Questions 35-36 refer to the experiment described below. H2 gas and N2 gas were placed in a rigid ve — Equilibrium Chemistry Question

STEPS:

1. Identify the Change (Stress) to the System: At time $t_2$, more $NH_3$ gas is added to the vessel. This increases the concentration of the product in the equilibrium system: $3\ H_2(g) + N_2(g) \rightleftharpoons 2\ NH_3(g)$.
2. Apply Le Chatelier’s Principle: According to this principle, if a chemical system at equilibrium is disturbed by a change in concentration, the system will shift its equilibrium position in a direction that tends to counteract that disturbance.
3. Predict the Direction of the Shift: Because the concentration of a product ($NH_3$) was increased, the system will attempt to consume the excess $NH_3$ by shifting to the left (toward the reactants).
4. Determine the Effect on Reactants: As the reaction shifts to the left, $NH_3$ molecules will decompose to produce more $N_2$ and $H_2$ gas. This results in an increase in the amount (and concentration) of both $N_2$ and $H_2$.
5. Conclusion: Statement D correctly identifies that the amount of $N_2$ will increase as the system works to restore equilibrium.

WHY_OTHERS_WRONG:

• A & B: The value of the equilibrium constant ($K$) is a function of temperature only. Since the problem specifies that the temperature is held constant, the value of $K$ will not increase or decrease regardless of changes in concentration or pressure.
• C: Adding more gas molecules ($NH_3$) to a rigid container immediately increases the total pressure. Furthermore, the subsequent shift to the left converts 2 moles of gas ($NH_3$) into 4 moles of gas ($3\ H_2 + 1\ N_2$), which would cause the pressure to increase even further.
• E: As explained in the steps, the shift to the left consumes $NH_3$ and produces $H_2$. Therefore, the amount of $H_2$ will increase, not decrease.