Chemical Kinetics

Chemical kinetics builds in a strict logical sequence, and this article follows the NCERT chapter's own order.

• Rate of a chemical reaction — average and instantaneous rate, with correct units and the stoichiometric factor.
• Factors influencing rate — concentration, temperature and catalyst, leading to the rate law.
• Rate law, rate constant and order — the experimental relationship between rate and concentration.
• Molecularity, elementary and complex reactions — colliding species, the rate-determining step, and how molecularity differs from order.
• Integrated rate equations — concentration-time equations for zero and first order, including gas-phase pressure forms.
• Half-life — the characteristic time for concentration to fall by half, for zero and first order kinetics.
• Pseudo first order reactions — how excess of one reactant makes a higher-order reaction look first order.
• Temperature dependence — the Arrhenius equation, activation energy, and the Maxwell-Boltzmann distribution.
• Effect of a catalyst — an alternate, lower-energy pathway without altering thermodynamics.
• Collision theory — collision frequency, effective collisions, orientation, and the steric factor.

Average and instantaneous rate of a reaction

Speed, for a moving car, is distance covered per unit time. For a chemical reaction the equivalent idea is change in concentration of a reactant or product per unit time. For the simplest reaction $R \to P$ in a system of constant volume, with $[R]_1$ and $[R]_2$ the reactant concentrations at times $t_1$ and $t_2$, the average rate over that interval is defined using the decrease in reactant concentration or, equivalently, the increase in product concentration:

where: the minus sign in front of $\Delta[R]/\Delta t$ is essential — $\Delta[R]$ is negative because the reactant is being consumed, and multiplying by $-1$ makes the rate a positive quantity. Units follow directly: concentration in $\text{mol L}^{-1}$ and time in seconds give rate as $\text{mol L}^{-1}\text{s}^{-1}$; for gas-phase reactions monitored by partial pressure, units become $\text{atm s}^{-1}$. Average rate is only ever as good as the interval over which it is calculated, and a reaction's rate is not constant — it falls as reactants are used up. Shrinking the interval to zero gives the instantaneous rate:

where: this is the slope of the tangent to the concentration-time curve at the instant of interest. NCERT's hydrolysis of butyl chloride example shows the difference sharply: the average rate over successive $50$-second windows falls from $1.90\times10^{-4}$ to $0.40\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$, while the instantaneous rate at $t = 600\ \text{s}$, read from a tangent, gives a single sharp value of $5.12\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$. When stoichiometric coefficients differ from one, each species' rate of change must be divided by its own coefficient before they agree — in $2\text{HI}(g) \to \text{H}_2(g) + \text{I}_2(g)$, HI disappears twice as fast as $\text{H}_2$ or $\text{I}_2$ appears. The general definition for $aA + bB \to cC + dD$ is

where: dividing by the stoichiometric coefficient of each species is what makes "the rate of the reaction" a single unambiguous number, regardless of which species you happen to be monitoring in the lab.

Rate law, rate constant and order of a reaction

For the general reaction $aA + bB \to cC + dD$, experiments show that the rate depends on the concentration of one or more reactants, raised to powers that must be found by experiment. This relationship is the rate law:

where: $k$ is the rate constant, a proportionality factor characteristic of the reaction at a given temperature, and $x$, $y$ are the powers to which $[A]$ and $[B]$ must be raised — the order with respect to $A$ and $B$. The overall order is the sum $x + y$, and it can be zero, an integer, or a fraction; it is never predicted from the balanced equation, only measured. The classic demonstration is $2\text{NO}(g) + \text{O}_2(g) \to 2\text{NO}_2(g)$. Doubling $[\text{NO}]$ at fixed $[\text{O}_2]$ quadruples the rate ($0.096 \to 0.384\ \text{mol L}^{-1}\text{s}^{-1}$), showing dependence on the square of $[\text{NO}]$; doubling $[\text{O}_2]$ alone simply doubles the rate, showing first-order dependence on $[\text{O}_2]$. The rate law is $\text{Rate} = k[\text{NO}]^2[\text{O}_2]$, third order overall, and here the exponents happen to match the stoichiometric coefficients. That match is a coincidence, not a rule: for $\text{CHCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{HCl}$ the measured rate law is $\text{Rate} = k[\text{CHCl}_3][\text{Cl}_2]^{1/2}$, a fractional order no equation-reading could predict. The rate law must always come from experiment. The units of $k$ change with the overall order $n$, since both sides of the rate law must stay dimensionally consistent:

