13.1 Reaction Rates, Rate Laws, and Order of Reaction

13.1 Reaction Rates, Rate Laws, and Order of Reaction

Chemical reaction engineering combines chemical kinetics with physical transport phenomena to design, size, and optimize chemical reactors. In the NCEES PE Chemical Reference Handbook, chemical kinetics forms the foundation of reactor design. Before sizing an isothermal or non-isothermal vessel, the chemical engineer must establish the correct mathematical rate law, identify the true reaction order, and understand the physical significance and units of the kinetic rate constant.

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1. Stoichiometric Definitions of Reaction Rate

For a homogeneous chemical reaction occurring in a fluid volume $V$, the rate of reaction of species $i$ ($r_i$) is an intensive property defined as the moles of species $i$ formed (or consumed) per unit volume per unit time:

$r_i = \frac{1}{V} \frac{dn_i}{dt}$

Where:

• $n_i$ = moles of species $i$ in the reacting system ($\text{mol}$ or $\text{kmol}$).
• $V$ = volume of the reacting fluid mixture ($\text{L}$ or $\text{m}^3$).
• $t$ = time ($\text{s}$, $\text{min}$, or $\text{h}$).
• $r_i$ = rate of formation of species $i$ (positive for products, negative for reactants, with units of $\text{mol}/(\text{L}\cdot\text{s})$ or $\text{kmol}/(\text{m}^3\cdot\text{h})$).

By convention in chemical engineering, the rate of disappearance (or consumption) of a reactant $A$ is written as a positive quantity: $-r_A = -\frac{1}{V} \frac{dn_A}{dt}$.

Stoichiometric Relative Rates

Consider the general chemical reaction:

$a A + b B \longrightarrow c C + d D$

Using the generalized stoichiometric equation $\sum \nu_j A_j = 0$, where stoichiometric coefficients $\nu_j$ are negative for reactants ($\nu_A = -a, \nu_B = -b$) and positive for products ($\nu_C = +c, \nu_D = +d$), the single, species-independent rate of reaction $r_{rxn}$ is:

$r_{rxn} = \frac{r_i}{\nu_i} = \frac{-r_A}{a} = \frac{-r_B}{b} = \frac{r_C}{c} = \frac{r_D}{d}$

Consequently, the relative rates of consumption and formation are directly linked by stoichiometry:

$-r_B = \frac{b}{a} (-r_A), \qquad r_C = \frac{c}{a} (-r_A), \qquad r_D = \frac{d}{a} (-r_A)$

[!IMPORTANT] Stoichiometric Rate Constants:
When a problem provides a rate law such as $-r_A = k_A C_A^2$, the rate constant $k_A$ is defined with respect to reactant $A$. If the rate of consumption of reactant $B$ is asked for a reaction $2A + B \to C$, then $-r_B = \frac{1}{2}(-r_A) = \frac{1}{2} k_A C_A^2 = k_B C_A^2$. Therefore, $k_B = \frac{1}{2} k_A$. Forgetting this stoichiometric factor is a frequent source of lost points on the PE exam.

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2. Power-Law Rate Expressions and Reaction Order

For many homogeneous reactions, the rate of consumption of reactant $A$ can be parameterized by an empirical power-law rate law:

$-r_A = k C_A^\alpha C_B^\beta \dots C_M^\gamma$

Where:

• $k$ = reaction rate constant (a strong function of temperature via the Arrhenius equation, but independent of reactant concentrations).
• $C_i$ = molar concentration of species $i$ ($\text{mol/L}$ or $\text{kmol/m}^3$).
• $\alpha, \beta, \dots, \gamma$ = partial reaction orders with respect to components $A, B, \dots, M$.
• $n = \alpha + \beta + \dots + \gamma$ = overall reaction order.

