Using Initial Rates to Find a Rate Equation for H2O2 Decomposition
A student studies the decomposition of hydrogen peroxide, 2H2O2(aq) → 2H2O(l) + O2(g), using potassium iodide as a catalyst. In each run, the total volume of the reaction mixture is made up to 50.0 cm3 with distilled water and kept at 25 °C. Oxygen gas is collected in a gas syringe, and the initial rate is measured from the tangent to the volume-time graph at t = 0.
Run A: 20.0 cm3 of 0.20 mol dm-3 H2O2, 10.0 cm3 of 0.10 mol dm-3 KI, 20.0 cm3 water. Initial rate = 2.0 cm3 min-1. Run B: 20.0 cm3 of 0.40 mol dm-3 H2O2, 10.0 cm3 of 0.10 mol dm-3 KI, 10.0 cm3 water. Initial rate = 4.0 cm3 min-1. Run C: 20.0 cm3 of 0.20 mol dm-3 H2O2, 10.0 cm3 of 0.20 mol dm-3 KI, 20.0 cm3 water. Initial rate = 4.0 cm3 min-1.
A classmate says, 'Run B has double the concentration of Run A, so the reaction is second order in H2O2.'
1. Deduce the order with respect to H2O2 and with respect to I-, and write the rate equation. Show your reasoning.
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Because the total volume is 50.0 cm3 in every run, concentration is proportional to the amount added. In A, [H2O2] = 20.0 × 0.20/50.0 = 0.080 mol dm-3 and [I-] = 10.0 × 0.10/50.0 = 0.020 mol dm-3. In B, [H2O2] = 20.0 × 0.40/50.0 = 0.160 mol dm-3 and [I-] is still 0.020 mol dm-3. Doubling [H2O2] at constant [I-] doubles the initial rate from 2.0 to 4.0 cm3 min-1, so the reaction is first order in H2O2. In C, [H2O2] is again 0.080 mol dm-3 but [I-] = 10.0 × 0.20/50.0 = 0.040 mol dm-3. Doubling [I-] at constant [H2O2] doubles the rate from 2.0 to 4.0 cm3 min-1, so the reaction is first order in I-. The rate equation is therefore rate = k[H2O2][I-], with overall order 2.
2. Critique the classmate's statement that the reaction is second order in H2O2 because Run B has double the concentration of Run A.
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The observation is correct but the conclusion confuses two different ideas. Doubling [H2O2] while keeping [I-] constant doubles the rate. That means the rate is proportional to [H2O2] to the first power, so the reaction is first order with respect to H2O2, not second order. The reaction is second order overall because it is first order in H2O2 and first order in I-, and 1 + 1 = 2. If the reaction were second order in H2O2, doubling [H2O2] with [I-] constant would quadruple the rate. To test for second order in H2O2, the student would need a run in which doubling [H2O2] changes the rate by a factor of four.
3. The student also compares the total volume of oxygen collected after 5.0 min in each run. Explain why this could lead to a wrong conclusion about the reaction order.
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The total volume after a fixed time depends on how the rate changes as the reaction proceeds, not only on the initial rate. As H2O2 is used up, its concentration falls and the rate falls, so the volume-time curve is not a straight line. Runs with different initial concentrations will have different average rates and may have reacted to different extents by 5.0 min. A run with a high initial concentration may produce more gas partly because it has not yet slowed as much, not simply because the order is higher. The initial rate is measured at t = 0, when the concentrations are known and before products accumulate, so it is the appropriate quantity for finding the order. Using a single total volume at a fixed time mixes up order, rate constant and extent of reaction.