Practice a College Board-style short-answer question on Equilibrium. Write your response, then reveal the model answer to see exactly how each part earns its point.
Short Answer Question · Unit 7 · Le Châtelier's Principle & Solubility
A sealed, rigid container holds the following gas-phase reaction at equilibrium: 2NO₂(g) ⇌ N₂O₄(g), where the forward reaction is exothermic. Separately, a saturated aqueous solution of PbCl₂ is at equilibrium according to: PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq).
A
Predict and justify the direction the NO₂/N₂O₄ equilibrium will shift if the temperature of the container is increased.
✓ Model answer (earns the point)
The equilibrium will shift toward reactants (toward NO₂). Since the forward reaction is exothermic, the reverse reaction is endothermic; increasing temperature favors the endothermic direction, which absorbs the added heat. By Le Châtelier's principle, the system shifts to partially counteract the temperature increase, favoring the reverse (NO₂-forming) reaction.
Why it scores: Correctly predicts the direction (toward reactants) AND justifies it by connecting the exothermic/endothermic nature of each direction to which one absorbs the added heat.
B
Explain how the value of K for the NO₂/N₂O₄ equilibrium changes (if at all) as a result of the temperature increase in Part A, and contrast this with how K would change if instead more NO₂ gas were added at constant temperature.
✓ Model answer (earns the point)
The value of K decreases when temperature increases, since the equilibrium shifts toward reactants (less N₂O₄, more NO₂ relative to before), changing the ratio that defines K. In contrast, if more NO₂ were added at constant temperature, the equilibrium would shift toward products to partially consume the added NO₂, but K would NOT change — only concentration or volume/pressure changes shift the position of equilibrium without altering K; temperature changes are unique in that they actually change the value of K itself.
Why it scores: States that temperature changes K (with correct direction) AND explicitly contrasts this with the concentration-change scenario where K stays constant — showing understanding of the key distinction between these two types of stress.
C
A student adds solid Pb(NO₃)₂ to the saturated PbCl₂ solution. Predict what happens to the molar solubility of PbCl₂, and explain your reasoning in terms of the common-ion effect.
✓ Model answer (earns the point)
The molar solubility of PbCl₂ will decrease. Adding Pb(NO₃)₂ introduces additional Pb²⁺ ions, which is a common ion already present in the PbCl₂ dissolution equilibrium. This increases the reaction quotient Q above Ksp momentarily, so by Le Châtelier's principle, the equilibrium shifts toward the solid (toward PbCl₂), causing some of the dissolved Pb²⁺ and Cl⁻ to precipitate back out — decreasing the amount of PbCl₂ that can remain dissolved.
Why it scores: Correctly predicts decreased solubility AND identifies Pb²⁺ specifically as the common ion AND explains the mechanism (Q exceeds Ksp, equilibrium shifts toward the solid).
How to score points on SAQs
Always connect the direction of a shift to the underlying reason. "Shifts toward reactants" alone doesn't score as well as "shifts toward reactants because the reverse reaction is endothermic and absorbs the added heat."
Know the difference between stresses that change K and stresses that don't. Only temperature changes K; concentration and volume/pressure changes shift position but leave K unchanged.
Name the specific common ion when discussing the common-ion effect. "A shared ion is added" is vaguer than "Pb²⁺ is a common ion already present in the equilibrium."
Use Le Châtelier's principle explicitly as your reasoning tool. Reference "shifts to partially counteract" language to show you understand the mechanism, not just the result.
Keep it tight. 2–4 sentences per part is plenty. Long answers don't score higher; they just waste exam time.