Dimensional Units of the Kinetic Rate Constant ($k$)

The units of the rate of reaction are always $[\text{concentration}] \cdot [\text{time}]^{-1}$. Equating dimensions across the power-law equation reveals how the units of $k$ depend fundamentally on overall order $n$:

$[-r_A] = [k] [C]^n \implies [k] = \frac{[\text{concentration}]}{[\text{time}] \cdot [\text{concentration}]^n} = [\text{concentration}]^{1-n} \cdot [\text{time}]^{-1}$

[!TIP] Quick Unit Check for Reaction Order:
On multiple-choice PE exam questions, inspecting the units of $k$ immediately reveals the overall reaction order without performing any algebra. If $k$ has units of $\text{h}^{-1}$, the reaction is undeniably first order ($n = 1$). If $k$ is in $\text{m}^3/(\text{kmol}\cdot\text{min})$, it is second order ($n = 2$).

Gas-Phase Reactions in Partial Pressures

For ideal gas mixtures where partial pressure $p_i = C_i R T$, rate expressions are often written in terms of partial pressures:

$-r_A = k_p p_A^\alpha p_B^\beta$

Where the pressure-basis rate constant $k_p$ relates to the concentration-basis constant $k_c$ via:

$k_c = k_p (R T)^n$

Here, $n = \alpha + \beta$ is the overall order, and $R$ is the universal gas constant in compatible units.

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3. Elementary vs. Non-Elementary Kinetics & Mechanisms

A central distinction in chemical kinetics is whether a reaction proceeds in a single molecular step or through a sequence of intermediate steps.

Elementary Reactions

An elementary reaction occurs exactly as written at the molecular level, in a single event where colliding molecules break and form bonds simultaneously. For elementary reactions, and only for elementary reactions, the partial reaction orders are identically equal to the stoichiometric coefficients:

• Monomolecular: $A \longrightarrow \text{Products} \implies -r_A = k C_A$
• Bimolecular: $A + B \longrightarrow \text{Products} \implies -r_A = k C_A C_B$
• Bimolecular (self): $2 A \longrightarrow \text{Products} \implies -r_A = k C_A^2$
• Termolecular: $2 A + B \longrightarrow \text{Products} \implies -r_A = k C_A^2 C_B$

Non-Elementary Reactions and Heterogeneous Catalysis

Most industrial processes are non-elementary, proceeding through complex multi-step reaction networks involving free radicals, ionic intermediates, or chemisorbed surface species. Their empirical reaction orders may be zero, fractional, or negative, and often vary with concentration.

A classic example is catalytic surface kinetics governed by the Langmuir-Hinshelwood-Hougen-Watson (LHHW) mechanism or enzymatic kinetics governed by the Michaelis-Menten model:

$-r_A = \frac{k_1 C_A}{1 + k_2 C_A}$

• Low reactant concentration ($k_2 C_A \ll 1$): The denominator approaches $1$, and the rate reduces to pseudo-first-order kinetics: $-r_A \approx k_1 C_A$.
• High reactant concentration ($k_2 C_A \gg 1$): Catalyst active sites become fully saturated. The $k_2 C_A$ term dominates the denominator, causing the concentration terms to cancel: $-r_A \approx \frac{k_1 C_A}{k_2 C_A} = \frac{k_1}{k_2} = k_{apparent}$. The reaction shifts to pseudo-zero-order kinetics.

Reversible Reactions and Thermodynamic Consistency

For a reversible elementary reaction $a A \rightleftharpoons b B$, the net rate of disappearance of $A$ is the difference between the forward and reverse rates:

$-r_A = k_f C_A^a - k_r C_B^b = k_f \left( C_A^a - \frac{C_B^b}{K_c} \right)$

At thermodynamic equilibrium, the macroscopic rate vanishes ($-r_A = 0$):

$k_f C_{A,eq}^a = k_r C_{B,eq}^b \implies K_c = \frac{k_f}{k_r} = \frac{C_{B,eq}^b}{C_{A,eq}^a}$

This relationship enforces strict thermodynamic consistency between kinetic rate constants and the chemical equilibrium constant $K_c$.

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4. Differential and Integral Methods of Kinetic Data Analysis

To determine the reaction order and rate constant from laboratory batch reactor data, two primary approaches are utilized: the differential method and the integral method.

