Question

Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) When 0.17 moles of Cl2(g) are added to the equilibrium system at constant temperature: The value of Kc The value of Qc Kc. The reaction must run in the forward direction to restablish equilibrium. run in the reverse direction to restablish equilibrium. remain the same. It is already at equilibrium. The concentration of PCl3 will

Answer 1

Answer:

K remains the same;

Q < K;

The reaction must run in the forward direction to reestablish the equilibrium;

The concentration of $PCl_3$ will decrease.

Explanation:

In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.

Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.

The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.

As a result, since $PCl_3$ is also a reactant, its concentration will decrease.