10.1 Reaction Rates, Rate Laws, and Integrated Rate Equations

Reaction Rates, Rate Laws, and Integrated Rate Equations

Chemical kinetics investigates the rates of chemical processes, the physical and chemical variables that alter these rates, and the microscopic molecular mechanisms connecting reactants to products. While chemical thermodynamics determines the spontaneous direction and theoretical equilibrium yield of a reaction ($\Delta G < 0$), it reveals nothing about the timescale required to attain equilibrium. On the PRC Chemist Licensure Examination, mastery of chemical kinetics demands rigorous proficiency with stoichiometric rate definitions, differential and integrated rate laws, graphical transformations, half-life relationships, experimental rate-order evaluations, and radioactive decay processes.

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1. Stoichiometric Definition of Reaction Rate

For a general homogeneous reaction occurring in a closed system at constant volume:

$aA + bB \longrightarrow cC + dD$

the rate of disappearance of reactants and rate of appearance of products are stoichiometrically coupled. To avoid ambiguous rates that depend arbitrarily on which chemical species is monitored, the IUPAC defines the rate of reaction ($v$) as the time derivative of the extent of reaction ($\xi$) divided by system volume ($V$), or the rate of change of concentration divided by the stoichiometric coefficient ($\nu_J$, taken as negative for reactants and positive for products):

$v \equiv \frac{1}{\nu_J} \frac{d[J]}{dt}$

$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$

Instantaneous vs. Average Rates

• Average Rate: Calculated over a finite macroscopic time interval $\Delta t = t_2 - t_1$: $\text{Rate}_{avg} = -\frac{1}{a} \frac{[A]_2 - [A]_1}{t_2 - t_1} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t}$
• Instantaneous Rate: The derivative evaluated at an infinitesimal instant $t$ (the tangent slope of the concentration-time curve at that specific point): $\text{Rate}_{inst} = -\frac{1}{a} \lim_{\Delta t \to 0} \frac{\Delta[A]}{\Delta t} = -\frac{1}{a}\frac{d[A]}{dt}$
• Initial Rate ($v_0$): The instantaneous rate at the exact commencement of the reaction ($t = 0$), prior to any significant depletion of initial reactants or interference from reverse reactions and product inhibition.

Stoichiometric Rate Conversion Example

Consider the gas-phase decomposition of dinitrogen pentoxide: $2\text{N}_2\text{O}_5(g) \longrightarrow 4\text{NO}_2(g) + \text{O}_2(g)$ If the rate of decomposition of $\text{N}_2\text{O}_5$ is measured as $-\frac{d[\text{N}_2\text{O}_5]}{dt} = 0.040\text{ mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$, the rates of appearance of products are: $\frac{d[\text{NO}_2]}{dt} = -\frac{4}{-2}\frac{d[\text{N}_2\text{O}_5]}{dt} = 2 \times (0.040) = 0.080\text{ mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$ $\frac{d[\text{O}_2]}{dt} = -\frac{1}{-2}\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{2} \times (0.040) = 0.020\text{ mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$ $\text{Overall Reaction Rate } v = 0.020\text{ mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$

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2. Differential Rate Laws and Reaction Orders

A differential rate law expresses the instantaneous reaction rate as a mathematical function of the instantaneous concentrations of reacting species:

$v = k [A]^m [B]^n [C]^p$

• Rate Constant ($k$): A proportionality constant specific to a given reaction at a specified temperature. It is strictly independent of reactant concentrations, but varies exponentially with temperature and ionic strength.
• Partial Orders ($m, n, p$): The powers to which individual concentrations are raised. The partial order with respect to component $A$ is $m$, with respect to $B$ is $n$. Reaction orders are strictly experimental quantities; they cannot be deduced from the stoichiometric coefficients of an overall balanced chemical equation (unless the reaction is known to proceed via a single elementary step).
• Overall Reaction Order ($n_{tot}$): The algebraic sum of all partial orders: $n_{tot} = m + n + p$ Reaction orders can be positive integers ($0, 1, 2, 3$), negative numbers (indicating inhibition), or fractional values (common in free-radical gas reactions, e.g., the thermal synthesis of phosgene $\text{CO} + \text{Cl}_2 \rightarrow \text{COCl}_2$ follows $v = k [\text{CO}][\text{Cl}_2]^{3/2}$ with overall order $2.5$).

