At 250ºC, the equilibrium constant, K_P, for the reaction PCl₅(g) PCl₃(g) + Cl₂(g) is 1.80. Sufficient PCl₅ is put into a reaction vessel to give an initial pressure of 2.74 atm at 250ºC. What is the partial pressure of PCl₅ after the system has reached equilibrium.

Question

Quizplus · Accepted Answer

The balanced chemical equation for the given reaction is: PCl5(g) ↔ PCl3(g) + Cl2(g) The equilibrium constant expression is: K_P = [PCl3][Cl2]/[PCl5] At equilibrium, let the partial pressure of PCl5 be x atm. Then, the partial pressures of PCl3 and Cl2 will be (2.74 - x) atm and (2.74 - x) atm, respectively (since 1 mole of PCl5 gives 1 mole each of PCl3 and Cl2). Substituting these values into the equilibrium constant expression, we get: 1.8 = [(2.74 - x)(2.74 - x)]/x Simplifying the equation, we get: 1.8x = (2.74 - x)(2.74 - x) 1.8x = 7.5076 - 5.48x + x^2 x^2 - 7.28x + 7.5076 = 0 Solving for x using the quadratic formula, we get: x = 1.24 or 6.04 The value of x cannot be 6.04 atm because it is greater than the initial pressure of PCl5. Therefore, the partial pressure of PCl5 at equilibrium is 1.24 atm. Hence, the answer is B.

At 250ºC, the Equilibrium Constant, KP, for the Reaction PCl5(g)

At 250ºC, the Equilibrium Constant, K_P, for the Reaction PCl₅(g)