13.1 Reaction Rates, Rate Laws & Determination of Reaction Order

13.1 Reaction Rates, Rate Laws & Determination of Reaction Order

Quick Summary: Chemical kinetics studies the velocities of chemical reactions and the molecular pathways through which transformations occur. Reaction rate is defined as the change in concentration of a reactant or product per unit time, linked stoichiometrically by $\text{Rate} = -\frac{1}{a}\frac{d[\text{A}]}{dt} = \frac{1}{c}\frac{d[\text{C}]}{dt}$. Differential rate laws take the form $\text{Rate} = k[\text{A}]^m[\text{B}]^n$, where reaction orders $m$ and $n$ must be deduced empirically and cannot be inferred from balanced equations. The units of the rate constant $k$ reflect the overall reaction order, and the method of initial rates provides a rigorous experimental approach to determine orders and rate constants.

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1. Definition of Chemical Reaction Rate & Stoichiometric Equivalence

The rate of a chemical reaction quantifies how rapidly reactants convert into products. In solution or homogeneous gas phases, rate is expressed as the change in molar concentration ($\text{M} = \text{mol/L}$) per unit time (typically seconds, $\text{s}$): $\text{Rate} = -\frac{\Delta[\text{Reactant}]}{\Delta t} = +\frac{\Delta[\text{Product}]}{\Delta t}$ A negative sign is assigned to reactants because their concentrations decrease over time ($\Delta[\text{Reactant}] < 0$), ensuring that the reported reaction rate is always a positive quantity.

Because reactants and products participate in reactions according to defined stoichiometric ratios, the rates of consumption and formation are directly coupled. For the generalized reaction: $a\text{ A} + b\text{ B} \longrightarrow c\text{ C} + d\text{ D}$ the universally standardized reaction rate is formulated by dividing each species rate by its corresponding stoichiometric coefficient: $\text{Rate} = -\frac{1}{a}\frac{d[\text{A}]}{dt} = -\frac{1}{b}\frac{d[\text{B}]}{dt} = +\frac{1}{c}\frac{d[\text{C}]}{dt} = +\frac{1}{d}\frac{d[\text{D}]}{dt}$

Stoichiometric Rate Comparison

Consider the gas-phase decomposition of dinitrogen pentoxide: $2\text{ N}_2\text{O}_5(g) \longrightarrow 4\text{ NO}_2(g) + \text{O}_2(g)$ The stoichiometric rate equivalence expression is: $\text{Rate} = -\frac{1}{2}\frac{d[\text{N}_2\text{O}_5]}{dt} = +\frac{1}{4}\frac{d[\text{NO}_2]}{dt} = +\frac{d[\text{O}_2]}{dt}$ This relationship establishes that nitrogen dioxide ($\text{NO}_2$) forms at four times the rate at which oxygen ($\text{O}_2$) is generated, and at twice the rate at which dinitrogen pentoxide ($\text{N}_2\text{O}_5$) decomposes: $\frac{d[\text{NO}_2]}{dt} = 4\frac{d[\text{O}_2]}{dt} = -2\frac{d[\text{N}_2\text{O}_5]}{dt}$

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2. Average, Instantaneous, and Initial Rates

Kinetic rates can be evaluated across different temporal windows:

1. Average Rate: The concentration change observed over a finite macroscopic time interval, $\text{Rate}_{\text{avg}} = -\frac{\Delta[\text{A}]}{\Delta t} = -\frac{[\text{A}]_2 - [\text{A}]_1}{t_2 - t_1}$. Because reactant concentrations decrease as the reaction proceeds, molecular collision frequencies decline, causing the average rate to diminish progressively over time.
2. Instantaneous Rate: The rate of reaction at a single specific instant in time ($t$). Mathematically, it is the derivative of concentration with respect to time, evaluated as the slope of the tangent line drawn to the concentration-versus-time curve at that specific point: $\text{Rate}_{\text{inst}} = \lim_{\Delta t \to 0} \left(-\frac{\Delta[\text{A}]}{\Delta t}\right) = -\frac{d[\text{A}]}{dt}$
3. Initial Rate ($\text{Rate}_0$): The instantaneous rate measured at the very beginning of the reaction ($t = 0$), immediately after mixing reactants. The initial rate is especially valuable in laboratory kinetics because product concentrations are essentially zero, which completely prevents reverse reactions from interfering with the forward kinetic measurement.

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3. The Differential Rate Law & Reaction Orders

The differential rate law describes how the instantaneous reaction rate depends on the concentrations of reactants present in the system: $\text{Rate} = k[\text{A}]^m[\text{B}]^n$ In this formulation:

• $k$ is the rate constant, a proportionality constant that is unique to each reaction and strongly dependent on temperature and catalysts, but independent of reactant concentrations.
• Exponents $m$ and $n$ represent the reaction orders with respect to reactants A and B.
• The sum $(m + n)$ represents the overall reaction order.

