Chemical Kinetics

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 3.

Why does a diamond last forever when it "should" turn into graphite?

Thermodynamics says that at room temperature graphite is the more stable form of carbon, so a diamond ring is, in principle, on its way to becoming pencil lead. Nobody has ever seen it happen. Meanwhile, milk left on a warm kitchen counter turns sour within hours, and a pressure cooker cooks lentils in a fraction of the usual time.

The difference is speed. Thermodynamics tells you whether a reaction can happen. Chemical kinetics tells you how fast it happens, what controls the speed, and how the molecules actually get from reactants to products. The whole chapter is about three questions: how do we measure speed, how does speed depend on concentration, and how does it depend on temperature and catalysts.

Rate of a reaction

The rate of reaction is the change in the concentration of a reactant or a product per unit time. A reactant is used up, so its concentration falls, and we put a minus sign in front to keep the rate positive:

$$\text{rate} = -\dfrac{\Delta[R]}{\Delta t} = +\dfrac{\Delta[P]}{\Delta t}$$

The unit is mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹). For gases, when concentration is replaced by partial pressure, the unit becomes atm s⁻¹.

When coefficients are not 1. Different species change at different speeds, so each rate of change is divided by its coefficient, which gives one number for the reaction as a whole. For 2HI → H₂ + I₂:

$$\text{rate} = -\dfrac{1}{2}\dfrac{\Delta[\text{HI}]}{\Delta t} = \dfrac{\Delta[\text{H}_2]}{\Delta t} = \dfrac{\Delta[\text{I}_2]}{\Delta t}$$

Worked example: rate and stoichiometry. For 2N₂O₅ → 4NO₂ + O₂, the rate of formation of NO₂ is $4.0 \times 10^{-3}$ mol L⁻¹ s⁻¹. Find the rate of reaction, the rate of disappearance of N₂O₅, and the rate of formation of O₂.

The common wrong step: quoting the rate of NO₂ formation as "the rate of reaction". Always divide by the coefficient to get the single rate of the reaction.

Rate law, order and the rate constant

What makes a reaction fast? Mostly concentration, temperature and the presence of a catalyst. The first is expressed by the rate law, an equation worked out from experiment that links the rate to the concentrations of reactants, each raised to a power:

$$\text{rate} = k\,[A]^x\,[B]^y$$

Here $k$ is the rate constant. It does not depend on concentration, but it does depend on temperature and on the catalyst. The order of the reaction is the sum of the powers, $x + y$. The powers may or may not match the coefficients in the balanced equation, because the rate law comes from experiment, never from the equation. Order can be 0, 1, 2, 3 or a fraction.

The unit of $k$ depends on the order, because $k = \text{rate}/[\text{conc}]^n$ and the rate always has units of mol L⁻¹ s⁻¹.

For order $n$, the unit of $k$ is $(\text{mol L}^{-1})^{1-n}\ \text{s}^{-1}$. A favourite CUET question hands you the unit and asks for the order.

Worked example: finding the rate law from data. For A + B → products, the initial rates are measured.

The common wrong step: reading the order from the balanced equation. Here the equation has one A and one B, yet the rate depends on [A] squared.

Molecularity versus order. Molecularity is the number of reacting species that collide at the same moment in a single step (an elementary reaction). It is 1 (unimolecular), 2 (bimolecular) or, very rarely, 3 (termolecular), because a collision of four particles at once is almost impossible.

Most reactions are complex: they happen in several steps (a mechanism), and the slowest step is the rate determining step, which controls the overall rate. The overall rate law follows that slowest step. A classic case is the decomposition of hydrogen peroxide with iodide ions:

Notice that IO⁻ is made in step 1 and used up in step 2, so it does not appear in the overall equation. Iodide is regenerated, so it is a catalyst. The rate law uses only step 1 (the slow one), so the order is 2 even though the overall equation shows two H₂O₂.

Pause and predict: for a zero order reaction, what happens to the rate if you double the concentration of the reactant?

