At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia

l pressure of SO2 is 35.0 atm and that of O2 is 15.9 atm. The partial pressure of SO3 is ________ atm. At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partial pressure of SO2 is 35.0 atm and that of O2 is 15.9 atm. The partial pressure of SO3 is ________ atm. 192 6.20 × 10-4 4.21 × 10-3 40.2 82.0

Chemistry

1 answer:

lorasvet [2.7K] · Answer 1 · 2026-03-04T13:53:57+03:00

Answer:

The partial pressure of SO₃ is measured at 82.0 atm.

Explanation:

The equilibrium constant Kp is defined as the ratio of the equilibrium pressures of the gaseous products, each raised to the power of their respective coefficients in the reaction, divided by the pressures of the gaseous reactants raised to their coefficients.

For the given reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

$Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm$