At 900.0 K, the equilibrium constant (K_p) for the following reaction is 0.345 bar^-1: 2SO₂ + O₂(g) → 2SO₃(g)
At equilibrium, the partial pressure of SO₂ is 35.0 bar and that of O₂ is 15.9 bar. The partial pressure of SO₃ is ________ bar.

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:K_p = (P_SO3)² / (P_SO2)²(P_O2)Substituting the given values into the expression, we get:0.345 bar^-1 = (P_SO3)² / (35.0 bar)²(15.9 bar)Solving for P_SO3, we get:P_SO3 = 82.0 barTherefore, the partial pressure of SO₃ is 82.0 bar.

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

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