Maharashtra State Board · Class 12 · Chemistry · Chapter 6

Chemical Kinetics — Formula Sheet

Board Formulas

11 formulas

1.Rate of Reaction
$Rate = - \frac{1}{a} \frac{d [ A ]}{d t} = \frac{1}{b} \frac{d [ B ]}{d t}$
For aA → bB. Sign is chosen so that rate is positive.
2.Rate Law★
$Rate = k [A]^{m} [B]^{n}$
m, n are determined experimentally, NOT from stoichiometry. m + n = order.
3.Zero-Order Integrated Rate
$[A] = [A]_{0} - k t$
Concentration decreases linearly with time. k has units mol·L⁻¹·s⁻¹.
4.First-Order Integrated Rate★
$k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}$
Log form. Slope of log[A] vs t gives −k/2.303.
5.First-Order Half-Life★
$t_{1/2} = \frac{0.693}{k}$
Independent of initial concentration. Signature of first-order kinetics.
6.Second-Order Integrated Rate
$\frac{1}{[ A ]} - \frac{1}{[ A ] _{0}} = k t$
For a single reactant with rate = k[A]². Plot 1/[A] vs t gives straight line.
7.Second-Order Half-Life
$t_{1/2} = \frac{1}{k [ A ] _{0}}$
Depends on initial concentration — distinguishes it from first-order.
8.Arrhenius Equation★
$k = A e^{- E_{a} / R T}$
A = frequency factor, E_a = activation energy. Fundamental temperature dependence of rate.
9.Log Form of Arrhenius Equation
$lo g k = lo g A - \frac{E _{a}}{2.303 R T}$
Straight line: log k vs 1/T with slope −E_a/(2.303R).
10.Ratio of Rate Constants at Two Temperatures★
$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$
Two-point form of Arrhenius equation. Useful when calculating E_a.
11.Temperature Coefficient
$μ = \frac{k _{T + 10}}{k _{T}} \approx 2 to 3$
Rate roughly doubles for every 10 K rise near room temperature.