Dimensional Analysis of the Rate Constant $k$

Because reaction rate always carries units of concentration per unit time (typically $\text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$ or $\text{M}\cdot\text{s}^{-1}$), the dimensional units of $k$ depend directly on the overall reaction order $n_{tot}$:

$\text{Units of } k = (\text{concentration})^{1 - n_{tot}} \cdot (\text{time})^{-1} = \left(\frac{\text{mol}}{\text{L}}\right)^{1 - n_{tot}} \text{s}^{-1} = \text{M}^{1 - n_{tot}} \text{s}^{-1}$

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3. Integrated Rate Equations and Half-Life Derivations

Integrated rate laws express reactant concentration directly as an explicit function of continuous reaction time $t$. Let us examine the three fundamental classical rate laws for reactions of the type $A \longrightarrow \text{Products}$.

1. Zero-Order Reactions

A zero-order reaction proceeds at a constant velocity that is entirely independent of reactant concentration (common in surface-catalyzed heterogeneous reactions where active sites are fully saturated, and enzyme reactions under substrate saturation):

$-\frac{d[A]}{dt} = k$

Integrating from initial concentration $[A]_0$ at $t = 0$ to $[A]_t$ at time $t$:

$\int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt \implies [A]_t - [A]_0 = -kt$

$[A]_t = [A]_0 - kt$

• Linear Graphical Plot: Plotting $[A]_t$ versus $t$ yields a straight line with slope $m = -k$ and $y$-intercept $= [A]_0$.
• Half-Life ($t_{1/2}$): The time required for the concentration to drop to half its initial value ($[A]_{t_{1/2}} = [A]_0 / 2$): $\frac{[A]_0}{2} = [A]_0 - k t_{1/2} \implies k t_{1/2} = \frac{[A]_0}{2}$ $t_{1/2} = \frac{[A]_0}{2k}$ Crucial Diagnostic: For a zero-order reaction, the half-life is directly proportional to the initial reactant concentration. Successive half-lives become progressively shorter ($t_{1/2}, \frac{1}{2}t_{1/2}, \frac{1}{4}t_{1/2}$). The reaction terminates completely in finite time $t_{completion} = [A]_0 / k = 2 t_{1/2}$.

2. First-Order Reactions

A first-order reaction has a rate directly proportional to the first power of concentration:

$-\frac{d[A]}{dt} = k[A]$

Separating variables and integrating:

$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_{0}^{t} dt \implies \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$

$\ln[A]_t = \ln[A]_0 - kt \iff [A]_t = [A]_0 e^{-kt}$

Converting to base-10 logarithms: $\log_{10}[A]_t = \log_{10}[A]_0 - \frac{k}{2.3026} t$

• Linear Graphical Plot: Plotting $\ln[A]_t$ versus $t$ yields a straight line with slope $m = -k$ and $y$-intercept $= \ln[A]_0$. (Or plotting $\log_{10}[A]_t$ vs $t$ with slope $-k/2.303$).
• Half-Life ($t_{1/2}$): Substituting $[A]_{t_{1/2}} = [A]_0 / 2$: $\ln\left(\frac{[A]_0 / 2}{[A]_0}\right) = \ln\left(\frac{1}{2}\right) = -\ln 2 = -kt_{1/2}$ $t_{1/2} = \frac{\ln 2}{k} = \frac{0.69315}{k}$ Crucial Diagnostic: For a first-order process, the half-life is strictly independent of the initial concentration. Every successive half-life takes the exact same duration ($t_{1/2}$). After $n$ half-lives, the remaining fraction is $(1/2)^n$.