Check your answer: nothing. Rate $= k$, a constant, so concentration does not matter. This happens when something else limits the speed, such as the surface of a catalyst already saturated with reactant molecules.

Beyond the textbook. For the exam: NH₃ decomposing on hot platinum is zero order because the metal surface is saturated, so the rate does not depend on the NH₃ concentration. In reality: the same saturation idea explains enzymes. When there is plenty of substrate, every enzyme molecule is busy, and adding more substrate does not speed things up, so enzyme reactions behave as zero order at high substrate concentration. Why it matters: the "odd" zero order idea turns out to be how living cells work.

Integrated rate equations: how concentration changes with time

The rate law gives speed at one moment. Integrating it gives the concentration at any time.

Zero order. The rate is constant, so the concentration falls by the same amount every second, a straight line down:

$$[R] = [R]0 - kt \qquad t{1/2} = \dfrac{[R]_0}{2k}$$

First order. The rate is proportional to the amount left, so as the reactant is used up the reaction slows, giving an exponential decay:

$$\ln\dfrac{[R]_0}{[R]} = kt \quad\text{or}\quad k = \dfrac{2.303}{t}\log\dfrac{[R]0}{[R]} \qquad t{1/2} = \dfrac{0.693}{k}$$

Why first order half life is constant. Each half life removes half of whatever is left, so the time to go from 100% to 50% equals the time to go from 50% to 25%. After $n$ half lives, a fraction $(1/2)^n$ remains: 75% complete is 2 half lives, 87.5% is 3, 93.75% is 4. For zero order, a bigger starting amount takes longer to halve, because the same amount is removed each second.

Worked example: first order. A first order reaction is 75% complete in 40 minutes. Find $k$, the half life, and the time for 99% completion.

The common wrong step: using $[R]_0/[R] = 75$ for "75% complete". The ratio compares starting amount to the amount that is left, which is 25%.

Worked example: zero order. A zero order reaction starts at 0.80 mol L⁻¹ and $k = 2.0 \times 10^{-2}$ mol L⁻¹ s⁻¹. Find the half life and the time for completion.

The common wrong step: applying the first order idea that "the reaction never quite finishes". A zero order reaction finishes in a finite time, which is exactly twice its half life.

Gas phase first order. For A(g) → B(g) + C(g) at constant volume, you cannot measure concentration easily but you can measure total pressure. If $p_i$ is the starting pressure of A and $x$ is the pressure of A that has reacted, then $p_A = p_i - x$ and $p_t = (p_i - x) + x + x = p_i + x$. Hence $p_A = 2p_i - p_t$ and

$$k = \dfrac{2.303}{t}\log\dfrac{p_i}{2p_i - p_t}$$

Worked example: pressure data. A(g) → B(g) + C(g) starts at 0.50 atm. After 100 s the total pressure is 0.70 atm. Find $k$.

The common wrong step: using $p_t$ in place of $2p_i - p_t$. The total pressure includes the products, not just the reactant.

Pseudo first order reactions. Some reactions are second order but look first order because one reactant is in huge excess and its concentration hardly changes. In the acid catalysed hydrolysis of an ester, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, water is the solvent. The rate $= k[\text{ester}][\text{H}_2\text{O}]$ becomes $k'[\text{ester}]$, where $k' = k[\text{H}_2\text{O}]$ is constant. Inversion of cane sugar, C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose), behaves the same way.

Beyond the textbook. For the exam: radioactive decay is first order, and its half life does not depend on how much material you start with. In reality: this is why carbon dating works. Carbon-14 has a half life of about 5730 years, so after one half life half of it is left, after two a quarter, and the fraction remaining gives the age of a sample, whatever its size. Why it matters: a constant half life is what makes a quantity usable as a clock.

Pause and predict: you double the starting concentration of a reactant. What happens to the half life of a zero order reaction and a first order reaction?

Check your answer: for zero order, $t_{1/2} = [R]0/2k$ doubles. For first order, $t{1/2} = 0.693/k$ does not change.

Temperature and the Arrhenius equation

For most reactions, the rate (and $k$) roughly doubles for every 10 K rise in temperature, and sometimes triples. Heating does not make molecules "better". It changes how many have enough energy.

The energy barrier. Reactants must pass through a high-energy, unstable arrangement called the activated complex before turning into products. The activation energy, $E_a$, is the extra energy the reactants need to climb to that peak. Not every collision supplies it. At any temperature molecules have a spread of energies (the Maxwell–Boltzmann distribution): a few very slow, a few very fast, and most near a most probable energy. Raising the temperature shifts the peak to a higher energy and flattens and broadens the curve, so a much larger area lies beyond $E_a$. That fraction of molecules with $E \geq E_a$ roughly doubles for a 10 K rise, so the rate does too.

Arrhenius put this into one equation:

$$k = A\,e^{-E_a/RT}$$

where $A$ is the frequency (pre-exponential) factor, related to how often molecules collide in the right orientation, $R$ is the gas constant and $T$ is the temperature in kelvin. A larger $A$ or a smaller $E_a$ gives a larger $k$. In logarithmic form:

$$\ln k = \ln A - \dfrac{E_a}{RT} \qquad \log k = \log A - \dfrac{E_a}{2.303\,RT}$$

So a graph of $\ln k$ against $1/T$ is a straight line with slope $-E_a/R$ and intercept $\ln A$. Comparing two temperatures, the $A$ cancels:

$$\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{T_2 - T_1}{T_1 T_2}\right)$$

Worked example: activation energy. The rate of a reaction doubles when the temperature rises from 300 K to 310 K. Find $E_a$. ($R = 8.314$ J K⁻¹ mol⁻¹)

The common wrong step: using temperatures in °C inside the product $T_1T_2$, or quoting the answer in joules while the question wants kJ. Use kelvin throughout.

Beyond the textbook. For the exam: the rate doubles (or triples) for every 10 K rise. In reality: "doubles" is not a law. The factor depends on $E_a$ and on the temperature. A reaction with $E_a$ of about 54 kJ mol⁻¹ doubles near room temperature, but with $E_a$ = 100 kJ mol⁻¹ the same 10 K rise (300 K to 310 K) raises the rate about 3.6 times. Why it matters: a reaction with a high barrier is far more sensitive to temperature, which is why refrigeration slows spoilage so effectively.

Catalysts. A catalyst speeds up a reaction without being used up, by giving the reactants an alternative route with a lower activation energy. On the energy profile, the peak is lower.

A catalyst cannot change $\Delta G$, so it cannot make a reaction possible that was not already possible. It only changes the speed.

Collision theory: what the Arrhenius factors mean

Collision theory pictures molecules as hard spheres that react when they collide. The number of collisions per second per unit volume is the collision frequency, $Z$. But only a small fraction of collisions lead to products, for two reasons. The colliding molecules need enough energy, equal to or greater than the threshold energy, and they need the right orientation. A collision with the wrong orientation simply bounces apart. The orientation requirement is the steric (probability) factor, $P$.

$$\text{rate} = P\,Z_{AB}\,e^{-E_a/RT}$$

Threshold energy = activation energy + the average energy the reacting molecules already have. Do not treat the two as equal. Collisions that combine sufficient energy with correct orientation are called effective collisions.

Pause and predict: two molecules collide with more than enough energy but the wrong orientation. Is it an effective collision?

Check your answer: no. Energy alone is not enough. The molecules must also meet in the way that lets the right bonds break and form, which is what the factor $P$ takes into account.

Formula sheet

Exam traps

60-second revision

Where this chapter connects

• Before this: Electrochemistry tells you which redox reactions are possible, and kinetics tells you how fast they run. Thermodynamics and Equilibrium give $\Delta G$ and $K$, which a catalyst leaves untouched.
• Next: The d and f Block Elements explains why transition metals and their ions are such good catalysts.
• Forward link: Differential Equations is the maths behind integrated rate laws, since rate $= -d[R]/dt$ is a differential equation whose solution is the exponential decay you used here.