A sample of 79.3 g of hydrogen bromide is reacted with 80.0 g of barium hydroxide to produce barium bromide and water. Using the
2 answers:
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Answer:
118.22 atm
Explanation:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
KP = 0.13 =
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: