Half-Lives, Activation Energy and a Catalyst Claim
A student studies the first-order gas-phase isomerisation of compound X in a constant-volume flask. Starting from 0.100 mol dm⁻³ of X each time, she measures a half-life of 57.8 min at 300 K and 7.22 min at 320 K. She needs the activation energy and the time for which a 0.100 mol dm⁻³ sample must be held at 320 K so that no more than 0.025 mol dm⁻³ of X remains. A supplier then offers a catalyst claimed to lower the activation energy to 40 kJ mol⁻¹ and to make the reaction 'a thousand times faster at 300 K'. The student has only the same flask and a stopwatch, no calorimeter and no temperature-jump apparatus. Data: R = 8.314 J K⁻¹ mol⁻¹, ln 2 = 0.693, ln 4 = 1.386, ln 10 = 2.303.
1. Obtain the rate constants at both temperatures and hence the activation energy.
Read the worked answer
For a first-order reaction k = ln2/t½.
• At 300 K: t½ = 57.8 min = 3468 s → k₁ = 0.693/3468 = 2.00 × 10⁻⁴ s⁻¹ • At 320 K: t½ = 7.22 min = 433 s → k₂ = 0.693/433 = 1.60 × 10⁻³ s⁻¹
Arrhenius: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
ln(1.60 × 10⁻³ / 2.00 × 10⁻⁴) = ln 8.00 = 2.08
1/300 − 1/320 = (320 − 300)/(300 × 320) = 2.083 × 10⁻⁴ K⁻¹
Ea = (2.08 × 8.314)/(2.083 × 10⁻⁴) = 17.3/2.083 × 10⁻⁴ ≈ 8.3 × 10⁴ J mol⁻¹ ≈ 83 kJ mol⁻¹.
The roughly eight-fold rise in k for a 20 K rise is consistent with an activation energy of this order, not with a diffusion-limited or barrierless process.
2. How long must the sample be held at 320 K to leave at most 0.025 mol dm⁻³ of X, and why is the initial concentration irrelevant?
Read the worked answer
First-order integrated law: ln([X]₀/[X]) = kt, with k = 1.60 × 10⁻³ s⁻¹ at 320 K.
[X]₀/[X] = 0.100/0.025 = 4, so t = ln 4 / k = 1.386/(1.60 × 10⁻³) = 866 s ≈ 14.4 min.
Equivalently, 75 % conversion is two half-lives: 2 × 7.22 min = 14.4 min - the same answer, which is a useful cross-check.
Why [X]₀ does not matter: for a first-order reaction the fraction remaining after time t is e^(−kt), independent of starting concentration. A 0.500 mol dm⁻³ sample would also need 14.4 min to fall to one quarter of its value; only the absolute amount left changes.
3. Assess the supplier's claim quantitatively. What half-life would the student actually observe at 300 K for each of the two scenarios, and what does that imply for testing the claim?
Read the worked answer
A 1000-fold rate increase at 300 K requires:
ΔEa = RT ln 1000 = 8.314 × 300 × 6.908 ≈ 17.2 kJ mol⁻¹,
so the activation energy would have to be about 83 − 17 = 66 kJ mol⁻¹, not 40 kJ mol⁻¹.
If Ea really were 40 kJ mol⁻¹:
k_cat/k = exp[(83 000 − 40 000)/(8.314 × 300)] = exp(17.2) ≈ 3 × 10⁷.
That is tens of millions of times faster, not a thousand times, so the two parts of the claim are mutually inconsistent. Either the '40 kJ mol⁻¹' figure or the 'thousand times' figure is wrong.
What she would see at 300 K: • Claim as stated (1000×): t½ = 57.8 min/1000 ≈ 3.5 s - just measurable with a stopwatch. • Ea = 40 kJ mol⁻¹ version: t½ ≈ 57.8 min/(3 × 10⁷) ≈ 10⁻⁴ s - far too fast for manual timing; she would see immediate completion.
Judgement: the experiment can distinguish the two hypotheses, so the claim is testable, but the supplier's statement cannot be accepted as written.