2.2 Rate of chemical change

Choose two well-separated points on the tangent, not on the curve, to calculate its gradient. A reactant concentration has a negative gradient, so report its disappearance rate as a positive magnitude unless a signed change is requested.

Worked tangent example: on a concentration–time graph for $\ce{HCl}$, two points on the tangent at $t=0$ are $(0\,\mathrm{s},0.250\,\mathrm{mol\,dm^{-3}})$ and $(14\,\mathrm{s},0.100\,\mathrm{mol\,dm^{-3}})$. The tangent gradient is $(0.100-0.250)/(14-0)=-0.0107\,\mathrm{mol\,dm^{-3}\,s^{-1}}$. For $\ce{Mg + 2HCl -> MgCl2 + H2}$, divide the positive disappearance-rate magnitude by the HCl coefficient: $v=0.0107/2=0.0054\,\mathrm{mol\,dm^{-3}\,s^{-1}}$.

At one temperature, only the fraction of collisions above Ea and with a productive orientation can react. Raising temperature increases that fraction, not the energy of every particle by the same amount. Use collision frequency to explain concentration or pressure effects and the energy distribution to explain the stronger temperature effect.

Powdered CaCO₃ reacts faster than equal-mass chips because more surface sites are exposed, not because its particles have higher kinetic energy. For each changed condition, identify exactly what changes—collision frequency, energy distribution or pathway—and hold other variables constant in a fair comparison.

Two Maxwell–Boltzmann curves for the same number of particles have equal total area. At higher T the curve is broader with a lower peak and a larger area to the right of a fixed Ea line; the peak does not move to Ea and no particle count is lost.

At fixed temperature a catalyst does not change the Maxwell–Boltzmann distribution; it moves the threshold to a lower Ea, increasing the area beyond it. On an energy profile it changes the pathway and peak, not ΔH, reactant/product energies or the equilibrium constant.

Add all elementary steps and cancel intermediates to recover the overall equation. Then derive the rate dependence expected from the slow step, eliminating an intermediate when necessary, and compare with experiment. A catalyst consumed early and regenerated later cancels from the overall equation but is not an intermediate.

For each step, measure Ea upward from its own preceding reactant or intermediate valley to the next peak. The rate-determining barrier is the largest of those step barriers, which need not be the peak with the greatest absolute height above the page baseline.

A step A + 2B → products is termolecular because three reacting particles meet in that elementary event. Molecularity is always a positive whole-number description of one proposed step; reaction order is experimental and can be zero, fractional or unrelated to overall stoichiometric coefficients.

If doubling [A] while holding [B] constant quadruples rate, the order in A is 2; if doubling [B] leaves rate unchanged, the order in B is 0, giving rate = k[A]². Choose trial pairs with only one changed concentration before combining exponents.

Worked initial-rate deduction: doubling $[\ce{Fe^{3+}}]$ from $1.00\times10^{-2}$ to $2.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ at constant $[\ce{I^-}]$ doubles rate, so the order in $\ce{Fe^{3+}}$ is 1. Doubling $[\ce{I^-}]$ at constant $[\ce{Fe^{3+}}]$ increases rate from $3.24\times10^{-5}$ to $1.30\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$, approximately fourfold, so its order is 2. Therefore $v=k[\ce{Fe^{3+}}][\ce{I^-}]^2$, third order overall.

Use multiple representations as cross-checks: zero-order rate is independent of concentration and [A] falls linearly; first-order rate is proportional to [A] and gives exponential decay; second-order rate curves upward on a rate-versus-concentration plot. Do not infer order from one balanced equation.

Derive rather than memorize the units: first-order k has units s⁻¹, while an overall second-order law written with mol dm⁻³ and seconds gives dm³ mol⁻¹ s⁻¹. Concentration changes rate but does not change k at fixed temperature.

Worked $k$ example: for $v=k[\ce{Fe^{3+}}][\ce{I^-}]^2$, use $v=1.62\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ and both concentrations $1.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}$. $k=\frac{1.62\times10^{-5}}{(1.00\times10^{-2})(1.00\times10^{-2})^2}=16.2\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$. Substitution returns the measured rate, and the units cancel those of a third-order concentration term.

A steeper negative ln k versus 1/T gradient means a larger activation energy. When two temperatures are given, subtracting the two linear equations removes ln A and lets Ea be found without knowing the frequency factor.

Worked Arrhenius-plot example: using line points $(0.00302\,\mathrm{K^{-1}},-6.8)$ and $(0.00334\,\mathrm{K^{-1}},-10.8)$, the gradient is $[-6.8-(-10.8)]/(0.00302-0.00334)=-1.25\times10^4\,\mathrm{K}$. Since gradient $=-E_a/R$, $E_a=-(-1.25\times10^4)(8.31)=1.04\times10^5\,\mathrm{J\,mol^{-1}}=104\,\mathrm{kJ\,mol^{-1}}$. The negative plot gradient therefore produces a positive activation energy.

From a fitted line, intercept b gives A = eᵇ, while gradient m gives Ea = −mR. Check the pair against k = Ae^(−Ea/RT) at one data temperature. A has the same units as k for the stated rate law; it is not an activation energy or a universal constant.

Worked intercept example: for the same first-order Arrhenius line, use $\ln k=-6.02$, $1/T=0.00296\,\mathrm{K^{-1}}$ and gradient $-E_a/R=-12500\,\mathrm{K}$. From $\ln k=(-E_a/R)(1/T)+\ln A$, $-6.02=(-12500)(0.00296)+\ln A$, so $\ln A=30.98$ and $A=e^{30.98}=2.85\times10^{13}\,\mathrm{s^{-1}}$. The unit is $\mathrm{s^{-1}}$ because this reaction is first order and $A$ has the same units as $k$.