where: for zero order ($n=0$), $k$ carries units $\text{mol L}^{-1}\text{s}^{-1}$; for first order ($n=1$), $k$ is simply $\text{s}^{-1}$; for second order ($n=2$), $k$ carries $\text{mol}^{-1}\text{L s}^{-1}$. Reading units backward instantly identifies order — $k = 2.3\times10^{-5}\ \text{L mol}^{-1}\text{s}^{-1}$ can only be second order, $k = 3\times10^{-4}\ \text{s}^{-1}$ only first.

Molecularity, elementary and complex reactions

A balanced chemical equation almost never shows how a reaction actually happens at the molecular level, because most reactions proceed through a sequence of simpler steps rather than in one leap. A reaction occurring in a single step is called an elementary reaction; a sequence of elementary steps that together accomplish the overall transformation is a mechanism, and the overall reaction is then called a complex reaction.

Molecularity is defined only for a single elementary step: the number of reacting species (atoms, ions or molecules) that must collide simultaneously. Decomposition of ammonium nitrite, $\text{NH}_4\text{NO}_2 \to \text{N}_2 + 2\text{H}_2\text{O}$, is unimolecular; dissociation of hydrogen iodide, $2\text{HI} \to \text{H}_2 + \text{I}_2$, is bimolecular; and $2\text{NO} + \text{O}_2 \to 2\text{NO}_2$ is trimolecular. Simultaneous collisions of four or more molecules are so improbable that higher molecularities are essentially never observed — a reaction that looks like it needs more colliding species, such as $\text{KClO}_3 + 6\text{FeSO}_4 + 3\text{H}_2\text{SO}_4 \to \text{KCl} + 3\text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}$, must actually proceed through several simpler steps, even though this reaction is experimentally only second order overall. For a multi-step mechanism, the overall rate is set entirely by the slowest step, the rate-determining step (RDS) — exactly as a relay team's speed is set by its slowest runner. The iodide-catalysed decomposition of hydrogen peroxide, $2\text{H}_2\text{O}_2 \xrightarrow{\text{I}^-} 2\text{H}_2\text{O} + \text{O}_2$, is experimentally first order in both $\text{H}_2\text{O}_2$ and $\text{I}^-$, matching a two-step mechanism: $\text{H}_2\text{O}_2 + \text{I}^- \to \text{H}_2\text{O} + \text{IO}^-$ (slow, the RDS), then $\text{H}_2\text{O}_2 + \text{IO}^- \to \text{H}_2\text{O} + \text{I}^- + \text{O}_2$ (fast). $\text{IO}^-$ is an intermediate, formed and consumed but absent from the overall equation. Order is experimental, can be zero or fractional, and applies to elementary and complex reactions alike; molecularity is theoretical, never zero or fractional, and meaningful only for a single elementary step. For a complex reaction, overall order equals the molecularity of its rate-determining step.

Integrated rate equations: zero order reactions

The differential rate law is awkward to use directly, since it requires a tangent slope at every instant. Integrated rate equations relate concentration directly to elapsed time, so a single concentration measurement at any moment is enough to extract $k$. For a zero order reaction, $R \to P$ with $\text{Rate} = k[R]^0 = k$, integration (worked in full below) gives

where: $[R]_0$ is the initial concentration at $t=0$. This is a straight line of $[R]$ against $t$, slope $-k$, intercept $[R]_0$ — the diagnostic test for zero order kinetics. Zero order behaviour is uncommon, appearing under saturating conditions: decomposition of gaseous ammonia on a hot platinum surface, $2\text{NH}_3(g) \xrightarrow{\text{Pt}} \text{N}_2(g) + 3\text{H}_2(g)$, is zero order at high pressure because the catalyst surface is already saturated, so raising $[\text{NH}_3]$ further cannot speed the reaction up.

