H2(g) + I2(g) ⇄ 2 HI(g) Hydrogen gas reacts with iodine gas at constant temperature in a sealed rigi — Equilibrium Chemistry Question

STEPS:

1. Understand the state of dynamic equilibrium:
* Initially, the reaction system $H_2(g) + I_2(g) \rightleftharpoons 2\ HI(g)$ is at a state of dynamic chemical equilibrium.
* At dynamic equilibrium, the rate of the forward reaction is exactly equal to the rate of the reverse reaction ($\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}}$).

2. Analyze the immediate physical change to the system:
* When additional $I_2(g)$ is introduced into the rigid reaction vessel, the concentration and partial pressure of reactant iodine molecules immediately increase.

3. Apply collision theory to the forward reaction:
* Collision theory dictates that the rate of a chemical reaction depends on how frequently reactant molecules collide.
* Increasing the concentration of $I_2(g)$ molecules means there is a higher density of reactant particles in the container.
* This physical change increases the likelihood and frequency of collisions between $I_2(g)$ and $H_2(g)$ molecules, which is the necessary step for the forward reaction to take place.
* Consequently, the rate of the forward reaction immediately increases.

4. Analyze the immediate effect on the reverse reaction:
* Immediately after adding $I_2(g)$, the concentration of the product, $HI(g)$, has not yet had time to change.
* Since the concentration of $HI$ is initially unchanged, the frequency of collisions between $HI$ molecules remains constant, meaning the rate of the reverse reaction does not immediately change.

5. Compare the reaction rates to determine the net direction:
* Because the forward rate has increased while the reverse rate is initially unchanged, the rate of the forward reaction becomes greater than the rate of the reverse reaction ($\text{Rate}_{\text{forward}} > \text{Rate}_{\text{reverse}}$).
* This establishes Option C as the correct answer. Over time, this inequality will drive a net forward shift to produce more products and re-establish equilibrium.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Only the rate of the forward reaction increases because of the heightened concentration of reactant molecules. Neither rate decreases immediately; the reverse rate remains initially constant, and the forward rate increases.
• Option B is incorrect: The reaction rates do not remain unchanged. Because additional reactant gas molecules are introduced, the collision frequency of reactants must increase, which directly increases the forward reaction rate.
• Option D is incorrect: This choice describes a situation where the reverse reaction is favored over the forward reaction (which would occur if a product like $HI$ were added). Because a reactant was added, the forward rate must become greater than the reverse rate to consume the excess reactant.