3. Second-Order Reactions (Single Reactant: $2A \longrightarrow \text{Products}$)

When the reaction rate depends on the square of a single reactant concentration:

$-\frac{d[A]}{dt} = k [A]^2$

Separating variables and integrating:

$\int_{[A]_0}^{[A]_t} -\frac{d[A]}{[A]^2} = k \int_{0}^{t} dt \implies \left[ \frac{1}{[A]} \right]_{[A]_0}^{[A]_t} = kt$

$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

• Linear Graphical Plot: Plotting $\frac{1}{[A]_t}$ versus $t$ yields a straight line with positive slope $m = +k$ and $y$-intercept $= 1/[A]_0$.
• Half-Life ($t_{1/2}$): Substituting $[A]_{t_{1/2}} = [A]_0 / 2$: $\frac{2}{[A]_0} = \frac{1}{[A]_0} + kt_{1/2} \implies \frac{1}{[A]_0} = kt_{1/2}$ $t_{1/2} = \frac{1}{k [A]_0}$ Crucial Diagnostic: For a second-order reaction, the half-life is inversely proportional to the initial concentration. Each successive half-life is double the preceding one ($t_{1/2}, 2t_{1/2}, 4t_{1/2}, 8t_{1/2}$). As the reactant dilutes, the collision probability falls quadratically, drastically prolonging the decay.

4. Second-Order Reactions (Two Reactants: $A + B \longrightarrow \text{Products}$)

When the rate law is first-order in $A$ and first-order in $B$ ($v = k [A][B]$) with unequal initial concentrations ($[A]_0 \ne [B]_0$):

Let $x$ be the concentration of $A$ and $B$ that has reacted at time $t$, such that $[A]_t = [A]_0 - x$ and $[B]_t = [B]_0 - x$: $\frac{dx}{dt} = k ([A]_0 - x)([B]_0 - x)$ Integrating by partial fractions yields the integrated rate law: $\frac{1}{[A]_0 - [B]_0} \ln\left( \frac{[B]_0 [A]_t}{[A]_0 [B]_t} \right) = kt$ Plotting $\ln\left(\frac{[B]_0 [A]_t}{[A]_0 [B]_t}\right)$ versus $t$ yields a straight line with slope $k([A]_0 - [B]_0)$.

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4. Comprehensive Comparison of Integrated Rate Laws

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5. Experimental Methods of Determining Reaction Order

1. Initial Rates Method (Method of Initial Rates)

In this classic quantitative method, multiple kinetic trials are executed with systematically varied initial reactant concentrations. Because the rates are evaluated at $t = 0$, reaction products have not yet accumulated, eliminating reverse reactions.

$\frac{v_{0, 1}}{v_{0, 2}} = \frac{k [A]_{0, 1}^m [B]_{0, 1}^n}{k [A]_{0, 2}^m [B]_{0, 2}^n} = \left(\frac{[A]_{0, 1}}{[A]_{0, 2}}\right)^m \quad \text{(holding } [B]_0 \text{ constant)}$ Taking natural logarithms: $m = \frac{\ln(v_{0, 1} / v_{0, 2})}{\ln([A]_{0, 1} / [A]_{0, 2})}$

Worked Step-by-Step Problem: Initial Rates Analysis

Experimental Data: Determine the rate law and rate constant for $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$ at $298\text{ K}$:

• Run 1: $[\text{NO}]_0 = 0.0100\text{ M}$, $[\text{O}_2]_0 = 0.0100\text{ M}$, $v_0 = 2.50 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}$
• Run 2: $[\text{NO}]_0 = 0.0200\text{ M}$, $[\text{O}_2]_0 = 0.0100\text{ M}$, $v_0 = 1.00 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}$
• Run 3: $[\text{NO}]_0 = 0.0100\text{ M}$, $[\text{O}_2]_0 = 0.0200\text{ M}$, $v_0 = 5.00 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}$

