Dynamic equilibrium of gases

Question

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$ The equilibrium constant for the reaction above is $0.13$ at $\pu{830^\circ C}$. In one experiment $\pu{2.00 mol}$ of $\ce{SO2}$ and $\pu{2.00 mol}$ of $\ce{O2}$ were initially present in a flask until it reaches dynamic equilibrium. What must the total pressure of equilibrium be in order to have an 80.0% yield of $\ce{SO3}$?

How do I approach this type of problem? Is it correct to assume that the total pressure of the equilibrium would be the pressure from the both sides of the equation (pressure of $\ce{SO2}$ + pressure of $\ce{O2}$ + pressure of $\ce{SO3}$)? And how do I find these different pressures?

I am thinking of using the method: $K_\mathrm{p}=0.3=\Large\frac{p^2_\ce{SO3}}{p^2_\ce{SO2}\times p_\ce{O2}}$

Hence, to find the respective pressure of the gasses, I attempted the RICE table, (Reaction, initial mol, change in mol, equilibrium point). How can I proceed further ?

Answer 1

For an ideal gas mixture the total pressure is the sum of the partial pressures of all the gases in the mixture. To make intuitive sense of this, consider that since it is a gas mixture, all the gases essentially have access to the whole container, and so will homogenously fill it. Each gas wil the exert a certain pressure on the container walls. The sum the pressures from all the contributing gases will then equal the total pressure the container walls experience.

We also have the relationship $$p_\text{x}/p_\text{total}=n_\text{x}/n_\text{total}$$

The potential yield for the rxn is 2 mol ($\ce{SO2}$ limiting reactant). To get a 80% yield, we therefore have 1.6 mol $\ce{SO3}$ at equilibrium. Therefore there will be 0.4 mol $\ce{SO2}$ and 1.2 mol $\ce{O2}$ left at equilibrium. $n_\text{total}=3.2$

$n_\ce{SO2}/n_\text{total}=0.4/3.2=0.125$

$n_\ce{SO3}/n_\text{total}=1.6/3.2=0.5$

$n_\ce{O2}/n_\text{total}=1.2/3.2=0.375$

To find the partial pressures of the individual gases, we combine the ratios above with relationship 1

$p_\ce{SO2}=0.125\times p_\text{total}$
$p_\ce{SO3}=0.5\times p_\text{total}$
$p_\ce{O2}=0.375\times p_\text{total}$

As you know, the equilibrium can be expressed with partial pressures, and is constant for all changes in pressure. $K_\mathrm{p}=p_\ce{SO_3}^2/p_\ce{SO_2}^2p_\ce{O_2}=0.13$ at 830 Celsius

Plugging in the partial pressures and solving for p(total):

$(0.5x)^2/((0.125x)^2\times0.375x)=0.13$ $\implies x=328.205$ (The unit can be given by the units of the eq. constant.)