Defending a First-Order Claim from Titration Data
A student at a Muzaffarpur school follows the decomposition of hydrogen peroxide in alkaline medium by withdrawing exactly 10.0 mL of the reaction mixture at fixed intervals and titrating it against standard KMnO4 in acid medium. The burette readings were 25.0 mL at t = 0, 20.0 mL at 10 min and 16.0 mL at 20 min. A classmate objects that the reaction cannot be first order, because the titre falls by 5.0 mL in the first ten minutes but only by 4.0 mL in the next ten - 'if the order were one, the fall should stay constant'. The teacher asks the pair to settle the question from the same data and then to extend the work: when the reaction is repeated at 35 °C it is found to be exactly twice as fast as at 25 °C. They must also say what the units of the rate constant reveal and, from the activation energy, how the stock bottle of H2O2 should be stored.
1. Decide from the data whether the reaction is first order, and expose the flaw in the classmate's argument.
Read the worked answer
The titre of KMnO4 is proportional to the H2O2 still present, so the volume at time t can stand in for [H2O2] in the integrated rate law: k = (2.303/t)·log(V₀/Vt).
- 0 → 10 min: k = (2.303/10)·log(25.0/20.0) = 0.2303 × 0.0969 = 2.23 × 10⁻² min⁻¹.
- 0 → 20 min: k = (2.303/20)·log(25.0/16.0) = 0.1151 × 0.1938 = 2.23 × 10⁻² min⁻¹.
The constant k over two very different extents of reaction is the test of first order, and the reaction passes.
The classmate compared amounts decomposed, but first order demands that equal fractions decompose in equal times. Check the fractions remaining: 20.0/25.0 = 0.80 and 16.0/20.0 = 0.80. Each ten-minute window leaves 80% of whatever was present, which is exactly exponential decay. A constant absolute fall per ten minutes would be zero-order behaviour, and that is contradicted by the data because the fall shrinks as the concentration falls. A good answer also notes the assumption being made: OH⁻ is present in large excess, so the reaction is pseudo-first-order in H2O2, and the measured k is k′ = k[OH⁻].
2. Using the mean rate constant, calculate the half-life and the time needed for 90% decomposition.
Read the worked answer
Take k = 2.23 × 10⁻² min⁻¹.
Half-life: t½ = 0.693/k = 0.693/2.23 × 10⁻² = 31.1 min. A useful cross-check is that this is consistent with the data: after 20 min, 64% of the peroxide remains (0.80 × 0.80), and 64% lies between 50% at one half-life and 100% at zero, as it must.
Time for 90% decomposition means 10% remains: t = (2.303/k)·log(100/10) = (2.303/2.23 × 10⁻²) × 1 = about 103 min (roughly 3.3 half-lives, which matches the rule that 90% completion takes between three and four half-lives).
Reasoning point worth stating: the half-life of a first-order reaction is independent of the starting concentration, so the answer would be the same whether the student began with 0.1 M or 0.5 M peroxide - provided the alkali is still in excess. If the solution were so concentrated that OH⁻ became depleted, the pseudo-first-order treatment would fail and t½ would no longer be constant; that is the limit of validity of this calculation.
3. Calculate the activation energy from the temperature data and use it to recommend storage conditions, explaining your reasoning.
Read the worked answer
Use the two-temperature form of the Arrhenius equation: log(k₂/k₁) = Ea/(2.303R) × (T₂ − T₁)/(T₁T₂), with k₂/k₁ = 2, T₁ = 298 K, T₂ = 308 K.
log 2 = 0.3010 = Ea/(2.303 × 8.314) × (10)/(298 × 308) 0.3010 = Ea/19.147 × 1.090 × 10⁻⁴ 0.3010 = Ea × 5.69 × 10⁻⁶ Ea ≈ 5.3 × 10⁴ J mol⁻¹ ≈ 53 kJ mol⁻¹.
Interpretation: a doubling of rate for a 10 K rise corresponds to an activation energy of about 53 kJ mol⁻¹, a moderate barrier typical of a reaction that is slow but far from inert at room temperature. Because the rate is exponentially sensitive to temperature, every 10 K of warmth roughly halves the shelf life. Practical recommendation: store hydrogen peroxide in a cool, dark place (a refrigerator or a shaded cupboard) in a stoppered, vented container, because light and dust catalyse the same decomposition and the evolved oxygen will pressurise a sealed bottle. Also avoid storing it near transition-metal salts or alkali, which catalyse the decomposition by providing an alternative low-Ea pathway. The judgement here is the link between the exponential term and a storage instruction - the number alone is not the answer.