Physical Significance of Reaction Orders

• Zero Order ($m = 0$): The rate is entirely independent of reactant concentration ($\text{Rate} = k[\text{A}]^0 = k$). Doubling $[\text{A}]$ produces no change in reaction rate.
• First Order ($m = 1$): The rate is directly proportional to concentration ($\text{Rate} = k[\text{A}]^1$). Doubling $[\text{A}]$ doubles the rate ($2^1 = 2$).
• Second Order ($m = 2$): The rate is proportional to the square of concentration ($\text{Rate} = k[\text{A}]^2$). Doubling $[\text{A}]$ quadruples the rate ($2^2 = 4$).

Critical General Chemistry Principle: Reaction orders cannot be deduced from the stoichiometric coefficients of an overall balanced chemical equation. An overall equation describes net mass balance, not the individual molecular collisions occurring along the reaction pathway. Reaction orders must be determined strictly through experimental measurement.

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4. Dimensional Analysis and Units of the Rate Constant ($k$)

Because the reaction rate always possesses units of $\text{M}\cdot\text{s}^{-1}$ (or $\text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$), the dimensional units of the rate constant $k$ must adjust to balance the overall reaction order ($p = m + n$): $\text{units of } k = \frac{\text{Rate}}{[\text{Concentration}]^p} = \frac{\text{M}\cdot\text{s}^{-1}}{\text{M}^p} = \text{M}^{1-p}\cdot\text{s}^{-1}$

Summary of Rate Constant Units by Overall Order

Identifying the units of $k$ on an examination immediately reveals the overall order of the reaction without requiring any additional information.

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5. The Method of Initial Rates: Systematic Quantitative Deduction

The method of initial rates isolates the individual effect of each reactant on the initial reaction velocity. A series of experiments is performed in which the concentration of one reactant is systematically varied while holding all other initial concentrations constant.

Worked Example: Kinetic Analysis of Nitric Oxide Oxidation

Consider hypothetical initial-rate data for the gas-phase oxidation of nitric oxide at one fixed temperature: $2\text{ NO}(g) + \text{O}_2(g) \longrightarrow 2\text{ NO}_2(g)$ Three experimental runs yield the following initial rate data:

Step 1: Determine the Partial Order with Respect to NO ($m$)

Select two trials where $[\text{O}_2]$ remains constant while $[\text{NO}]$ changes (Trials 1 and 2): $\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[\text{NO}]_2^m[\text{O}_2]_2^n}{k[\text{NO}]_1^m[\text{O}_2]_1^n} \implies \frac{1.00 \times 10^{-4}\text{ M/s}}{2.50 \times 10^{-5}\text{ M/s}} = \left(\frac{0.020\text{ M}}{0.010\text{ M}}\right)^m$ $4.0 = (2.0)^m \implies 2^2 = 2^m \implies m = 2$ The reaction is second order in NO.

Step 2: Determine the Partial Order with Respect to $\text{O}_2$ ($n$)

Select two trials where $[\text{NO}]$ remains constant while $[\text{O}_2]$ changes (Trials 1 and 3): $\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[\text{NO}]_3^m[\text{O}_2]_3^n}{k[\text{NO}]_1^m[\text{O}_2]_1^n} \implies \frac{5.00 \times 10^{-5}\text{ M/s}}{2.50 \times 10^{-5}\text{ M/s}} = \left(\frac{0.020\text{ M}}{0.010\text{ M}}\right)^n$ $2.0 = (2.0)^n \implies 2^1 = 2^n \implies n = 1$ The reaction is first order in $\text{O}_2$.

Step 3: Formulate the Complete Rate Law

$\text{Rate} = k[\text{NO}]^2[\text{O}_2]$ The overall reaction order is $2 + 1 = 3$ (third order overall).

Step 4: Calculate the Numerical Value and Units of $k$

Substitute data from Trial 1 into the rate law: $2.50 \times 10^{-5}\text{ M}\cdot\text{s}^{-1} = k(0.010\text{ M})^2(0.010\text{ M}) = k(1.0 \times 10^{-6}\text{ M}^3)$ $k = \frac{2.50 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}}{1.0 \times 10^{-6}\text{ M}^3} = 25\text{ M}^{-2}\cdot\text{s}^{-1} = 25\text{ L}^2\cdot\text{mol}^{-2}\cdot\text{s}^{-1}$ The rate constant is $25\text{ M}^{-2}\cdot\text{s}^{-1}$ at the temperature of these trials.