The Differential Method

In a constant-volume batch reactor, the rate of reaction is $-r_A = -\frac{dC_A}{dt}$. Taking the natural logarithm of a power-law rate law $-r_A = k C_A^n$:

$\ln\left( -\frac{dC_A}{dt} \right) = \ln(k) + n \ln(C_A)$

By measuring concentration $C_A$ over time, determining numerical derivatives $dC_A/dt$, and plotting $\ln(-dC_A/dt)$ versus $\ln(C_A)$:

• The slope of the resulting straight line equals the reaction order $n$.
• The y-intercept equals $\ln(k)$.

The Method of Initial Rates

A powerful variant of the differential method widely tested on the PE exam is the method of initial rates. Multiple experimental runs are initiated at varying starting concentrations ($C_{A0}, C_{B0}$), and the initial slope $(-dC_A/dt)_{t=0}$ is measured before significant conversion occurs or product inhibition develops. Taking the ratio of rates between two runs where only one concentration is varied isolates that species' partial reaction order:

$\frac{(-r_{A0})_2}{(-r_{A0})_1} = \left( \frac{C_{A0,2}}{C_{A0,1}} \right)^\alpha$

The Integral Method

The integral method postulates a tentative reaction order, integrates the differential rate expression analytically, and tests whether experimental $(C_A, t)$ data yield a linear plot:

1. Zero Order ($-r_A = k$): $-\int_{C_{A0}}^{C_A} dC_A = k \int_0^t dt \implies C_A = C_{A0} - k t$ Linear plot: $C_A$ vs. $t$ has slope $-k$ and intercept $C_{A0}$.

2. First Order ($-r_A = k C_A$): $-\int_{C_{A0}}^{C_A} \frac{dC_A}{C_A} = k \int_0^t dt \implies \ln\left( \frac{C_{A0}}{C_A} \right) = k t \quad \text{or} \quad C_A = C_{A0} e^{-k t}$ Linear plot: $\ln(C_A)$ vs. $t$ has slope $-k$ and intercept $\ln(C_{A0})$.

3. Second Order ($2A \to \text{Products}$, $-r_A = k C_A^2$): $-\int_{C_{A0}}^{C_A} \frac{dC_A}{C_A^2} = k \int_0^t dt \implies \frac{1}{C_A} - \frac{1}{C_{A0}} = k t$ Linear plot: $1/C_A$ vs. $t$ has slope $+k$ and intercept $1/C_{A0}$.

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5. Half-Life ($t_{1/2}$) Formulations and Reaction Order Diagnostics

The half-life ($t_{1/2}$) is the time required for the reactant concentration to drop to exactly half of its initial value ($C_A = C_{A0}/2$, corresponding to fractional conversion $X = 0.50$). The functional dependence of $t_{1/2}$ on initial concentration $C_{A0}$ provides an instantaneous diagnostic test for reaction order.

Analytical Derivations for Constant Volume

1. Zero Order: $C_A = \frac{C_{A0}}{2} = C_{A0} - k t_{1/2} \implies t_{1/2} = \frac{C_{A0}}{2 k}$ Behavior: Half-life is directly proportional to initial concentration. Doubling $C_{A0}$ doubles $t_{1/2}$.

2. First Order: $\ln\left( \frac{C_{A0}}{C_{A0}/2} \right) = \ln(2) = k t_{1/2} \implies t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.69315}{k}$ Behavior: Half-life is strictly independent of initial concentration. Every successive half-life takes the exact same amount of time.

3. Second Order ($2A \to \text{Products}$): $\frac{1}{C_{A0}/2} - \frac{1}{C_{A0}} = \frac{2}{C_{A0}} - \frac{1}{C_{A0}} = \frac{1}{C_{A0}} = k t_{1/2} \implies t_{1/2} = \frac{1}{k C_{A0}}$ Behavior: Half-life is inversely proportional to initial concentration. Doubling $C_{A0}$ cuts $t_{1/2}$ in half.