Integrated rate equations: first order reactions

For a first order reaction, $R \to P$ with $\text{Rate} = k[R]$, the same integration procedure (below, step by step) gives

where: plotting $\ln[R]$ (or $\log([R]_0/[R])$) against $t$ gives a straight line, slope $-k$ (or $k/2.303$) — the diagnostic test for first order kinetics. The equivalent exponential form is $[R] = [R]_0 e^{-kt}$, and between any two times $k = \dfrac{1}{t_2-t_1}\ln\dfrac{[R]_1}{[R]_2}$, without ever needing $[R]_0$. For a first order gas-phase reaction $A(g) \to B(g) + C(g)$ monitored by total pressure, if $p_i$ is the initial pressure and $x$ the drop in partial pressure of $A$ at time $t$, then $p_t = p_i + x$ (one mole each of $B, C$ appears per mole of $A$ lost), so $p_A = 2p_i - p_t$, giving

where: this is the concentration form of equation (7) in disguise, since partial pressure is proportional to concentration; it lets kinetics be followed by a manometer, convenient for decompositions such as $\text{N}_2\text{O}_5(g) \to 2\text{NO}_2(g) + \tfrac{1}{2}\text{O}_2(g)$. Hydrogenation of ethene and all radioactive decay also obey first order kinetics.

Half-life of a reaction

The half-life, $t_{1/2}$, is the time needed for a reactant's concentration to fall to exactly half its initial value. For a zero order reaction, substituting $[R] = [R]_0/2$ at $t = t_{1/2}$ into equation (6) gives

where: $t_{1/2}$ is directly proportional to starting concentration for a zero order reaction — halve $[R]_0$ and the half-life halves too. For a first order reaction, the same substitution into equation (7) gives the far more famous result

where: the half-life of a first order reaction is completely independent of initial concentration — it depends on $k$ alone. This is the fastest experimental test for first order behaviour: if repeated halvings always take the same time no matter where the clock starts, the reaction is first order.

Pseudo first order reactions

Some reactions that are genuinely second order (or higher) behave experimentally as first order because one reactant is present in such large excess that its concentration barely changes. Ethyl acetate hydrolysis, $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \to \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$, is truly second order, but with water as solvent in vast excess (say $10\ \text{mol}$ against $0.01\ \text{mol}$ ester), its concentration is essentially unchanged, and the rate law collapses to $\text{Rate} = k'[\text{CH}_3\text{COOC}_2\text{H}_5]$, apparently first order. Such reactions are pseudo first order reactions; acid-catalysed inversion of cane sugar into glucose and fructose is the other classic example, again first order only because water is in enormous excess.

Temperature dependence of the rate: the Arrhenius equation

Most reactions speed up sharply as temperature rises — a rule of thumb states that the rate constant roughly doubles for every $10\,^{\circ}\text{C}$ rise near room temperature. The precise relationship, first proposed by van't Hoff and given physical justification by the Swedish chemist Svante Arrhenius, is

where: $A$ is the Arrhenius factor or pre-exponential factor (related to collision frequency), $E_a$ is the activation energy in $\text{J mol}^{-1}$ — the minimum extra energy reactant molecules must acquire to react — $R$ is the gas constant, and $T$ is absolute temperature in kelvin. Reactant molecules must first form an unstable, high-energy activated complex (transition state) before breaking down into products; $E_a$ is the energy needed to reach that barrier's top, and $e^{-E_a/RT}$ is the fraction of molecules energetic enough to clear it.

Taking the natural log of equation (11) gives $\ln k = -E_a/RT + \ln A$, so a plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$ and intercept $\ln A$. Writing this at two temperatures and subtracting (full derivation below) gives the two-point form used constantly in numericals:

where: $k_1$ and $k_2$ are rate constants at absolute temperatures $T_1$ and $T_2$. This equation finds $E_a$ from any two measured rate constants, or predicts $k$ at a third temperature once $E_a$ is known. The reason temperature matters so much is statistical, not a property of any single molecule. Not every molecule carries the same kinetic energy; the fraction $N_E/N_T$ possessing energy $E$ follows a bell-shaped Maxwell-Boltzmann distribution. Raising temperature shifts the curve's peak higher and, crucially, broadens and flattens it, so a much larger fraction now lies above $E_a$ — the molecular reason the rate roughly doubles for a $10\ \text{K}$ rise: the shaded area beyond $E_a$ doubles.

