Introduction & Context

Reaction-order determination is the first step in quantifying how fast a chemical transformation proceeds. In process engineering, it underpins the design of reactors, the sizing of downstream separation units, and the optimisation of operating conditions. Typical applications include wastewater treatment (where pollutant abatement must meet discharge limits), pharmaceutical synthesis (where selectivity and batch time are critical), and combustion processes (where heat-release rates affect furnace safety).

The calculation takes a set of concentration–time data, assumes integer orders (0, 1, or 2), and identifies which order yields the most linear transformed plot. The slope of the best-fit straight line then gives the rate constant in the units required by that order. Because the method is algebraic and does not require iterative solvers, it is embedded in spreadsheet tools, control-system firmware, and regulatory spreadsheets for rapid field validation.

Methodology & Formulas

Data preparation
Convert temperature to absolute scale: \[ T_{\text{K}} = T_{\text{C}} + 273.15 \] Create discrete time abscissa: \[ t_{i} = t_{\text{min}},\; t_{\text{min}}+\Delta t,\; \ldots ,\; t_{\text{max}} \]

Linearisation for each candidate order

Order	Transformation	Ordinate	Slope Symbol	Rate-constant Relation
0	\( C\) vs \( t\)	\( C_{i}\)	\( m_{0}\)	\( k_{0}=-m_{0}\)
1	\( \ln C\) vs \( t\)	\( \ln C_{i}\)	\( m_{1}\)	\( k_{1}=-m_{1}\)
2	\( 1/C\) vs \( t\)	\( 1/C_{i}\)	\( m_{2}\)	\( k_{2}=m_{2}\)

Least-squares slope
For any transformed ordinate \( y_{i}\) (chosen from the table above) the slope is \[ m = \frac{n\sum t_{i}y_{i}-\sum t_{i}\sum y_{i}}{n\sum t_{i}^{2}-\left(\sum t_{i}\right)^{2}} \] with \( n\) the number of data points.
Coefficient of determination
Compute fitted values \( \hat{y}_{i}=y_{0}+m\,t_{i}\) and residuals: \[ R^{2}=1-\frac{\sum (y_{i}-\hat{y}_{i})^{2}}{\sum (y_{i}-\bar{y})^{2}},\qquad \bar{y}=\frac{1}{n}\sum y_{i} \] Repeat for each order; the order with the highest \( R^{2}\) is selected.

Validity regimes for the linearisation

Order	Conversion range	Warning threshold
0	\( C/C_{0}\)	<0.8
1	\( C/C_{0}\)	<0.1 or >1.0
2	\( C/C_{0}\)	<0.2 or >1.0

Outside these ranges, the underlying linearity assumption breaks down, and a more rigorous non-linear regression is advised.

Temperature consistency for Arrhenius use

Variable	Range	Unit
\( T_{\text{K}}\)	353–373	K

Data outside this interval should not be extrapolated with a single Arrhenius correlation without re-estimating activation energy.

Use the differential method when you have noisy data but many closely-spaced points; it tolerates scatter because it relies on local slopes.
Use the integral method when data are smooth and sparse; integrated rate laws reduce random error by averaging over the entire run.
If the reaction is complex (e.g., reversible or parallel steps), start with the integral form—its closed-form equations make regression more stable.

Zero-order: plot C_A vs t—look for a straight line.
First-order: plot ln C_A vs t—linearity indicates first order.
Second-order: plot 1/C_A vs t—linearity points to second order.
If none of these give a good R², assume a fractional or mixed order and switch to nonlinear regression.

Minimum of 5 points per decade for simple nth-order reactions.
If the rate constant must be known within ±5%, aim for 8–10 points.
Space points logarithmically (e.g., 100%, 75%, 50%, 25%, 10% conversion) to evenly weight the regression.

Measure replicate runs at low, mid, and high conversion to estimate σ_i.
Use weighted least squares with weights w_i = 1/σ_i²; most software accepts a weight column.
If replicates are impractical, assume variance ∝ C² (typical for GC analysis) and apply w_i = 1/C_i².

Record residual-sum-of-squares (RSS) for both models.
Perform an F-test: F = [(RSS_simple − RSS_complex)/(p_complex − p_simple)] / [RSS_complex/(n − p_complex)].
If the calculated F exceeds the tabulated F at 95% confidence, the additional parameter (fractional order) is justified.

Worked Example: Determining Reaction Order from Concentration–Time Data

A process engineer is validating the kinetics of a liquid-phase decomposition in a pilot reactor. A single run is performed at 90 °C with an initial concentration of 250 mmol L^-1. Samples are withdrawn every 5 min for 30 min and analysed off-line. The engineer needs to know the reaction order and the rate constant to scale the reactor for continuous operation.

Knowns

Initial concentration, \(C_0 = 250\) mmol L^-1
Temperature, \(T = 90\) °C (363 K)
Sampling interval, \(\Delta t = 5\) min
Final time, \(t_{\text{max}} = 30\) min
Seven data points: \(t = 0,\,5,\,10,\,15,\,20,\,25,\,30\) min
Measured concentrations (with analytical noise) provided in the data set

Step-by-Step Determination

Assume the rate law fits the general form \[ -\frac{\mathrm{d}C}{\mathrm{d}t}=kC^n \] and integrate for \(n = 0,\,1,\,2\).

For each assumed order, linearise the data:

Order	Linear form	Plot
0	\(C = C_0 - k_0 t\)	\(C\) vs \(t\)
1	\(\ln C = \ln C_0 - k_1 t\)	\(\ln C\) vs \(t\)
2	\(\frac{1}{C} = \frac{1}{C_0} + k_2 t\)	\(\frac{1}{C}\) vs \(t\)

Perform ordinary-least-squares on each linear form to obtain slopes and coefficients of determination \(R^2\):
- Zero-order: \(R^2_0 = 0.853\)
- First-order: \(R^2_1 = 0.999\)
- Second-order: \(R^2_2 = 0.923\)
Select the order with the highest \(R^2\). Here, the first-order fit is clearly superior.
Extract the rate constant from the slope of the first-order plot: \[ k = -\text{slope}_1 = 0.044\ \text{min}^{-1} \]

Final Answer

The decomposition follows first-order kinetics with a rate constant
\(k = 0.044\ \text{min}^{-1}\) at 90 °C.

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