contestada

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. 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

Respuesta :

dsdrajlin

Answer:

The partial pressure of SO₃ is 82.0 atm

Explanation:

The equilibrium constant Kp is equal to the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients.

For the reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

[tex]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[/tex]

Otras preguntas