Effect of a catalyst

A catalyst is a substance that speeds up a reaction without itself being permanently consumed, and must never be confused with simply adding more reactant. Manganese dioxide catalysing $2\text{KClO}_3 \xrightarrow{\text{MnO}_2} 2\text{KCl} + 3\text{O}_2$ is a standard board example. The accepted explanation is intermediate complex theory: the catalyst forms a temporary bond with a reactant, giving an intermediate complex that decomposes to release the product and regenerate the catalyst unchanged. The net effect, visible in the energy diagram above, is an alternate pathway with lower activation energy — and since equation (11) shows $k$ rises steeply as $E_a$ falls, even a modest drop in $E_a$ produces a large rate increase. Three catalyst facts are exam favourites: it does not alter $\Delta G$ and cannot make a non-spontaneous reaction spontaneous, it does not change the equilibrium constant (it accelerates forward and reverse reactions equally, so equilibrium is reached faster with unchanged composition), and a substance that decreases rate is not a catalyst at all but an inhibitor.

Collision theory of chemical reactions

Arrhenius's equation works well but does not explain why $A$ and $E_a$ take the values they do. Collision theory, developed by Max Trautz and William Lewis around 1916–1918 from the kinetic theory of gases, supplies that picture. Reactant molecules are treated as hard spheres, reacting only when they collide; the number of collisions per second per unit volume is the collision frequency, $Z$. For a bimolecular reaction $A + B \to \text{Products}$, collision theory gives

where: $Z_{AB}$ is the collision frequency of $A$ with $B$, and $e^{-E_a/RT}$ is the fraction of collisions energetic enough to exceed $E_a$ (comparing with equation (11), $A$ is essentially the collision frequency). This form works for atoms and small molecules but overestimates rate for larger ones, since not every energetic collision reacts — the molecules also need the correct orientation to let old bonds break and new bonds form. Collisions with both sufficient energy and correct orientation are effective collisions, and a steric factor $P$ captures the oriented fraction:

where: $P$ is typically less than one. Related is threshold energy = activation energy + energy already possessed by the reacting molecules — the total kinetic energy needed at collision, of which $E_a$ is only the extra amount above what molecules already carry. Collision theory has real limits, since treating molecules as featureless spheres ignores structure, but it bridges the empirical Arrhenius equation to a molecular picture of rate.

Board level - Rate law and order from initial-rate data (CBSE pattern)

For the reaction $2\text{NO}(g) + \text{O}_2(g) \to 2\text{NO}_2(g)$, the following initial rates were measured at constant temperature. *Experiment 1: $[\text{NO}] = 0.30\ \text{mol L}^{-1}$, $[\text{O}_2] = 0.30\ \text{mol L}^{-1}$, rate $= 0.096\ \text{mol L}^{-1}\text{s}^{-1}$. Experiment 2: $[\text{NO}] = 0.60\ \text{mol L}^{-1}$, $[\text{O}_2] = 0.30\ \text{mol L}^{-1}$, rate $= 0.384\ \text{mol L}^{-1}\text{s}^{-1}$. Experiment 3: $[\text{NO}] = 0.30\ \text{mol L}^{-1}$, $[\text{O}_2] = 0.60\ \text{mol L}^{-1}$, rate $= 0.192\ \text{mol L}^{-1}\text{s}^{-1}$. Find the order with respect to each reactant, the overall order, the rate law, and the rate constant with correct units.*

Step 1. Find the order with respect to NO by comparing Experiments 1 and 2, where $[\text{O}_2]$ is held fixed and $[\text{NO}]$ is doubled: the rate rises from $0.096$ to $0.384$, a factor of $0.384/0.096 = 4$. Since $2^x = 4$ gives $x = 2$, the reaction is second order in NO.

Step 2. Find the order with respect to $\text{O}_2$ by comparing Experiments 1 and 3, where $[\text{NO}]$ is held fixed and $[\text{O}_2]$ is doubled: the rate rises from $0.096$ to $0.192$, a factor of exactly $2$. Since $2^y = 2$ gives $y = 1$, the reaction is first order in $\text{O}_2$.

