Reference ID: MET-E5FC | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The half-life of a first-order reaction is the time required for the concentration of a reactant to decrease to one-half of its initial value. In process engineering, this metric is indispensable for designing reactors, storage vessels, and thermal treatments where degradation or conversion must be controlled—e.g., vitamin loss during pasteurisation, peroxide decomposition in polymer extrusion, or flavour decay in shelf-life studies. Because the reaction rate constant \(k\) is temperature-dependent via the Arrhenius equation, the half-life collapses temperature, activation energy, and pre-exponential factor into a single intuitive number that operators and designers can use directly for scale-up or safety assessments.
Methodology & Formulas
Absolute temperature
Convert the operating temperature from °C to kelvin:
\[ T_{\text{K}} = T_{\text{°C}} + 273.15 \]
Arrhenius rate constant
Compute the first-order rate constant \(k\) (units s-1):
\[ k = A\,\exp\left(-\frac{E}{R\,T_{\text{K}}}\right) \]
where
\(A\) is the pre-exponential factor (s-1)
\(E\) is the activation energy (kJ kmol-1)
\(R = 8.314\) kJ kmol-1 K-1
First-order half-life
For a first-order decay, the half-life is independent of initial concentration:
\[ t_{1/2} = \frac{\ln 2}{k} \]
To avoid division by zero or negative values, enforce \(k > 0\); otherwise, the calculation is physically meaningless.
Unit conversion (optional)
Convert seconds to hours when reporting:
\[ t_{1/2,\text{h}} = \frac{t_{1/2}}{3600} \]
Validity regimes
Parameter
Criterion
Interpretation
Temperature
\(T_{\text{K}} > 0\)
Below absolute zero is non-physical
Rate constant
\(k > 0\)
Non-positive \(k\) implies either numerical underflow or invalid Arrhenius parameters
The half-life (t½) for a first-order reaction is the time required for the concentration of the reactant to drop to 50% of its initial value. It is calculated from the rate constant k with the formula: t½ = ln(2)/k ≈ 0.693/k. Because k has units of reciprocal time (e.g., min−1 or h−1), t½ automatically comes out in the desired time unit.
No—half-life is independent of starting concentration for first-order kinetics. Whether you begin with 10% or 90% conversion, the time to reach the next 50% reduction remains the same. This makes first-order reactions especially predictable for scale-up and scheduling.
Use any concentration-vs-time data set that shows at least two points. A quick check:
Plot ln(C) versus time; if the slope is linear, the reaction is first-order.
Extract k from the slope (k = –slope).
Calculate t½ = 0.693/k.
On-line analyzers or grab-sample data logged every few minutes are usually sufficient for plant-grade accuracy.
Temperature is the dominant variable. The Arrhenius equation k = A e(–Ea/RT) shows that k increases exponentially with T, so t½ drops sharply as you heat the reactor. For every 10 °C rise, a rule-of-thumb is a two- to three-fold decrease in half-life, but always verify with plant data because activation energies differ by chemistry.
Yes. After n half-lives, the remaining fraction is (½)n. Examples:
After 1 half-life: 50% remains.
After 3 half-lives: 12.5% remains.
After 7 half-lives: <1% remains—often treated as “complete” for purge or change-over calculations.
This shortcut lets you schedule downstream steps without integrating the full rate equation each time.
Worked Example – Half-Life of a First-Order Decomposition in a CSTR Feed
A specialty-chemical plant feeds a heat-sensitive reagent to a continuous stirred-tank reactor. The reagent decomposes by a first-order reaction whose rate constant is not tabulated at the operating temperature. The process engineer must determine the reagent’s half-life under the chosen conditions to verify that conversion remains below 1% during the average residence time.
Compute the rate constant k from the Arrhenius equation:
\[ k = A\,e^{-E/(RT)} \]
\[ k = (1.0 \times 10^{12})\,\exp\!\left(\frac{-95,000}{8.314 \times 298.15}\right) \]
\[ k = 2.27 \times 10^{-5}\ \text{s}^{-1} \]
For a first-order reaction, the half-life is:
\[ t_{1/2} = \frac{\ln 2}{k} \]
\[ t_{1/2} = \frac{0.693}{2.27 \times 10^{-5}\ \text{s}^{-1}} \]
\[ t_{1/2} = 30,551\ \text{s} \]
Convert to hours for operational convenience:
\[ t_{1/2} = \frac{30,551}{3600} = 8.49\ \text{h} \]
Final Answer: The reagent’s half-life under the stated conditions is 8.49 h (30,551 s). Because this comfortably exceeds the planned 2-h residence time, less than 1% of the reagent will decompose in the reactor.
"Un projet n'est jamais trop grand s'il est bien conçu."— André Citroën
"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise."— Charles de Gaulle
