The equilibrium constant, K_c, for the reaction PCl₃(g) + Cl₂(g) PCl₅(g) is 49 at 230°C.
If 0.70 mol of PCl₃ is added to 0.70 mol of Cl₂ in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl₃ when equilibrium has been established?

Question

The equilibrium constant, K_c, for the reaction PCl₃(g) + Cl₂(g) The equilibrium constant, Kc, for the reaction PCl3(g) + Cl2(g) PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol of Cl2 in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl3 when equilibrium has been established? A) 0.049 M B) 0.11 M C) 0.35 M D) 0.59 M E) 0.83 M

The equilibrium constant, K<sub>c</sub>, for the reaction PCl<sub>3</sub>(g) + Cl<sub>2</sub>(g) PCl<sub>5</sub>(g) is 49 at 230°C. If 0.70 mol of PCl<sub>3</sub> is added to 0.70 mol of Cl<sub>2</sub> in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl<sub>3</sub> when equilibrium has been established? A) 0.049 M B) 0.11 M C) 0.35 M D) 0.59 M E) 0.83 M

PCl₅(g) is 49 at 230°C.
If 0.70 mol of PCl₃ is added to 0.70 mol of Cl₂ in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl₃ when equilibrium has been established?

Quizplus · Accepted Answer

The balanced equation for the reaction is: PCl₃(g) + Cl₂(g) 11eb6c5c_3a81_ad92_8d9d_6dd916f55ef3_TB8482_11 PCl₅(g) Let the initial amount of PCl₃ be x. Then, the initial amount of Cl₂ will also be x, since they are in a 1:1 mole ratio in the balanced equation. At equilibrium, the amount of PCl₃ that has reacted to form PCl₅ will be x - y, where y is the equilibrium amount of PCl₃. Similarly, the amount of Cl₂ that has reacted to form PCl₅ will also be x - y. The equilibrium concentration of PCl₃ can be calculated as follows: K_c = [PCl₅] / ([PCl₃] * [Cl₂]) 49 = y / (x - y)^2 Since the initial amount of both PCl₃ and Cl₂ is 0.70 mol, their initial concentrations are both 0.70 M. At equilibrium, the total amount of gas will be (0.70 - y) + (0.70 - y) + y = 1.40 mol. Therefore, the equilibrium concentration of PCl₅ will be y / 1.40 L. Substituting into the equilibrium constant expression and solving for y, we get: 49 = y / ((0.70 - y)^2) y = 0.245 M Therefore, the equilibrium concentration of PCl₃ is: y / 1.40 = 0.245 M / 1.40 = 0.11 M The correct answer is B.

The Equilibrium Constant, Kc, for the Reaction PCl3(g) + Cl2(g)

The Equilibrium Constant, K_c, for the Reaction PCl₃(g) + Cl₂(g)