Half-life data at two temperatures: can a 12-minute reactor meet the 90% specification?
A buffered aqueous intermediate decomposes in a plug-flow reactor. Its half-life is 69.3 min at 300 K and does not change when the initial concentration is raised from 0.020 mol L−1 to 0.080 mol L−1. At 310 K the half-life falls to 17.3 min. The process team wants to operate at 320 K because the existing heat exchanger can hold that temperature, and the reactor's residence time at 320 K is 12.0 min. The specification requires at least 90% decomposition per pass. A colleague argues that because the reaction is 'unimolecular' the activation energy must be small, so the rate constant can simply be scaled with temperature without further checking. Extract the order and rate constants from the half-lives, decide what activation energy the data support, and judge whether the 320 K proposal meets the specification or whether the reactor must be modified.
1. What is the order of reaction, and what are the rate constants at 300 K and 310 K? Justify the order from the data rather than from the word 'decomposition'.
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Order. The half-life is unchanged (69.3 min) when the initial concentration is quadrupled from 0.020 to 0.080 mol L−1. Only a first-order reaction has t1/2 independent of [A]0, because t1/2 = ln2/k. A zero-order reaction would halve its half-life on quadrupling the concentration, and a second-order reaction would double it.
Rate constants.
- At 300 K: k1 = 0.693/t1/2 = 0.693/69.3 min = 1.00 × 10−2 min−1.
- At 310 K: k2 = 0.693/17.3 min = 4.01 × 10−2 min−1.
So k roughly quadruples for a 10 K rise. That is a strong temperature response, which already contradicts the colleague's claim that the activation energy is small; a low-Ea process would show a change of a few per cent, not 300%. Units of min−1 also confirm first order.
2. Calculate the activation energy from the two rate constants and state the assumptions you had to make.
Read the worked answer
Use the two-point Arrhenius form: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2).
ln(4.01 × 10−2 / 1.00 × 10−2) = ln 4 = 1.386. 1/300 − 1/310 = (310 − 300)/(300 × 310) = 10/93000 = 1.075 × 10−4 K−1.
Ea = 1.386 × 8.314 / 1.075 × 10−4 ≈ 1.07 × 10^5 J mol−1 ≈ 107 kJ mol−1.
Assumptions and warnings. (i) k was measured at only two temperatures, so there is no check of Arrhenius linearity and no error estimate; a third temperature is needed. (ii) The mechanism and hence Ea must be unchanged over 300–310 K - if the buffer, dissolved CO2 or wall catalysis contributes differently at each temperature, the calculated Ea is an apparent value only. (iii) An Ea of about 107 kJ mol−1 is chemically reasonable for bond-breaking and confirms that temperature control is critical: a 10 K error in the set point changes k by a factor of 4.
3. Predict the conversion at 320 K with a 12.0 min residence time. Does it meet the 90% specification, and how much confidence do you place in the prediction?
Read the worked answer
Extrapolate k to 320 K: ln(k3/k1) = (Ea/R)(1/300 − 1/320) = (1.0718 × 10^5/8.314) × (20/96000) = 12890 × 2.083 × 10−4 ≈ 2.686. k3 = 1.00 × 10−2 × e^2.686 ≈ 1.47 × 10−1 min−1.
Conversion for first order in a plug-flow reactor: X = 1 − e^(−k t) = 1 − e^(−0.147 × 12.0) = 1 − e^(−1.764) ≈ 1 − 0.171 = 0.829.
So about 83% conversion - below the 90% specification. The required residence time would be t = ln(1/0.10)/k = 2.303/0.147 ≈ 15.7 min, i.e. roughly 16 min.
Judgement on reliability. The prediction is an extrapolation 10 K beyond the highest measured point, using only two data points and assuming constant Ea and unchanged mechanism. Side reactions (hydrolysis, oxidation) may grow faster than the main step at 320 K, and the buffer's pKa changes with temperature. Before rebuilding the reactor, run one experiment at 320 K; if the measured k is within about 10% of 0.147 min−1 the design is sound, otherwise re-fit the Arrhenius parameters.