Questions: Le Chatelier's Principle and Equilibrium Shifts

Question 1

A student adds a catalyst to the equilibrium reaction N₂ + 3H₂ ⇌ 2NH₃ to increase ammonia yield. What actually happens?

• A. The equilibrium shifts toward products, increasing the concentration of NH₃ at equilibrium
• B. The equilibrium shifts toward reactants, which then shift back, eventually settling at higher NH₃
• C. The equilibrium position is unchanged — the catalyst speeds up both forward and reverse reactions equally, so the system reaches the same equilibrium faster
• D. The equilibrium constant K increases, giving a higher yield of NH₃

Answer: (C)

A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount. Since both rates increase equally, the ratio of forward to reverse rate — and therefore the equilibrium constant K — is unchanged. The equilibrium position (the concentrations at equilibrium) does not shift. What does change is how quickly equilibrium is reached. This is a very common misconception: catalysts affect kinetics (rate), not thermodynamics (equilibrium position or K). To increase NH₃ yield, you would need to remove NH₃ as it forms, increase pressure, or lower the temperature.

Question 2

The Haber process (N₂ + 3H₂ ⇌ 2NH₃, exothermic) is run at elevated temperature in practice, even though this shifts equilibrium toward reactants. What does this reveal about Le Chatelier's principle?

• A. Le Chatelier's principle is wrong — temperature does not affect equilibrium position
• B. Le Chatelier's principle correctly predicts that yield decreases, but higher temperature is used anyway because the rate is too slow at low temperature — kinetics and equilibrium are separate considerations
• C. Le Chatelier's principle predicts that elevated temperature increases yield for exothermic reactions
• D. The Haber process is endothermic, so elevated temperature correctly shifts toward products

Answer: (B)

Le Chatelier's principle is correct: raising temperature for an exothermic reaction shifts equilibrium toward reactants, decreasing K and reducing equilibrium yield. In practice, Haber plants use elevated temperature (~400°C) because at lower temperatures the reaction is too slow to be commercially viable even with a catalyst. This illustrates a crucial point: Le Chatelier's principle describes equilibrium position (thermodynamics), while reaction rate is a kinetics question. Industrial chemistry often involves optimizing the trade-off between yield (favored by low temperature) and rate (favored by high temperature).

Question 3

Adding an inert gas like argon at constant volume to a gaseous equilibrium shifts the position of equilibrium toward the side with fewer moles of gas.

• T. True
• F. False

Answer: False

Adding an inert gas at constant volume does NOT shift the equilibrium. Le Chatelier's pressure argument applies when the total pressure changes the partial pressures of the reacting gases. An inert gas at constant volume increases total pressure but does not change the partial pressures of N₂, H₂, or NH₃ — those remain determined by their own concentrations, which haven't changed. Since the partial pressures of reactants and products are unchanged, Q = K still holds and there is no shift. The confusion arises from conflating total pressure with the partial pressures of reactive species.

Question 4

For an exothermic reaction at equilibrium, raising the temperature causes the equilibrium constant K to decrease.

• T. True
• F. False

Answer: True

Temperature is the one stress that actually changes K. For an exothermic reaction, heat is a product: reactants ⇌ products + heat. Raising temperature adds 'heat,' and the system shifts to consume it — toward reactants — lowering the concentration ratio of products to reactants and therefore decreasing K. This is fundamentally different from concentration or pressure changes, which shift the position of equilibrium (changing concentrations) without altering K. The van't Hoff equation (d ln K / dT = ΔH°/RT²) formalizes this: exothermic ΔH° < 0 means K decreases as T increases.

Question 5

How is a temperature change fundamentally different from a concentration change in its effect on chemical equilibrium? Why does this distinction matter?

Model answer: A concentration change shifts the position of equilibrium — it changes the equilibrium concentrations — but does not change the equilibrium constant K. The system reaches a new state where Q returns to the unchanged value of K. A temperature change actually changes K itself, because temperature affects the intrinsic thermodynamics of the reaction (the Gibbs energy of the system). For an exothermic reaction, raising temperature decreases K; for an endothermic reaction, it increases K. This matters because K determines the maximum possible yield — changing concentration can optimize around a fixed yield ceiling, but changing temperature moves the ceiling itself.

This distinction is practically important in industrial chemistry. You can drive a reaction forward by removing products (Le Chatelier: concentration stress, no change to K) — this is exploited in continuous-flow reactors. But to achieve a fundamentally different equilibrium yield, you must change the temperature, which comes at a kinetic cost. Understanding whether a stress changes K or merely Q is the key to distinguishing thermodynamic from kinetic optimization.