Step 3. Write the rate law and overall order: $\text{Rate} = k[\text{NO}]^2[\text{O}_2]^1$, so the overall order is $2 + 1 = 3$ — third order, even though the numerical exponents happen to match the stoichiometric coefficients here.

Step 4. Find $k$ using Experiment 1: $k = \dfrac{\text{Rate}}{[\text{NO}]^2[\text{O}_2]} = \dfrac{0.096}{(0.30)^2(0.30)} = \dfrac{0.096}{0.027} = 3.56\ \text{mol}^{-2}\text{L}^2\text{s}^{-1}$; Experiment 3 confirms the same value: $0.192/0.054 = 3.56$. Answer: order in NO $= 2$, order in $\text{O}_2 = 1$, overall order $= 3$, rate law $\text{Rate} = k[\text{NO}]^2[\text{O}_2]$, and $k = 3.56\ \text{mol}^{-2}\text{L}^2\text{s}^{-1}$.

Competitive level (JEE / NEET) - First order half-life and fractional completion

A first order reaction has rate constant $k = 1.15\times10^{-3}\ \text{s}^{-1}$. Find (a) the half-life of the reaction, (b) the time for the reaction to be $87.5\%$ complete, and (c) the time for a $5\ \text{g}$ sample of reactant to fall to $3\ \text{g}$.

Step 1. Half-life directly from equation (10): $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{1.15\times10^{-3}} = 602.6\ \text{s}$, about $603\ \text{s}$.

Step 2. For $87.5\%$ completion, exactly $12.5\% = 1/8$ of the reactant remains, i.e. $[R]/[R]_0 = (1/2)^3$ — precisely three half-lives. So $t = 3 \times 602.6 = 1808\ \text{s}$. Checking with equation (7): $t = \dfrac{2.303}{k}\log(8) = 2003 \times 0.903 = 1809\ \text{s}$, matching to rounding.

Step 3. For $5\ \text{g} \to 3\ \text{g}$, fraction remaining is $[R]/[R]_0 = 3/5 = 0.6$ (mass ratios substitute directly for concentration ratios at fixed volume). From equation (7): $t = 2003 \times \log(1.667) = 2003 \times 0.2218 = 444\ \text{s}$. Answer: $t_{1/2} \approx 603\ \text{s}$, time for $87.5\%$ completion $\approx 1808\ \text{s}$ (three half-lives), and time for $5\ \text{g}$ to fall to $3\ \text{g}$ is $444\ \text{s}$. Timing note: under ninety seconds once you recognise the shortcut. Any fraction remaining that is a clean power of one-half — $1/2$, $1/4$, $1/8$, $1/16$ — should immediately be read as $n$ half-lives ($n = 1, 2, 3, 4$) and multiplied by $t_{1/2}$, avoiding the log calculation entirely. Reach for the full logarithmic formula, equation (7), only when the fraction remaining is not a clean power of two, as in part (c).

Competitive level (JEE / NEET) - Arrhenius equation: finding Ea and predicting k at a new temperature

The rate constants of a reaction are $k_1 = 0.02\ \text{s}^{-1}$ at $T_1 = 500\ \text{K}$ and $k_2 = 0.07\ \text{s}^{-1}$ at $T_2 = 700\ \text{K}$. Take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$. Find (a) the activation energy $E_a$, and (b) the rate constant $k_3$ at $T_3 = 600\ \text{K}$.

Step 1. Apply equation (12): $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$. Compute the left side: $\log(0.07/0.02) = \log(3.5) = 0.544$. Compute the bracket: $1/500 - 1/700 = 0.002 - 0.001429 = 5.714\times10^{-4}\ \text{K}^{-1}$.

Step 2. Solve for $E_a$: $0.544 = \dfrac{E_a}{2.303 \times 8.314}\times 5.714\times10^{-4} = \dfrac{E_a \times 5.714\times10^{-4}}{19.15}$, so $E_a = \dfrac{0.544 \times 19.15}{5.714\times10^{-4}} = 18{,}231\ \text{J mol}^{-1} \approx 18.23\ \text{kJ mol}^{-1}$.

