⏱️ Chemical Kinetics & Catalysis

Chemical kinetics uncovers the reaction pathway, transition states, and molecularity behind thermodynamic transformations. Key competitive exam challenges include Steady State Approximation (SSA), parallel and consecutive rate differentials, Arrhenius graph slopes, and pseudo-first order kinetics.

1. 📈 Rate Laws & Integrated Rate Equations

1.1 Comparison of Reaction Orders

Reaction Order ( $n$ )	Differential Rate Law	Integrated Rate Equation	Half-Life ( $t_{1 / 2}$ )	Units of Rate Constant $k$	Linear Plot
0	$- \frac{d [A]}{d t} = k$	$[A]_{t} = [A]_{0} - k t$	$t_{1 / 2} = \frac{[A]_{0}}{2 k}$	$mol \cdot L^{- 1} \cdot s^{- 1}$	$[A]_{t}$ vs $t$ (slope $= - k$ )
1	$- \frac{d [A]}{d t} = k [A]$	$\ln (\frac{[A]_{0}}{[A]_{t}}) = k t ⟹ [A]_{t} = [A]_{0} e^{- k t}$	$t_{1 / 2} = \frac{\ln 2}{k} = \frac{0.693}{k}$	$s^{- 1}$	$\ln [A]_{t}$ vs $t$ (slope $= - k$ )
2	$- \frac{d [A]}{d t} = k [A]^{2}$	$\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}} = k t$	$t_{1 / 2} = \frac{1}{k [A]_{0}}$	$L \cdot {mol}^{- 1} \cdot s^{- 1}$	$\frac{1}{[A]_{t}}$ vs $t$ (slope $= + k$ )
$n$	$- \frac{d [A]}{d t} = k [A]^{n}$	$\frac{1}{n - 1} (\frac{1}{[A]_{t}^{n - 1}} - \frac{1}{[A]_{0}^{n - 1}}) = k t$	$t_{1 / 2} \propto \frac{1}{[A]_{0}^{n - 1}}$	$(mol/L)^{1 - n} s^{- 1}$	$[A]_{t}^{1 - n}$ vs $t$

Powerful First-Order Shortcuts

$t_{75 %} = 2 \times t_{50 %}$
$t_{87.5 %} = 3 \times t_{50 %}$
$t_{99 %} = \frac{2 \ln 10}{\ln 2} t_{50 %} \approx \frac{2 \times 2.303}{0.693} t_{50 %} \approx 6.64 \times t_{50 %}$
$t_{99.9 %} = 10 \times t_{50 %}$

2. 🧩 Complex Reaction Mechanisms & Steady State Approximation

2.1 Parallel / Competing Reactions

A \overset{k_{1}}{\to} B and A \overset{k_{2}}{\to} C

- \frac{d [A]}{d t} = (k_{1} + k_{2}) [A] ⟹ k_{effective} = k_{1} + k_{2}

Product Ratio: \frac{[B]_{t}}{[C]_{t}} = \frac{k_{1}}{k_{2}} at all times t

Fraction of Product B = \frac{k_{1}}{k_{1} + k_{2}} \times 100 %

Overall Activation Energy: E_{a, eff} = \frac{k_{1} E_{a 1} + k_{2} E_{a 2}}{k_{1} + k_{2}}

2.2 Consecutive / Sequential Reactions

A \overset{k_{1}}{\to} B \overset{k_{2}}{\to} C

[A]_{t} = [A]_{0} e^{- k_{1} t}

[B]_{t} = [A]_{0} \frac{k_{1}}{k_{2} - k_{1}} (e^{- k_{1} t} - e^{- k_{2} t})

Time for Maximum Concentration of Intermediate $B$ : $t_{max} = \frac{\ln (k_{1} / k_{2})}{k_{1} - k_{2}} = \frac{2.303 \log (k_{1} / k_{2})}{k_{1} - k_{2}}$
Maximum Concentration of $B$ : $[B]_{max} = [A]_{0} {(\frac{k_{2}}{k_{1}})}^{\frac{k_{2}}{k_{1} - k_{2}}}$

3. 🌡️ Temperature Dependence: Arrhenius Equation

k = A e^{- E_{a} / (R T)}

\ln k = \ln A - \frac{E_{a}}{R} (\frac{1}{T}) ⟹ \log_{10} k = \log_{10} A - \frac{E_{a}}{2.303 R} (\frac{1}{T})

\ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) = \frac{E_{a} (T_{2} - T_{1})}{R T_{1} T_{2}}

Slope of $\ln k$ vs $\frac{1}{T}$ : $- \frac{E_{a}}{R}$
Slope of $\log_{10} k$ vs $\frac{1}{T}$ : $- \frac{E_{a}}{2.303 R}$
$y$ -intercept: $\ln A$ (or $\log_{10} A$ )
Catalyst Effect: Lowers activation energy $E_{a}$ to $E_{a}^{'}$ : $\frac{k_{cat}}{k_{uncat}} = \exp (\frac{E_{a} - E_{a}^{'}}{R T})$

⏱️ Chemical Kinetics & Catalysis ​

1. 📈 Rate Laws & Integrated Rate Equations ​

1.1 Comparison of Reaction Orders ​

2. 🧩 Complex Reaction Mechanisms & Steady State Approximation ​

2.1 Parallel / Competing Reactions ​

2.2 Consecutive / Sequential Reactions ​

3. 🌡️ Temperature Dependence: Arrhenius Equation ​