4. General Order $n$ ($n \neq 1$): $t_{1/2} = \frac{2^{n-1} - 1}{(n - 1) k C_{A0}^{n-1}} \propto C_{A0}^{1-n}$ Taking logarithms yields a direct method to find $n$ from two half-life measurements: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{C_{A0,2}}{C_{A0,1}} \right)^{n-1} \implies n = 1 + \frac{\ln[(t_{1/2})_1 / (t_{1/2})_2]}{\ln[C_{A0,2} / C_{A0,1}]}$

   Reaction Order Diagnostic via Half-Life Progression:
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   Zero Order (n=0):     t1/2(1) = 40 min  --->  t1/2(2) = 20 min  (Halves as CA drops)
   First Order (n=1):    t1/2(1) = 40 min  --->  t1/2(2) = 40 min  (Constant)
   Second Order (n=2):   t1/2(1) = 40 min  --->  t1/2(2) = 80 min  (Doubles as CA drops)

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6. Summary Comparison Table: Kinetics by Reaction Order

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7. Step-by-Step Worked Numerical Example: Method of Initial Rates & Rate Law Determination

Problem Statement

An engineer is characterizing the liquid-phase synthesis of a specialty monomer governed by the stoichiometry:

$2 A + B \longrightarrow C + 2 D$

Four isothermal batch experiments are performed at $25.0^\circ\text{C}$ in a constant-volume laboratory reactor. The initial rates of disappearance of reactant $A$ ($(-r_{A0})$) are recorded as a function of initial concentrations:

Calculate:

1. The partial reaction orders $\alpha$ and $\beta$, and the overall reaction order $n$.
2. The numerical value and dimensional units of the reaction rate constant $k_A$ based on species $A$.
3. The reaction rate constant $k_B$ based on the consumption of species $B$, and $k_C$ based on the formation of product $C$.
4. The predicted initial rate of formation of product $D$ ($r_{D0}$) if a new run is initiated with $C_{A0} = 0.500\text{ mol/L}$ and $C_{B0} = 0.200\text{ mol/L}$.

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Step 1: Determine Partial Reaction Orders

The general power-law rate expression based on species $A$ is:

$-r_A = k_A C_A^\alpha C_B^\beta$

To find $\alpha$, compare Run 2 and Run 1, where $C_{B0}$ is held constant at $0.100\text{ mol/L}$ while $C_{A0}$ doubles:

$\frac{(-r_{A0})_2}{(-r_{A0})_1} = \frac{0.01600}{0.00400} = 4.00 = \left( \frac{0.400}{0.200} \right)^\alpha = (2.00)^\alpha \implies \mathbf{\alpha = 2}$

The reaction is second order with respect to reactant $A$.

To find $\beta$, compare Run 3 and Run 1, where $C_{A0}$ is held constant at $0.200\text{ mol/L}$ while $C_{B0}$ triples:

$\frac{(-r_{A0})_3}{(-r_{A0})_1} = \frac{0.01200}{0.00400} = 3.00 = \left( \frac{0.300}{0.100} \right)^\beta = (3.00)^\beta \implies \mathbf{\beta = 1}$

The reaction is first order with respect to reactant $B$.

The overall reaction order is:

$n = \alpha + \beta = 2 + 1 = \mathbf{3 \quad (\text{third order overall})}$

(Cross-check with Run 4: $(-r_{A0})_4 = k_A (0.400)^2 (0.300) = k_A (0.0480)$. Ratio to Run 1 is $0.04800 / 0.00400 = 12.0 = 2^2 \times 3^1 = 12.0$. Matches perfectly).

