Question

Consider the following system at equilibrium where H° = 10.4 kJ, and Kc = 1.80×10-2, at 698 K. 2HI(g) H2(g) + I2(g) When 0.40 moles of H2(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 I2 will

Answer 1

Answer:

- Kc remains the same.

- Qc increases.

- The reaction will move in the reverse direction.

- The concentration of iodine will decrease.

Explanation:

Hello,

In this case, given the equilibrium:

$2HI(g)\rightleftharpoons H_2(g) + I_2(g)$

- We first must state that Kc remains the same as it a function of the temperature only and temperature is not changing.

- Then since 0.4 moles of hydrogen are added, and the reaction quotient Qc is:

$Qc=\frac{[H_2][I_2]}{[HI]^2}$

By increasing the concentration of hydrogen, the reaction quotient increases as well due to their directly proportional relationship.

- Due to that addition, the reaction will move in the reverse direction, as more HI must be formed to reestablish the equilibrium.

- Finally, due to the addition of hydrogen, more iodine will be consumed to reestablish equilibrium, for that reason, the concentration of iodine will decrease.

Best regards.