Question 20 of 37advanced🔍 AnalyzeNumerical3 marks

The following results have been obtained during the kinetic studies of the reaction:

2A+B→C+D2A + B \rightarrow C + D

Experiment | [A]/mol L−1^{-1} | [B]/mol L−1^{-1} | Initial rate of formation of D/mol L−1^{-1} min−1^{-1} I | 0.1 | 0.1 | 6.0×10−36.0 \times 10^{-3} II | 0.3 | 0.2 | 7.2×10−27.2 \times 10^{-2} III | 0.3 | 0.4 | 2.88×10−12.88 \times 10^{-1} IV | 0.4 | 0.1 | 2.40×10−22.40 \times 10^{-2} Determine the rate law and the rate constant for the reaction.

Correct Answer

Rate Law: Rate = k[A][B]²

Rate Constant: k = 6.0 mol⁻² L² min⁻¹

Exercise: EXERCISES | Q: 3.11 | (Page 26)
For More Understanding

Explanation

This question requires determining the order of reaction with respect to each reactant by comparing experiments where one reactant concentration is held constant. The rate law expression follows the form Rate = k[A]^x[B]^y where x and y are the orders with respect to A and B respectively.

Solution Steps

  1. Step 1: Determine order with respect to A Compare Experiments I and IV where [B] is constant (0.1 mol L⁻¹). [A] ratio = 0.4/0.1 = 4 Rate ratio = (2.40×102.40 \times 10⁻²)/(6.0×106.0 \times 10⁻³) = 4 Since 4ˣ = 4, therefore x = 1 (first order in A).

  2. Step 2: Determine order with respect to B Compare Experiments II and III where [A] is constant (0.3 mol L⁻¹). [B] ratio = 0.4/0.2 = 2 Rate ratio = (2.88×102.88 \times 10⁻¹)/(7.2×107.2 \times 10⁻²) = 4 Since 2ʸ = 4, therefore y = 2 (second order in B).

  3. Step 3: Write the Rate Law Rate = k[A]¹[B]² = k[A][B]²

  4. Step 4: Calculate Rate Constant (k) Using Experiment I data: Rate = 6.0×106.0 \times 10⁻³ mol L⁻¹ min⁻¹ [A] = 0.1 mol L⁻¹, [B] = 0.1 mol L⁻¹

    Substituting in rate law: 6.0×106.0 \times 10⁻³ = k × (0.1) × (0.1)² 6.0×106.0 \times 10⁻³ = k × 0.001 k = 6.0 mol⁻² L² min⁻¹

PreviousNext