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Step 2: Calculate Value and Units of Rate Constant $k_A$

Using experimental Run 1:

$k_A = \frac{(-r_{A0})_1}{C_{A0,1}^2 C_{B0,1}} = \frac{0.00400\text{ mol}/(\text{L}\cdot\text{s})}{(0.200\text{ mol/L})^2 \cdot (0.100\text{ mol/L})} = \frac{0.00400}{0.0400 \times 0.100} = \frac{0.00400}{0.00400} = \mathbf{1.000\text{ L}^2/(\text{mol}^2\cdot\text{s})}$

Units verification for third order ($n=3$):

$[k] = [\text{concentration}]^{1-3} [\text{time}]^{-1} = \left(\frac{\text{mol}}{\text{L}}\right)^{-2} \text{s}^{-1} = \mathbf{\text{L}^2/(\text{mol}^2\cdot\text{s})}$

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Step 3: Relate Rate Constants by Stoichiometry

From the stoichiometry $2A + B \to C + 2D$:

$\frac{-r_A}{2} = \frac{-r_B}{1} = \frac{r_C}{1} = \frac{r_D}{2}$

Therefore:

$-r_B = \frac{1}{2}(-r_A) = \frac{1}{2} k_A C_A^2 C_B = k_B C_A^2 C_B \implies k_B = \frac{1}{2} k_A = \mathbf{0.500\text{ L}^2/(\text{mol}^2\cdot\text{s})}$ $r_C = \frac{1}{2}(-r_A) = \frac{1}{2} k_A C_A^2 C_B = k_C C_A^2 C_B \implies k_C = \frac{1}{2} k_A = \mathbf{0.500\text{ L}^2/(\text{mol}^2\cdot\text{s})}$

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Step 4: Calculate Predicted Initial Rate of Formation of Product $D$

For $C_{A0} = 0.500\text{ mol/L}$ and $C_{B0} = 0.200\text{ mol/L}$:

$-r_{A0} = k_A C_{A0}^2 C_{B0} = 1.000 \times (0.500)^2 \times (0.200) = 1.000 \times 0.250 \times 0.200 = 0.0500\text{ mol}/(\text{L}\cdot\text{s})$

Since $r_D = \frac{2}{2}(-r_A) = -r_A$:

$r_{D0} = -r_{A0} = \mathbf{0.0500\text{ mol}/(\text{L}\cdot\text{s})}$

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8. Critical PE Exam Traps & Pitfalls

Trap 1: Assuming Exponents Equal Stoichiometric Coefficients
Never infer the rate law directly from the balanced chemical equation unless the problem statement explicitly confirms that the reaction is elementary. For instance, the reaction $2 \text{NO}_2 + \text{F}_2 \to 2 \text{NO}2\text{F}$ has overall stoichiometry of order 3, but laboratory kinetics reveal it is first order in each reactant: $-r{\text{NO}2} = k C{\text{NO}2} C{\text{F}_2}$ (order 2 overall). Always rely on experimental data or an explicit "elementary" descriptor.

Trap 2: Forgetting to Convert Basis When Stating $k$
On exam questions involving $a A + b B \to c C$, the question may state that $k_A = 0.40\text{ L}/(\text{mol}\cdot\text{s})$ and ask for the rate of formation of product $C$. You must multiply by $c/a$. Failing to convert between $k_A$, $k_B$, and $k_C$ is one of the most common numerical distractors created by exam writers.

Trap 3: Inverting the Half-Life Dependence
Remember that $t_{1/2} \propto C_{A0}^{1-n}$. For a second-order reaction ($n = 2$), $1-n = -1$, meaning $t_{1/2} = 1/(k C_{A0})$. If an engineer doubles the starting concentration for a second-order reaction, the half-life is cut in half, NOT doubled. Conversely, for a zero-order reaction ($n = 0$), $1-n = +1$, so doubling concentration doubles the time required to react $50%$.

Trap 4: Unit Errors in Ideal Gas Rate Laws
When converting between concentration-based rate constants ($k_c$) and pressure-based rate constants ($k_p$), remember $k_c = k_p (R T)^n$. The exponent is the overall order $n$, not the partial order of $A$, and $R$ must be expressed in units matching pressure and concentration (e.g., $0.08206\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$ or $8.314\times 10^{-5}\text{ m}^3\cdot\text{bar}/(\text{mol}\cdot\text{K})$).