Question

The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant KP for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is 0.13 at 803°C. In one experiment, 2.00 mol SO2 and 2.00 mol O2 were initially present in a flask. What must be the total pressure at equilibrium in order to have an 71.0% yield of SO3?

Answer 1

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = $\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}$

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

• With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
• With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

• XSO₂ = 0.58/3.29 = 0.176

• XO₂ = 1.29/3.29 = 0.392

• XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

• p(SO₂) = 0.176 * PT

• p(O₂) = 0.392 * PT

• p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

$\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm$