Chemical kinetics is the branch of chemistry that studies how fast reactions occur and what controls their speed. Thermodynamics tells us whether a reaction can happen; kinetics tells us how quickly it actually does. Rusting of iron is thermodynamically favourable yet painfully slow, while the neutralisation of an acid by a base is over in an instant.
Rate of a reaction
The rate of a reaction is the change in concentration of a reactant or product per unit time. For a general reaction $aA + bB \rightarrow cC + dD$, every species changes at a rate tied to its stoichiometric coefficient, so the single rate of reaction is written as
$$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$$
The minus signs for reactants show that their concentration falls with time. The usual unit is $\text{mol L}^{-1}\text{s}^{-1}$.
Average and instantaneous rate
The average rate over a time interval is $r_{avg} = -\frac{\Delta[R]}{\Delta t}$, the change in concentration divided by the elapsed time. As we shrink that interval to an instant ($\Delta t \rightarrow 0$) we obtain the instantaneous rate $r = -\frac{d[R]}{dt}$, the slope of the tangent to the concentration-time curve at that moment. Because reactions slow down as reactants are used up, the instantaneous rate is largest at the start and falls towards zero as the reaction approaches completion.
Factors affecting the rate
- Concentration: higher reactant concentration usually means a faster rate, since collisions are more frequent.
- Temperature: raising the temperature sharply increases the rate (roughly doubling for every $10\ ^\circ\text{C}$ rise).
- Catalyst: provides an alternative path of lower activation energy, speeding the reaction without being consumed.
- Surface area of a solid reactant and the nature of the reactants also matter; powdered solids react faster than lumps.
Rate law and rate constant
The rate law (rate expression) relates rate to concentration: for $aA + bB \rightarrow$ products, $\text{Rate} = k[A]^x[B]^y$. The exponents $x$ and $y$ are found experimentally and need not equal the coefficients $a$ and $b$. The proportionality constant $k$ is the rate constant (specific reaction rate): it equals the rate when all concentrations are unity and depends on temperature and the catalyst, but not on concentration.
Order and molecularity
The order of a reaction is the sum of the concentration exponents in the rate law, $x + y$. It can be $0, 1, 2$, fractional or even negative, and is determined experimentally. Molecularity is the number of reacting species (atoms, ions or molecules) that collide simultaneously in an elementary step; it is a theoretical, positive whole number ($1, 2, 3$). For an elementary reaction (one step) order equals molecularity. A complex reaction proceeds through several elementary steps; its overall rate is governed by the slowest step (the rate-determining step), so order is decided by that step and can differ from the apparent molecularity.
Units of the rate constant
From $\text{Rate} = k[\text{conc}]^n$, $k = \dfrac{\text{Rate}}{[\text{conc}]^n}$ has units $(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$. So a zero-order $k$ is in $\text{mol L}^{-1}\text{s}^{-1}$, a first-order $k$ in $\text{s}^{-1}$, and a second-order $k$ in $\text{L mol}^{-1}\text{s}^{-1}$.
For the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$, the rate of disappearance of $N_2O_5$ is $6.25 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$. Find the rate of the reaction and the rate of formation of $NO_2$ and $O_2$.
Solution- Rate of reaction $= -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{2}(6.25\times10^{-3}) = 3.125\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$.
- Rate of formation of $NO_2 = 4 \times \text{Rate} = 4 \times 3.125\times10^{-3} = 1.25\times10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$.
- Rate of formation of $O_2 = 1 \times \text{Rate} = 3.125\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$.
Answer: Rate $= 3.125\times10^{-3}$; $\frac{d[NO_2]}{dt} = 1.25\times10^{-2}$; $\frac{d[O_2]}{dt} = 3.125\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$.
A reaction is second order in $A$ and first order in $B$. Write the rate law, state the overall order, and find how the rate changes when (i) $[A]$ is doubled, (ii) both $[A]$ and $[B]$ are doubled.
Solution- Rate law: $\text{Rate} = k[A]^2[B]$. Overall order $= 2 + 1 = 3$.
- (i) Doubling $[A]$: rate $\propto (2)^2 = 4$, so the rate becomes $4$ times.
- (ii) Doubling both: rate $\propto (2)^2(2)^1 = 8$, so the rate becomes $8$ times.
Answer: $\text{Rate}=k[A]^2[B]$, third order; rate increases $4\times$ in (i) and $8\times$ in (ii).
For a reaction the rate constant is $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$. What is the order of the reaction, and what is the unit of $k$ for a zero-order and a second-order reaction?
Solution- The unit $\text{s}^{-1}$ corresponds to $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$ with $1-n = 0$, i.e. $n = 1$: a first-order reaction.
- Zero order: $k$ has units $(\text{mol L}^{-1})^{1}\text{s}^{-1} = \text{mol L}^{-1}\text{s}^{-1}$.
- Second order: $k$ has units $(\text{mol L}^{-1})^{-1}\text{s}^{-1} = \text{L mol}^{-1}\text{s}^{-1}$.
Answer: First order; zero-order $k$ in $\text{mol L}^{-1}\text{s}^{-1}$, second-order $k$ in $\text{L mol}^{-1}\text{s}^{-1}$.
The rate law for $2NO + O_2 \rightarrow 2NO_2$ is found to be $\text{Rate} = k[NO]^2[O_2]$. State the order with respect to each reactant and overall, and the units of $k$.
Solution- Order with respect to $NO = 2$; with respect to $O_2 = 1$.
- Overall order $= 2 + 1 = 3$.
- Units of $k$: $(\text{mol L}^{-1})^{1-3}\text{s}^{-1} = (\text{mol L}^{-1})^{-2}\text{s}^{-1} = \text{L}^2\text{mol}^{-2}\text{s}^{-1}$.
Answer: Second order in $NO$, first in $O_2$, third overall; $k$ in $\text{L}^2\text{mol}^{-2}\text{s}^{-1}$.
Initial-rate data for $A + B \rightarrow$ products: when $[A]$ is doubled (with $[B]$ fixed) the rate doubles; when $[B]$ is doubled (with $[A]$ fixed) the rate is unchanged. Determine the rate law and the overall order.
Solution- Doubling $[A]$ doubles rate: rate $\propto [A]^1$, so order in $A$ is $1$.
- Doubling $[B]$ leaves rate unchanged: rate $\propto [B]^0$, so order in $B$ is $0$.
- Rate law: $\text{Rate} = k[A]^1[B]^0 = k[A]$. Overall order $= 1 + 0 = 1$.
Answer: $\text{Rate} = k[A]$; first order overall (zero order in $B$).
The decomposition of $NH_3$ on a hot platinum surface follows $\text{Rate} = k[NH_3]^0$. Identify the order, explain why it is zero order, and give the units of $k$.
Solution- The exponent of $[NH_3]$ is $0$, so the reaction is zero order; rate $= k$, independent of $[NH_3]$.
- Reason: the platinum surface is fully covered (saturated) with $NH_3$, so adding more $NH_3$ cannot increase the number of molecules reacting on the surface.
- Units of $k$: $(\text{mol L}^{-1})^{1-0}\text{s}^{-1} = \text{mol L}^{-1}\text{s}^{-1}$.
Answer: Zero order; rate is surface-limited; $k$ in $\text{mol L}^{-1}\text{s}^{-1}$.