Concentration against time
Energy profile
Collision theory
For molecules to react they must collide, with enough energy along the line of their centres (the head-on part of the impact; a glancing blow does not count) to overcome the activation barrier $E_a$, and in the right orientation. The rate constant therefore contains three factors:
$Z$ is the collision frequency factor (it grows with concentration, molecular size and speed), $p$ is the steric (orientation) factor and $e^{-E_a/RT}$ is the fraction of collisions with energy above $E_a$. In the chamber, a reaction happens exactly when a collision meets all three conditions. The "Collision energies" view shows the measured distribution, which is exponential, $f(E) = \frac{1}{RT}e^{-E/RT}$, so the shaded fraction above $E_a$ is $e^{-E_a/RT}$. Textbook diagrams usually show the Maxwell–Boltzmann curve of molecular energies, which rises to a peak before its long tail. This view plots the energies of collisions in a two-dimensional chamber, which has no peak. The conclusion is the same: raising $T$ or lowering $E_a$ increases the area beyond $E_a$.
Activation energy, enthalpy and catalysts
A reaction-coordinate diagram (the Energy profile in the lab) plots energy as the reactants turn into products. $\Delta H$ is the energy difference between products and reactants: negative means exothermic (heat released), positive means endothermic (heat absorbed). The reverse barrier is $E_{a,r} = E_{a,f} - \Delta H$. A catalyst provides an alternative pathway with a lower barrier. It lowers both barriers by the same amount, so it speeds up the forward and reverse reactions equally: equilibrium is reached sooner, but the position of equilibrium and $K$ do not change.
Dynamic equilibrium
As A and B are used up, the forward rate falls; as C and D accumulate, the reverse rate rises. At equilibrium the rates are equal and concentrations stop changing, although particles keep reacting in both directions. For the elementary step A + B ⇌ C + D (a single-collision step, so its rate law can be read straight from the equation):
The reaction quotient $Q$ has the same form but uses the current concentrations. If $Q < K$ the net reaction goes forward; if $Q > K$ it goes backward. In this chamber all particles have the same mass and size, so $K = e^{-\Delta H/RT}$ (the entropy change is zero by construction); the van ’t Hoff equation then says $\ln K$ falls with $T$ for an exothermic reaction and rises for an endothermic one.
Le Chatelier's principle
- Concentration: adding a reactant or removing a product makes $Q < K$; the system shifts right until $Q = K$ again. $K$ itself is unchanged.
- Temperature: the only stress that changes $K$. Heating favours the endothermic direction.
- Pressure / volume: only matters if the number of gas molecules changes. Here $\Delta n = 0$, so compressing changes every concentration by the same factor and $Q$ is unchanged: no shift, although rates increase. For the Haber process ($\Delta n = -2$) compression shifts the equilibrium to the right.
- Catalyst: no shift; equilibrium is just reached sooner.
Integrated rate laws
For a reaction A → products with rate $= k[\mathrm A]^n$, integrating gives a different linear plot for each order. Choose an order and watch which of the three plots is a straight line.
[A] against t
ln[A] against t
1/[A] against t
| Order | Rate law | Integrated form | Linear plot | Gradient | Half-life |
|---|---|---|---|---|---|
| 0 | $k$ | $[\mathrm A] = [\mathrm A]_0 - kt$ | [A] vs t | $-k$ | $[\mathrm A]_0/2k$ |
| 1 | $k[\mathrm A]$ | $\ln[\mathrm A] = \ln[\mathrm A]_0 - kt$ | ln[A] vs t | $-k$ | $\ln 2/k$ (constant) |
| 2 | $k[\mathrm A]^2$ | $1/[\mathrm A] = 1/[\mathrm A]_0 + kt$ | 1/[A] vs t | $+k$ | $1/(k[\mathrm A]_0)$ |
Identify the order
A kinetics run gave the data below (with realistic measurement scatter). Plot it in your head or on paper: which order is it, and what is k?
