The rate of a reaction is the change in concentration of a reactant or product per unit time. It can be expressed as average rate (over a finite time interval) or instantaneous rate (at a particular moment, found as the slope of the concentration-vs-time curve at that point).
A negative sign is used for reactants (since their concentration decreases) to keep the rate a positive quantity.
For $aA + bB \rightarrow cC + dD$:
$$\text{Rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = -\dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt}$$Dividing by the coefficient ensures the rate is the same value no matter which species is used to measure it.
For $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$, if $\dfrac{d[\text{O}_2]}{dt} = 2.5 \times 10^{-3}\ \text{M/s}$, find the rate of the reaction and the rate of disappearance of NβOβ .
Rate $= \dfrac{1}{1}\dfrac{d[\text{O}_2]}{dt} = 2.5 \times 10^{-3}\ \text{M/s}$
Rate of disappearance of NβOβ $= 2 \times \text{Rate} = 2 \times 2.5\times10^{-3} = 5.0 \times 10^{-3}\ \text{M/s}$
The rate law expresses the rate of a reaction as a function of the molar concentrations of reactants, each raised to some power (determined experimentally, not from the stoichiometric coefficients). For $aA + bB \rightarrow \text{Products}$: $\text{Rate} = k[A]^x[B]^y$, where $x$ and $y$ are the orders with respect to A and B, found only through experiment.
The overall order of a reaction is the sum of the powers of concentration terms in the rate law ($x + y$). Order can be zero, a positive integer, a negative number, or even a fraction β unlike molecularity, which must be a positive integer.
The rate law cannot be predicted just by looking at the balanced chemical equation β it must be determined experimentally, since it depends on the actual reaction mechanism, which may proceed through several elementary steps.
The rate law for a reaction $2A + B \rightarrow C$ is found experimentally to be $\text{Rate} = k[A][B]^2$. What is the order with respect to A, with respect to B, and the overall order?
Order w.r.t. A $= 1$; order w.r.t. B $= 2$; overall order $= 1 + 2 = 3$ (third order overall) β note this does not match the stoichiometric coefficients (2 and 1), confirming rate laws must come from experiment.
| Molecularity | Order of Reaction |
|---|---|
| Number of reacting species (atoms/ions/molecules) colliding simultaneously in an elementary step | Sum of powers of concentration terms in the experimentally determined rate law |
| Always a whole number (1, 2, or rarely 3) | Can be zero, a fraction, or a whole number |
| Theoretical concept, applies only to a single elementary step/reaction | Experimental quantity, applies to the overall reaction (which may have multiple steps) |
| Has meaning only for elementary reactions, not for the overall complex reaction | Always has a definite experimental value for the overall reaction |
Rate is independent of reactant concentration: $\text{Rate} = k[A]^0 = k$. Such reactions are common when the rate is limited by something other than reactant concentration β e.g., a fixed amount of catalyst surface or fixed light intensity in photochemical reactions.
A plot of $[A]$ vs $t$ gives a straight line with slope $-k$ and y-intercept $[A]_0$.
A zero-order reaction has $k = 0.05\ \text{mol/L/s}$ and $[A]_0 = 1.0\ \text{M}$. Calculate [A] after 10 seconds.
$[A] = [A]_0 - kt = 1.0 - (0.05)(10) = 1.0 - 0.5 = 0.5\ \text{M}$
Rate is directly proportional to the concentration of one reactant: $\text{Rate} = k[A]$. Radioactive decay and many decomposition reactions follow first-order kinetics.
Equivalently: $\ln[A] = \ln[A]_0 - kt$. A plot of $\log[A]$ vs $t$ gives a straight line with slope $-k/2.303$.
A first-order reaction has $[A]_0 = 0.8\ \text{M}$, and after 20 minutes, $[A] = 0.2\ \text{M}$. Calculate the rate constant.
$k = \dfrac{2.303}{20}\log\dfrac{0.8}{0.2} = \dfrac{2.303}{20}\log(4) = \dfrac{2.303}{20}(0.602) = 0.0693\ \text{min}^{-1}$
The half-life ($t_{1/2}$) is the time required for the concentration of a reactant to fall to half its initial value.
The key distinguishing feature: for a first-order reaction, $t_{1/2}$ is independent of initial concentration β it's always the same value, a hallmark used to identify first-order kinetics experimentally.
A first-order reaction has a rate constant of $2.31 \times 10^{-3}\ \text{s}^{-1}$. Calculate its half-life.
$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{2.31\times10^{-3}} = 300\ \text{s}$
A pseudo first-order reaction is actually higher order (e.g., second order) but behaves like a first-order reaction under specific conditions β typically when one reactant is present in a huge excess, so its concentration remains essentially constant throughout the reaction.
Classic example: acid hydrolysis of an ester, $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$. Water is the solvent and present in huge excess, so its concentration barely changes β the rate law simplifies from $\text{Rate} = k'[\text{Ester}][\text{H}_2\text{O}]$ to $\text{Rate} = k[\text{Ester}]$, where $k = k'[\text{H}_2\text{O}]$ is treated as constant.
For most reactions, the rate constant roughly doubles or triples for every 10Β°C rise in temperature β a much steeper increase than can be explained just by increased molecular collision frequency, and is instead explained by the sharp increase in the fraction of molecules with energy exceeding the activation energy.
where $A$ = frequency (pre-exponential) factor, $E_a$ = activation energy, $R$ = gas constant, $T$ = absolute temperature.
Used to calculate activation energy from rate constants measured at two different temperatures.
The minimum extra energy that colliding reactant molecules must possess (over and above their average energy) to successfully form an activated complex/transition state and proceed to products. A lower Ea means a faster reaction at a given temperature.
The rate constant of a reaction is $2 \times 10^{-3}\ \text{s}^{-1}$ at 300 K and $8 \times 10^{-3}\ \text{s}^{-1}$ at 320 K. Calculate the activation energy. ($R = 8.314\ \text{J/mol K}$)
$\log\dfrac{k_2}{k_1} = \log\dfrac{8\times10^{-3}}{2\times10^{-3}} = \log(4) = 0.602$
$0.602 = \dfrac{E_a}{2.303 \times 8.314}\left(\dfrac{1}{300} - \dfrac{1}{320}\right) = \dfrac{E_a}{19.15}(2.08\times10^{-4})$
$E_a = \dfrac{0.602 \times 19.15}{2.08\times10^{-4}} \approx 55{,}440\ \text{J/mol} \approx 55.4\ \text{kJ/mol}$
A catalyst works by lowering $E_a$ (providing an alternate pathway), which β according to the Arrhenius equation β increases $k$ exponentially even at the same temperature, explaining why even small reductions in activation energy can produce large increases in rate.
Collision theory explains reaction rates in terms of molecules colliding with each other. For a reaction to occur, colliding molecules must have (i) energy equal to or greater than the activation energy, and (ii) correct orientation at the moment of collision β both conditions must be met simultaneously.
where $Z_{AB}$ = collision frequency, and $P$ = the steric (probability/orientation) factor, accounting for the fraction of collisions with proper molecular orientation.
Simply having enough energy isn't sufficient β molecules must also collide with the correct geometry for bonds to break and form properly. This is why the steric factor P is often much less than 1 for reactions involving complex molecules, where only a small fraction of orientations are effective.