At 900.0 K,the equilibrium constant (K_p) 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.

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:K_p = (P_SO3)^2 / (P_SO2)^2 * P_O2Substituting the given values into the expression, we get:0.345 = (P_SO3)^2 / (35.0 atm)^2 * 15.9 atmSolving for P_SO3, we get:P_SO3 = 82.0 atmTherefore, the partial pressure of SO₃ is 82.0 atm.

At 9000 K,the Equilibrium Constant (Kp)for the Following Reaction Is 0

At 9000 K,the Equilibrium Constant (K_p)for the Following Reaction Is 0