Step 1: Determine order with respect to $\text{NO}$ (using Runs 1 and 2): $\frac{v_{0, 2}}{v_{0, 1}} = \frac{1.00 \times 10^{-4}}{2.50 \times 10^{-5}} = 4.00 = \left(\frac{0.0200}{0.0100}\right)^m = (2.00)^m \implies 2^m = 4 \implies m = 2$

Step 2: Determine order with respect to $\text{O}_2$ (using Runs 1 and 3): $\frac{v_{0, 3}}{v_{0, 1}} = \frac{5.00 \times 10^{-5}}{2.50 \times 10^{-5}} = 2.00 = \left(\frac{0.0200}{0.0100}\right)^n = (2.00)^n \implies 2^n = 2 \implies n = 1$

Step 3: State rate law and overall order: $\text{Rate} = k [\text{NO}]^2 [\text{O}_2], \quad n_{tot} = 2 + 1 = 3 \quad \text{(Third-order overall)}$

Step 4: Compute rate constant $k$ with units using Run 1: $k = \frac{v_0}{[\text{NO}]_0^2 [\text{O}_2]_0} = \frac{2.50 \times 10^{-5}\text{ M}\cdot\text{s}^{-1}}{(0.0100\text{ M})^2 (0.0100\text{ M})} = \frac{2.50 \times 10^{-5}}{1.00 \times 10^{-6}\text{ M}^3} = 25.0\text{ M}^{-2}\cdot\text{s}^{-1}$

2. Graphical Method

Experimental concentration-versus-time data are transformed into three parallel graphs:

1. $[A]_t$ vs. $t$
2. $\ln[A]_t$ vs. $t$
3. $1/[A]_t$ vs. $t$ The plot exhibiting strict linearity (highest correlation coefficient $R^2$) identifies the true order ($0, 1,$ or $2$), and the slope gives the numerical magnitude of $k$.

3. Half-Life Dependence Method

For any reaction of arbitrary overall order $n$ (where $n \ne 1$): $t_{1/2} \propto \frac{1}{[A]_0^{n - 1}} = [A]_0^{1 - n}$ Comparing two distinct initial concentrations $[A]_{0, 1}$ and $[A]_{0, 2}$ with measured half-lives $t_{1/2, 1}$ and $t_{1/2, 2}$: $\frac{t_{1/2, 1}}{t_{1/2, 2}} = \left( \frac{[A]_{0, 1}}{[A]_{0, 2}} \right)^{1 - n} \implies 1 - n = \frac{\ln(t_{1/2, 1} / t_{1/2, 2})}{\ln([A]_{0, 1} / [A]_{0, 2})} \implies n = 1 - \frac{\ln(t_{1/2, 1} / t_{1/2, 2})}{\ln([A]_{0, 1} / [A]_{0, 2})}$

4. Isolation Method (Flood Method) and Pseudo-Order Kinetics

When a reaction involves multiple reactants (e.g., $v = k [A]^m [B]^n [C]^p$), measuring individual orders simultaneously is challenging. In the isolation method, reactants $B$ and $C$ are added in vast stoichiometric excess relative to $A$ ($[B]_0, [C]_0 \gg [A]_0$, typically $50\times$ to $100\times$ excess). As reactant $A$ is consumed completely, the concentrations of $B$ and $C$ change by an imperceptible percentage and remain effectively constant:

$v = k [A]^m [B]_0^n [C]_0^p = k_{obs} [A]^m \quad \text{where } k_{obs} = k [B]_0^n [C]_0^p$

The reaction behaves as a pseudo-$m$-order process. For instance, acid-catalyzed ester hydrolysis in dilute aqueous solution involves water as both solvent and reactant ($[\text{H}_2\text{O}] \approx 55.5\text{ M}$). Because $[\text{H}_2\text{O}]$ is effectively invariant, the second-order process $v = k [\text{ester}][\text{H}_2\text{O}]$ collapses to pseudo-first-order kinetics with $v = k_{obs}[\text{ester}]$, where $k_{obs} = k[\text{H}_2\text{O}]$.

