The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant \(K_{P}\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment \(2.00 \mathrm{~mol} \mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol} \mathrm{O}_{2}\) were initially present in a flask. What must the total pressure at equilibrium be in order to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

First, write the equilibrium constant expression for the given reaction. The equilibrium constant is the ratio of the concentrations or pressures of products over reactants raised to their stoichiometric coefficients. For the reaction \(2SO_{2} (g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\), the equilibrium constant in terms of pressure \(K_{p}\) is defined as: \[K_{p} = \frac{(P_{SO_{3}})^{2}}{(P_{SO_{2}})^{2} \cdot (P_{O_{2}})\]

The problem tells that the yield of \(SO_{3}\) is 80%. So it means 80% of \(SO_{2}\) is converted to \(SO_{3}\). At the equilibrium, the moles of \(SO_{3}\) is 80% of \(SO_{2}\), so: \[n_{SO_{3}} = 0.8 \cdot 2.00 = 1.60~ moles\] For \(SO_{2}\), 80% of its initial moles was transformed into \(SO_{3}\), so: \[n_{SO_{2}} = (1 - 0.8) \cdot 2.00 = 0.40~ moles\] For \(O_{2}\), half of the \(SO_{2}\) consumed was \(O_{2}\) because of their stoichiometric ratio, so: \[n_{O_{2}} = 2.00 - 0.5 \cdot 0.40 = 1.80~ moles\]

In the final step, substitute the number of moles and the value of \(K_{p}\) into the expression for \(K_{p}\) derived in step 1 to get the total pressure. Let \(P\) be the total pressure. Then \[P_{SO_{3}} = \frac{n_{SO_{3}}}{n_{total}} \cdot P = \frac{1.60}{1.60 + 0.40 + 1.80} \cdot P\] Similarly, \[P_{SO_{2}} = \frac{n_{SO_{2}}}{n_{total}} \cdot P = \frac{0.40}{1.60 + 0.40 + 1.80} \cdot P\] and \[P_{O_{2}} = \frac{n_{O_{2}}}{n_{total}} \cdot P = \frac{1.80}{1.60 + 0.40 + 1.80} \cdot P\]. Substitute these into the expression for \(K_{p}\) and solve for \(P\): \[0.13 = \frac{(\frac{1.60}{1.60 + 0.40 + 1.80})^2 \cdot P^2}{(\frac{0.40}{1.60 + 0.40 + 1.80})^2 \cdot (\frac{1.80}{1.60 + 0.40 + 1.80}) \cdot P^2}\] From this, we can solve to find the value of the total pressure \(P\).