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 29.5 atm and that of O₂ is 13.4 atm.The partial pressure of SO₃is ________ 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 / (29.5)^2 * 13.4Solving for P_SO3, we get:P_SO3 = 63.4 atmTherefore, the correct answer is A.

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