The Arrhenius equation
A plot of $\ln k$ against $1/T$ is a straight line with gradient $-E_a/R$ and intercept $\ln A$. Two measurements are enough to find $E_a$: $\ln(k_2/k_1) = \frac{E_a}{R}\left(\frac1{T_1}-\frac1{T_2}\right)$.
Measure Eₐ from two experiments
The initial-rates method
For rate $= k[\mathrm A]^m[\mathrm B]^n$, compare experiments in which only one concentration changes. If doubling [A] (with [B] fixed) doubles the rate, $m = 1$; if it quadruples the rate, $m = 2$; if nothing changes, $m = 0$. In general $m = \log(\text{rate ratio})/\log(\text{concentration ratio})$. Then substitute any row to find $k$.
Orders must be found by experiment. They need not match the coefficients in the balanced equation: the 2 in 2A says nothing about $m$. Only for a single elementary step can the rate law be read from the equation. In a multi-step mechanism, the slowest step (the rate-determining step) sets the rate law.
Your results: 2A + B → products, 25 °C
Industrial example: the Haber process
Four moles of gas become two and the reaction is exothermic. Le Chatelier says high pressure and low temperature give the highest equilibrium yield, but at low temperature the reaction (even over an iron catalyst) is far too slow, and very high pressures are expensive and dangerous. Plants therefore compromise at about 400–450 °C and 150–250 atm and recycle unreacted gas, converting roughly 15–25 % of it per pass.
Example 1: Equilibrium in the chamber (model scale)
Example 2: Fraction of successful collisions (model scale)
Example 3: First-order half-life
Example 4: Activation energy from two rate constants
Example 5: Orders from initial rates
| Exp | [A] / M | [B] / M | Rate / M s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0×10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0×10⁻⁴ |
| 3 | 0.10 | 0.30 | 6.0×10⁻⁴ |
1→2: [A] ×2, rate ×4, so $m = 2$. 1→3: [B] ×3, rate ×3, so $n = 1$. Rate $= k[\mathrm A]^2[\mathrm B]$; $k = 2.0\times10^{-4}/(0.10^2\times0.10) = $ 0.20 M⁻² s⁻¹.
Example 6: Haber equilibrium at 450 °C and 200 atm
Challenges (checked automatically)
Guided experiments
- Rates meet. Reset and watch the forward and reverse rates. Note the time at which they become equal and compare with when the concentration curves level off.
- Measure K. At equilibrium, read [A], [B], [C], [D] and calculate $K$. Compare with $e^{-\Delta H/RT}$. Repeat at 250 K and 600 K. Expect about 5.6 at 350 K, 11 at 250 K and 2.7 at 600 K: for an exothermic reaction, heating shifts the equilibrium back towards A and B. Counts of a few dozen particles fluctuate, so average several readings.
- Catalyst race. Reset with and without the catalyst at the same temperature. Which reaches equilibrium first? Are the final concentrations different? Keep temperature and starting amounts fixed and record the time to equilibrium. Expect the catalysed run to arrive sooner (its barrier is 60 % of the uncatalysed one) but with the same final amounts, because a catalyst lowers both barriers equally and leaves $K$ unchanged.
- Turn orientation off. Untick “Orientation must be right (steric factor)”. How much faster does equilibrium arrive? The steric factor in this model is 1/3, so expect every rate to triple and the time to equilibrium to fall to roughly a third, with no change in the final amounts.
- Insulated exothermic reaction. Tick “Insulated chamber (no thermostat)” and reset the exothermic preset. Why does the temperature climb, and why does the climb slow down? Expect T to rise because the heat released stays in the box, and the rise to level off as the forward and reverse rates meet and net conversion stops; the higher T also lowers $K$, so slightly less product forms than at constant temperature.
- Compress. At equilibrium, compress the chamber. Do the amounts of C and D change? What happens to the rates? Expect the amounts to stay the same, because there are two particles on each side of the equation, while both rates rise together because collisions become more frequent.