U7.10 · Reaction Quotient and Le Châtelier's PrinciplePractice · 用 Q 解释勒夏特列原理
Lessons covered:U7.9 Introduction to Le Châtelier's PrincipleU7.10 Reaction Quotient and Le Châtelier's Principle
- 1Multiple choice
H₂(g) + I₂(g) ⇌ 2HI(g) is at equilibrium in a rigid container. Some more H₂ is added. At that instant, how does Q compare with K, and which way does the equilibrium shift?
ExplanationQ = [HI]² ÷ ([H₂][I₂]). Adding H₂ makes the denominator larger, so Q immediately drops below K. There is too little product, the forward rate exceeds the reverse rate, and the equilibrium shifts forward until Q equals K again. Equal numbers of gas molecules on each side only mean a volume change has no effect; adding a substance still does.
Review: U7.10 Reaction Quotient and Le Châtelier's Principle
- 2Numeric
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is at equilibrium. At constant temperature, the volume is suddenly halved. At that instant, Q is how many times K?
ExplanationHalving the volume doubles every concentration. Q = [NH₃]² ÷ ([N₂][H₂]³): the top is multiplied by 2² = 4 and the bottom by 2 × 2³ = 16, so Q becomes 4 ÷ 16 = 0.25 of its value. Q < K, so the equilibrium shifts forward, toward fewer gas molecules.
Review: U7.10 Reaction Quotient and Le Châtelier's Principle
- 3Multiple choice
An exothermic reaction is at equilibrium. At the instant the temperature is raised, what happens to Q and to K, and which way does the equilibrium then shift?
ExplanationAt the instant of heating, no concentrations have changed yet, so Q is the same; what changes is K. For an exothermic forward reaction, heating lowers K, so Q > K and the reaction runs in reverse until Q falls to the new K. Changing concentrations or volume changes Q; changing temperature changes K.
Review: U7.10 Reaction Quotient and Le Châtelier's Principle
- 4Numeric
At some temperature, K = 2.0 for A(aq) ⇌ B(aq), and at equilibrium [A] = 0.10 M and [B] = 0.20 M. At constant temperature and volume, A is added so that [A] instantly becomes 0.40 M. What is [B] once equilibrium is restored?
MExplanationQ = 0.20 ÷ 0.40 = 0.5 < K, so it shifts forward. Let A fall by x: (0.20 + x) ÷ (0.40 − x) = 2.0, so 0.20 + x = 0.80 − 2.0x and x = 0.200. [B] = 0.400 M and [A] = 0.200 M. A ends up above its old equilibrium value (0.10) but below the value just after adding (0.40): the shift only partly offsets the change.
Review: U7.10 Reaction Quotient and Le Châtelier's Principle
- 5Multiple choice
A syringe holds N₂O₄(g) ⇌ 2NO₂(g) at equilibrium (NO₂ is reddish brown, N₂O₄ colorless). The plunger is pushed in quickly, compressing the gas. The color first darkens, then fades somewhat. How does the final color compare with the color before compression?
ExplanationCompression instantly raises every concentration, so the color darkens. Q = [NO₂]²/[N₂O₄] doubles (for a halved volume), so Q > K and the equilibrium shifts in reverse, toward N₂O₄, and the color fades a little. But at the new equilibrium [NO₂]² = K[N₂O₄] with a larger [N₂O₄], so [NO₂] is also higher than at the start, and the color is darker than before. Le Châtelier: a shift reduces a change but never fully cancels it.
Review: U7.10 Reaction Quotient and Le Châtelier's Principle
- 6Multiple choice
is at equilibrium in a rigid container. At 10 min one change is made (the temperature stays the same), and the concentrations change as shown. What change was made?
ExplanationLook at 10 min: only [H₂] jumps (0.022 → 0.072 M), while I₂ and HI don’t, so H₂ was added. Then Q < K and the equilibrium shifts right: H₂ and I₂ fall and HI rises, ending near 0.179 M. Removing HI would make HI drop suddenly; halving the volume would make all three jump at once; and the temperature didn’t change.
Review: U7.10 Reaction Quotient and Le Châtelier's PrincipleU7.9 Introduction to Le Châtelier's Principle
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