Notes
Reaction rates
Rate is the change in concentration over time, like $\displaystyle -\frac{\Delta[A]}{\Delta t}$. Rates increase with higher concentration, higher temperature, more surface area, and catalysts.
Rate laws
$\text{rate} = k[A]^m[B]^n$. The orders $m$ and $n$ come from experiments, not from coefficients (unless the step is elementary).
- Method of initial rates: if doubling [A] doubles the rate, first order; quadruples, second order; no change, zero order.
- Overall order $= m + n$. Units of $k$ depend on the overall order.
Integrated rate laws
| Order | Linear plot | Slope | Half-life |
|---|---|---|---|
| 0 | [A] vs $t$ | $-k$ | depends on [A]₀ |
| 1 | ln[A] vs $t$ | $-k$ | $\displaystyle t_{1/2} = \frac{0.693}{k}$ (constant) |
| 2 | $\displaystyle \frac{1}{[A]}$ vs $t$ | $+k$ | depends on [A]₀ |
- First-order reactions, like radioactive decay, have a constant half-life.
Mechanisms
- A mechanism is a series of elementary steps. Their rate laws do follow the coefficients.
- The rate-determining step (slowest step) sets the overall rate law.
- Intermediates are made then used up; catalysts are used then remade.
- The steps must add up to the overall equation.
Collision theory and energy
- Molecules react only if they collide with enough energy (at least the activation energy, $E_a$) and the right orientation.
- Higher temperature means more molecules have energy above $E_a$ (the Maxwell–Boltzmann distribution shifts right).
- A catalyst provides a new pathway with lower $E_a$. It doesn't change $\Delta H$ or the equilibrium.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$2^n = 4$ gives $n = 2$.
$1 + 2 = 3$.
First-order reactions give a linear ln[A] vs $t$ plot.
Three half-lives: $\displaystyle \left(\frac{1}{2}\right)^3$.
It changes the mechanism, not the energies of reactants and products.
The slow first step is elementary and determines the rate.
It's made in step 1 and used in step 2.
The fraction of molecules above $E_a$ grows with temperature.
Free response
Write your answer first, then open the scoring guide.
Free-response question
For $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$, initial rate data: Exp 1: [NO] = 0.010, [O₂] = 0.010, rate = $2.5 \times 10^{-5}$. Exp 2: [NO] = 0.020, [O₂] = 0.010, rate = $1.0 \times 10^{-4}$. Exp 3: [NO] = 0.010, [O₂] = 0.020, rate = $5.0 \times 10^{-5}$ (M/s).
(a) Determine the order with respect to NO and to $\mathrm{O_2}$.
Show the answer and scoring guide
Exp 1 → 2: [NO] doubles, rate ×4, so second order in NO. Exp 1 → 3: [O₂] doubles, rate ×2, so first order in O₂.
Check off the points you earned:
(b) Write the rate law and calculate $k$ with units.
Show the answer and scoring guide
rate $= k[NO]^2[O_2]$. $\displaystyle k = \frac{2.5 \times 10^{-5}}{(0.010)^2(0.010)} = 25\ M^{-2}s^{-1}$.
Check off the points you earned: