U5.2 · Introduction to Rate LawPractice · 速率定律入门
Lessons covered:U5.2 Introduction to Rate LawU5.5 Collision Model
- 1Multiple choice
With everything else held constant, halving [A] cuts the rate to 1/4. What is the order in A?
ExplanationRate ∝ : , so m = 2, second order. First order would halve the rate; zero order would leave it unchanged.
Review: U5.2 Introduction to Rate Law
- 2Numeric
A reaction has rate = k[A]²[B] with k = 0.45 M⁻² s⁻¹. What is the rate when [A] = 0.20 M and [B] = 0.30 M?
× 10M/sExplanationRate = 0.45 × (0.20)² × 0.30 = 0.45 × 0.040 × 0.30 = M/s. Square [A] first.
Review: U5.2 Introduction to Rate Law
- 3Multiple choice
For rate = k[A]²[B], with concentrations in M and time in s, what are the units of k?
ExplanationThe overall order is 2 + 1 = 3. k = rate ÷ ([A]²[B]), so its units are (M/s) ÷ M³ = M⁻² s⁻¹. In general, for overall order n, k has units of .
Review: U5.2 Introduction to Rate Law
- 4Multiple choice
The table gives initial-rate data for X + Y → Z (concentrations in M, rates in M/s). What is the rate law?
Trial [X] [Y] Initial rate 1 0.10 0.10 2 0.20 0.10 3 0.20 0.30 ExplanationTrials 1 and 2: [Y] is constant and doubling [X] doubles the rate, so first order in X. Trials 2 and 3: [X] is constant and tripling [Y] leaves the rate unchanged, so zero order in Y, and Y drops out of the rate law. Rate = k[X].
Review: U5.2 Introduction to Rate Law
- 5Multiple choice
A reaction is run at 25 °C and at 35 °C with the same initial concentrations. The initial rate at 35 °C is twice that at 25 °C. Why?
ExplanationRate = k × the concentration terms. With the same concentrations, doubling the rate means k doubled. k rises with temperature because a larger fraction of collisions reach the activation energy; the orders come from the mechanism and don’t change with temperature.
Your missed questions were saved automatically to your mistake notebook · 0
Your best score counts only full runs through the set, not retries of missed questions.