Question 19 of 37advanced🔍 AnalyzeNumerical3 marks

In a reaction between A and B, the initial rate of reaction (r0r_0) was measured for different initial concentrations of A and B as given below: A/ mol L−1^{-1}: 0.20, 0.20, 0.40 B/ mol L−1^{-1}: 0.30, 0.10, 0.05 r0r_0/mol L−1^{-1}s−1^{-1}: 5.07×10−55.07 \times 10^{-5}, 5.07×10−55.07 \times 10^{-5}, 1.43×10−41.43 \times 10^{-4} What is the order of the reaction with respect to A and B?

Correct Answer

Order with respect to A = 1.5 (or 3/2); Order with respect to B = 0

Exercise: EXERCISES | Q: 3.10 | (Page 26)
For More Understanding

Explanation

This question tests the understanding of determining order of reaction from experimental data. The order with respect to a reactant is determined by comparing experiments where other reactant concentrations remain constant. The context explains that order can be 0, 1, 2, 3 or even a fraction, and must be determined experimentally.

Solution Steps

  1. Step 1: Determine order with respect to B Comparing experiments 1 and 2: [A]₁ = [A]₂ = 0.20 mol L⁻¹ (constant) [B]₁ = 0.30 mol L⁻¹, [B]₂ = 0.10 mol L⁻¹ Rate₁ = Rate₂ = 5.07×105.07 \times 10⁻⁵ mol L⁻¹s⁻¹

    Since the rate remains unchanged when [B] is reduced by a factor of 3, the rate is independent of [B]. Order with respect to B = 0

  2. Step 2: Determine order with respect to A Comparing experiments 2 and 3: Since order w.r.t. B = 0, rate depends only on [A]. [A]₂ = 0.20 mol L⁻¹, [A]₃ = 0.40 mol L⁻¹ r₂ = 5.07×105.07 \times 10⁻⁵ mol L⁻¹s⁻¹, r₃ = 1.43×101.43 \times 10⁻⁴ mol L⁻¹s⁻¹

    Using rate law: r = k[A]ˣ r₃/r₂ = ([A]₃/[A]₂)ˣ (1.43×101.43 \times 10⁻⁴)/(5.07×105.07 \times 10⁻⁵) = (0.40/0.20)ˣ 2.82 ≈\approx 2ˣ

    Taking logarithm: x × log(2) = log(2.82) x × 0.301 ≈\approx 0.450 x ≈\approx 1.5

    Order with respect to A = 1.5 (or 3/2)

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