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The results given in the below table were obtained during kinetic studies of the following reaction:
2A + B → C + D
Experiment | [A]/mol L⁻¹ | [B]/mol L⁻¹ | Initial rate/mol L⁻¹ min⁻¹
I | 0.1 | 0.1 | 6.00 × 10⁻³
II | 0.1 | 0.2 | 2.40 × 10⁻²
III | 0.2 | 0.1 | 1.20 × 10⁻²
IV | X | 0.2 | 7.20 × 10⁻²
V | 0.3 | Y | 2.88 × 10⁻¹
X and Y in the given table are respectively:
- 0.3, 0.4
- 0.4, 0.3
- 0.4, 0.4
- 0.3, 0.3
Correct answer: 0.3, 0.4
Solution
Comparing I and III gives order 1 in A; comparing I and II gives order 2 in B, so rate = k[A][B]^2 with k = 6e-3/((0.1)(0.1)^2) = 6. For IV: 7.2e-2 = 6*X*(0.2)^2 -> X = 0.3. For V: 2.88e-1 = 6*(0.3)*Y^2 -> Y^2 = 0.16 -> Y = 0.4. So X,Y = 0.3, 0.4.
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