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6. Radioactive Decay Kinetics and Radiocarbon Dating

Nuclear transmutation and radioactive decay processes are pure first-order kinetic phenomena occurring at the atomic nucleus, completely independent of chemical environment, pressure, or temperature.

Governing Equations

Let $N(t)$ be the number of undecayed radioactive nuclei at time $t$:

$-\frac{dN}{dt} = \lambda N \implies N(t) = N_0 e^{-\lambda t}$ $\ln\left(\frac{N(t)}{N_0}\right) = -\lambda t$

where $\lambda$ is the decay constant (equivalent to first-order $k$), related to half-life by: $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{t_{1/2}}$

Activity ($A$)

The activity is the disintegration rate: $A(t) = -\frac{dN}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$

• Units: Becquerel ($1\text{ Bq} = 1\text{ disintegration per second}$) and Curie ($1\text{ Ci} = 3.700 \times 10^{10}\text{ Bq}$). Specific activity is typically expressed in disintegrations per minute per gram of total carbon ($\text{dpm/g C}$).

Radiocarbon ($^{14}\text{C}$) Dating Principles

• Atmospheric cosmic-ray neutrons react with atmospheric $^{14}\text{N}$: $^1_0 n + {^{14}_7\text{N}} \longrightarrow {^{14}_6\text{C}} + {^1_1 p}$
• Radioactive $^{14}\text{C}$ ($t_{1/2} = 5730\text{ years}$) oxidizes to $^{14}\text{CO}_2$ and incorporates into the biosphere via photosynthesis and the food chain, establishing a steady-state specific activity in living organisms of approximately $A_0 = 15.30\text{ dpm/g C}$.
• Upon death, carbon exchange ceases. The $^{14}\text{C}$ decays via beta emission (${^{14}_6\text{C}} \rightarrow {^{14}_7\text{N}} + \beta^-$), causing the residual specific activity to decline exponentially: $t = -\frac{1}{\lambda} \ln\left(\frac{A(t)}{A_0}\right) = -\frac{t_{1/2}}{\ln 2} \ln\left(\frac{A(t)}{A_0}\right) = \frac{t_{1/2}}{0.69315} \ln\left(\frac{A_0}{A(t)}\right)$

Worked Step-by-Step Calculation: Archaeological Artifact Dating

Scenario: A buried wood fragment recovered from a lake-sediment core has a measured $^{14}\text{C}$ specific activity of $3.825\text{ dpm/g C}$. If fresh wood has a specific activity of $15.30\text{ dpm/g C}$ and $t_{1/2} = 5730\text{ years}$, calculate the archaeological age of the wood.

Step 1: Compute fraction of remaining activity: $\frac{A(t)}{A_0} = \frac{3.825\text{ dpm/g C}}{15.30\text{ dpm/g C}} = 0.2500$ Notice that $0.2500 = (1/2)^2$, which corresponds to exactly two elapsed half-lives.

Step 2: Calculate age rigorously using the exponential decay equation: $\lambda = \frac{0.69315}{5730\text{ yr}} = 1.2097 \times 10^{-4}\text{ yr}^{-1}$ $t = \frac{\ln(A_0 / A(t))}{\lambda} = \frac{\ln(15.30 / 3.825)}{1.2097 \times 10^{-4}\text{ yr}^{-1}} = \frac{\ln(4.000)}{1.2097 \times 10^{-4}} = \frac{1.38629}{1.2097 \times 10^{-4}\text{ yr}^{-1}} = 11460\text{ years}$ Conclusion: The wood died approximately $11460\text{ years}$ ago.