

Therefore, the equilibrium partial pressures of SO₂, O₂, and SO₃ at 1100 K are 0.38 atm, 0.44 atm and 0.12 atm, respectively.