Step 3. Predict $k_3$ at $T_3 = 600\ \text{K}$ using $k_1$ and $E_a$ in the same two-point formula: $\log\dfrac{k_3}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_3}\right) = \dfrac{18{,}231}{19.15}\left(\dfrac{1}{500} - \dfrac{1}{600}\right) = 952.1 \times 3.333\times10^{-4} = 0.3174$.

Step 4. So $k_3/k_1 = 10^{0.3174} = 2.08$, giving $k_3 = 0.02 \times 2.08 = 0.0415\ \text{s}^{-1}$. Cross-check using $k_2$ and $T_2$: $\log(k_3/k_2) = 952.1\times(1/700-1/600) = -0.2267$, so $k_3 = 0.07\times10^{-0.2267} = 0.0415\ \text{s}^{-1}$ — the same answer both ways. Answer: $E_a \approx 18.23\ \text{kJ mol}^{-1}$ and $k_3 \approx 0.0415\ \text{s}^{-1}$ at $600\ \text{K}$. Timing note: about two minutes. The efficient path is Step 1 to Step 2 to get $E_a$ once, then reuse the same $E_a/2.303R$ constant (here $952.1$) for every further temperature prediction rather than recomputing it — and always sanity-check a predicted $k$ by verifying it lies between $k_1$ and $k_2$ when the new temperature lies between $T_1$ and $T_2$, which instantly catches a sign error in the bracket.

The single most common error in this chapter is treating the balanced equation as if it were the rate law — assuming the exponents in $\text{Rate} = k[A]^x[B]^y$ must equal the stoichiometric coefficients. They coincide only sometimes (as with $2\text{NO} + \text{O}_2$) and just as often do not (as with $\text{CHCl}_3 + \text{Cl}_2$, order $3/2$, or ester hydrolysis, apparent first order despite being bimolecular); the rate law must always be read off experimental data, never guessed. A close second is confusing order with molecularity. Order is experimental, can be zero or fractional, and describes the reaction as a whole; molecularity is theoretical, always a positive integer from $1$ to $3$, and has no meaning for a complex, multi-step reaction taken as a whole — only for its individual elementary steps. Students frequently drop the stoichiometric factor when coefficients are unequal, quoting the rate as simply $-d[\text{HI}]/dt$ for $2\text{HI} \to \text{H}_2 + \text{I}_2$ without dividing by $2$. Another frequent slip is forgetting that units of $k$ depend on order and must be derived, not memorised — writing $\text{s}^{-1}$ for a second order $k$ immediately signals a wrong order. For integrated rate laws, the most damaging mix-up is applying the first order half-life formula, $t_{1/2} = 0.693/k$, to a zero order reaction or vice versa — only the first order half-life is independent of starting concentration. In pseudo first order problems, students forget the true order still governs physically; $k'$ is only apparent, proportional to the near-constant concentration of the reactant in excess. In Arrhenius calculations, a very common slip is mixing units of $E_a$ (joules versus kilojoules) or forgetting $T$ must be in kelvin, not Celsius. Finally, students conflate threshold energy with activation energy: threshold energy is the total energy needed at collision (activation energy plus energy the molecules already possess), while $E_a$ is only the extra amount above what an average molecule already carries.

• Define average rate and instantaneous rate from a concentration-time data set, including the correct sign convention and units.
• Apply the stoichiometric factor correctly when reactant and product coefficients in a balanced equation are unequal.
• Write a rate law from experimental initial-rate data and determine the order with respect to each reactant and the overall order.
• Distinguish order from molecularity crisply, including explaining why molecularity has no meaning for a complex reaction as a whole.
• Derive and apply the integrated rate equations for zero order and first order reactions, including the gas-phase pressure form.
• Calculate and interpret half-life for both zero order (concentration-dependent) and first order (concentration-independent) reactions.
• Recognise a pseudo first order reaction and explain why its apparent order differs from its true order.
• Use the Arrhenius equation, both in single-temperature form and the two-temperature log form, to find $E_a$, $A$, or an unknown rate constant.
• Explain, using collision theory, why not every molecular collision produces product, distinguishing effective collisions, orientation and the steric factor $P$.
• Explain how a catalyst increases rate without changing $\Delta G$ or the equilibrium constant of a reaction.

Ask the tutor: if you want to work through more rate-law-from-data problems or practice Arrhenius numericals with three or more temperatures, just ask.