# Calculate P_{SO_2} at equilibrium at 298 K?

## For the following reaction run at 298 K; a mixture of equilibrium contains ${P}_{{O}_{2}} = 0.50$ atm, ${P}_{S {O}_{2}} = 2.0$ atm. Calculate ${P}_{S {O}_{2}}$ at equilibrium at 298 K? $2 S {O}_{2} \left(g\right) + {O}_{2} \left(g\right) \setminus r i g h t \le f t h a r p \infty n s 2 S {O}_{3} \left(g\right)$ \Delta G°_{f,SO_3(g)}=-371\frac{kJ}{mol} \Delta G°_{f,SO_2(g)}=-300\frac{kJ}{mol}

##### 1 Answer
Jul 28, 2018

Well, apparently all the ${\text{SO}}_{2}$ was consumed...

First, write the reaction and equilibrium constant:

$2 {\text{SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO}}_{3} \left(g\right)$

${K}_{P} = {P}_{S {O}_{3}}^{2} / \left({P}_{S {O}_{2}}^{2} {P}_{{O}_{2}}\right)$

What we want to solve for is the (always positive) partial pressure of ${\text{SO}}_{2}$ (not ${\text{SO}}_{3}$, apparently):

P_(SO_2) = sqrt(P_(SO_3)^2/(K_PP_(O_2))

I assume you gave ${P}_{{O}_{2}}$ and $\underline{\underline{{P}_{S {O}_{3}}}}$...

We know all of these quantities except for ${K}_{P}$. Remember that gas-phase reactions have $K \equiv {K}_{P}$, so:

$\Delta {G}_{r x n}^{\circ} \left(\text{gas-phase}\right) = - R T \ln {K}_{P}$

We already have enough information to find $\Delta {G}^{\circ}$, so we can find ${K}_{P}$ afterwards. These $\Delta {G}_{f}^{\circ}$ are correct and the reaction is balanced.

$\Delta {G}_{r x n}^{\circ} = \left[{\text{2 SO"_3(g) cdot DeltaG_f("SO"_3(g))^@] - ["2 SO"_2(g) cdot DeltaG_f("SO"_2(g))^@ + "1 O"_2(g) cdot "0 kJ/mol O}}_{2} \left(g\right)\right]$

$= \left[- \text{742 kJ/mol"] - [-"600 kJ/mol}\right]$

$= - \text{142 kJ/mol}$

Therefore,

${K}_{P} = {e}^{- \Delta {G}_{r x n}^{\circ} / R T}$

= e^(-(-"142 kJ/mol")//("0.008314 kJ/mol"cdot"K" cdot "298 K")

$= {e}^{57.3}$

As a result, we get the nearly zero partial pressure of ${\text{SO}}_{2}$:

${P}_{S {O}_{2}} = \sqrt{\left(\text{2.0 atm")^2/(e^(57.3)("0.50 atm}\right)}$

$=$ $\underline{1.02 \times {10}^{- 12} \text{atm}}$

i.e. basically zero...