Question 21 of 37intermediate🔧 ApplyNumerical3 marks

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment | [A]/mol L−1^{-1} | [B]/mol L−1^{-1} | Initial rate/mol L−1^{-1} min−1^{-1} I | 0.1 | 0.1 | 2.0×10−22.0 \times 10^{-2} II | – | 0.2 | 4.0×10−24.0 \times 10^{-2} III | 0.4 | 0.4 | – IV | – | 0.2 | 2.0×10−22.0 \times 10^{-2}

Correct Answer

Experiment II: [A] = 0.2 mol L⁻¹ Experiment III: Initial rate = 8.0×108.0 \times 10⁻² mol L⁻¹ min⁻¹ Experiment IV: [A] = 0.1 mol L⁻¹

Exercise: EXERCISES | Q: 3.12 | (Page 26)
For More Understanding

Explanation

This question tests understanding of rate law expressions. Since the reaction is first order with respect to A and zero order with respect to B, the rate law is Rate = k[A]¹[B]⁰ = k[A]. The concentration of B does not affect the rate. First, we determine the rate constant k from Experiment I, then use it to find the missing values in other experiments.

Solution Steps

  1. Step 1: Write the rate law expression. Since reaction is first order in A and zero order in B: Rate = k[A]¹[B]⁰ = k[A]

  2. Step 2: Calculate rate constant (k) from Experiment I. Rate = k[A] → 2.0×102.0 \times 10⁻² = k × 0.1 → k = 0.2 min⁻¹

  3. Step 3: Find [A] for Experiment II. Rate = k[A] → 4.0×104.0 \times 10⁻² = 0.2 × [A] → [A] = 0.2 mol L⁻¹

  4. Step 4: Find Initial rate for Experiment III. Rate = k[A] = 0.2×0.40.2 \times 0.4 = 8.0×108.0 \times 10⁻² mol L⁻¹ min⁻¹

  5. Step 5: Find [A] for Experiment IV. Rate = k[A] → 2.0×102.0 \times 10⁻² = 0.2 × [A] → [A] = 0.1 mol L⁻¹

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