U7.2 · Direction of Reversible ReactionsPractice · 可逆反应的方向
Lessons covered:U7.2 Direction of Reversible ReactionsU7.3 Reaction Quotient and Equilibrium Constant
- 1Multiple choice
For the elementary reaction A + B ⇌ C, kf = 0.5 M⁻¹ s⁻¹ and kr = 0.02 s⁻¹. At some moment [A] = [B] = 0.10 M and [C] = 0.40 M. Which conclusion is correct?
ExplanationFor an elementary reaction the rate law follows the coefficients: ratef = kf[A][B] = 0.5 × 0.10 × 0.10 = 5.0 × 10−3 M/s; rater = kr[C] = 0.02 × 0.40 = 8.0 × 10−3 M/s. The reverse rate is larger, so the net reaction runs in reverse: C decreases and A and B increase until the rates are equal. Comparing k values alone doesn’t work, because rates also depend on concentrations.
- 2Numeric
For the elementary reaction A + B ⇌ C, kf = 0.5 M⁻¹ s⁻¹ and kr = 0.02 s⁻¹. What is the equilibrium constant K?
ExplanationAt equilibrium ratef = rater: kf[A][B] = kr[C], which rearranges to K = [C] ÷ ([A][B]) = kf ÷ kr = 0.5 ÷ 0.02 = 25.
Review: U7.2 Direction of Reversible ReactionsU7.3 Reaction Quotient and Equilibrium Constant
- 3Multiple choice
At constant temperature, only the products of a reversible reaction are put into a sealed container. Which description of the time until equilibrium is correct?
ExplanationAt first there are no reactants, so the forward rate is zero, and the products are at their highest concentration, so the reverse rate is largest. The reverse reaction makes reactants, so the forward rate rises, while the products are used up and the reverse rate falls, until the two rates match. Equilibrium can be reached from either side.
- 4Numeric
For the elementary reaction 2A ⇌ B, kf = 0.2 M⁻¹ s⁻¹ and kr = 0.05 s⁻¹. At equilibrium [A] = 0.10 M. What is [B] at equilibrium?
MExplanationFor the elementary step 2A → B, the rate is proportional to [A]². At equilibrium ratef = rater: 0.2 × 0.10² = 0.05 × [B], so [B] = 0.002 ÷ 0.05 = 0.040 M. Forgetting the square gives 0.40 M.
- 5Multiple choice
At some moment, the forward rate of a reversible reaction is greater than the reverse rate. How does the reaction quotient Q compare with K at that moment?
ExplanationThe forward reaction is faster, so the net reaction runs forward: products build up and reactants fall, raising Q until it reaches K. So right now Q < K: there is less product than at equilibrium.
Review: U7.2 Direction of Reversible ReactionsU7.3 Reaction Quotient and Equilibrium Constant
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