First order reaction: rate $= k[A]$. Integrated law: $[A] = [A]_0 e^{-kt}$ or $\ln[A] = \ln[A]_0 - kt$. In terms of $\log$: $\log[A] = \log[A]_0 - kt/2.303$. Therefore: $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$. Half-life: $t_{1/2} = \dfrac{0.693}{k} = \dfrac{\ln 2}{k}$. Key: $k = 0.693/t_{1/2}$. Substituting in time formula: $t = \dfrac{2.303 \times t_{1/2}}{0.693}\log\dfrac{[A]_0}{[A]} = 3.322 \times t_{1/2} \times \log\dfrac{[A]_0}{[A]}$. For $x$% completion: $[A]/[A]_0 = (100-x)/100$. Time $= 3.322 \times t_{1/2} \times \log\dfrac{100}{100-x}$. For 99.9% ($x=99.9$): $t = 3.322 \times 1 \times \log(1000) = 3.322 \times 3 = 9.97 \approx 10$ min.

For first order reactions, the half-life method is often fastest for NEET. After $n$ half-lives, fraction remaining $= (1/2)^n$. Important values: 1 $t_{1/2}$: 50% remains. 2 $t_{1/2}$: 25% remains. 3 $t_{1/2}$: 12.5% remains. 4 $t_{1/2}$: 6.25% remains. 5 $t_{1/2}$: 3.125% remains. 6 $t_{1/2}$: $\approx$1.56%. 7 $t_{1/2}$: $\approx$0.78%. 8 $t_{1/2}$: $\approx$0.39%. 9 $t_{1/2}$: $\approx$0.195%. 10 $t_{1/2}$: $\approx$0.098% (very close to 0.1%). Shortcut: when the question gives a "nice" fraction (90%, 99%, 99.9%), check if it corresponds to an exact power of 2: $90\% \to 10\% = 1/10$ (not exact). $99\% \to 1\% \approx 1/100 \approx (1/2)^{6.64}$. $99.9\% \to 0.1\% \approx 1/1000 \approx (1/2)^{10}$. The approximation $(1/2)^{10} = 1/1024 \approx 1/1000$ is used here.

All radioactive decay processes follow first order kinetics. $N = N_0 e^{-\lambda t}$ where $\lambda$ = decay constant. $t_{1/2} = 0.693/\lambda$. Activity $A = \lambda N = A_0 e^{-\lambda t}$. Units: Becquerel (Bq) = 1 decay/s. Curie (Ci) = $3.7\times10^{10}$ Bq (activity of 1 g Ra). Important isotopes: $^{14}C$: $t_{1/2} = 5730$ yr (carbon dating — organisms stop incorporating $^{14}C$ at death; ratio $^{14}C/^{12}C$ decreases at known rate → age determination up to 50,000 yr). $^{238}U$: $t_{1/2} = 4.5 \times 10^9$ yr (uranium-lead dating for rocks/Earth age). $^{131}I$: $t_{1/2} = 8.02$ days (hyperthyroidism treatment). $^{99m}Tc$: $t_{1/2} = 6.01$ h (medical imaging — chosen because short half-life limits radiation dose). After 10 half-lives: radioactivity $= (1/2)^{10} \approx 0.1\%$ of original.

Pseudo-first order: one reactant in large excess. Rate $= k[A][B]$ becomes rate $= k'[A]$ when $[B] >> [A]$. $k' = k[B]$. Example: acid hydrolysis of ester CH3COOC2H5 + H2O → in dilute acid, $[H_2O] \approx$ 55 M $\approx$ constant. Apparent first order. Second order (same reactant): rate $= k[A]^2$. Integrated: $1/[A] = 1/[A]_0 + kt$. $t_{1/2} = 1/(k[A]_0)$ — depends on initial concentration! Each subsequent half-life is twice the previous one. Second order (two reactants): rate $= k[A][B]$. Complex integrated form unless $[A]_0 = [B]_0$. Method of initial rates determines individual orders: doubling $[A]$ while keeping $[B]$ constant: if rate doubles → first order in A; quadruples → second order in A; unchanged → zero order in A.

$k = Ae^{-E_a/RT}$. Taking log: $\log k = \log A - \dfrac{E_a}{2.303R} \cdot \dfrac{1}{T}$. Plot $\log k$ vs $1/T$: slope $= -E_a/(2.303R)$, intercept $= \log A$. Two-temperature version: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$. With $R = 8.314$ J/mol/K: for $E_a = 50$ kJ/mol, $T_1 = 300$ K, $T_2 = 310$ K: $\log(k_2/k_1) = \dfrac{50000}{2.303 \times 8.314}\left(\dfrac{1}{300} - \dfrac{1}{310}\right) = 2610 \times 1.075 \times 10^{-4} = 0.2806$. $k_2/k_1 = 10^{0.2806} \approx 1.9 \approx 2$. This confirms the approximate rule: rate doubles per 10°C rise (for $E_a \approx 50$ kJ/mol near 300 K).

Rate of reaction $= Z_{AB} \cdot p \cdot e^{-E_a/RT}$ where: $Z_{AB}$ = collision frequency (collisions per second per unit volume). $p$ = steric/probability factor ($0 < p \leq 1$). $e^{-E_a/RT}$ = fraction of collisions with energy $\geq E_a$. The frequency factor $A = Z_{AB} \cdot p$. Collision frequency $Z_{AB} \propto \sqrt{T}$ (from kinetic theory). But the exponential term $e^{-E_a/RT}$ changes much more rapidly with temperature than $\sqrt{T}$. So overall: rate increases strongly with $T$, mainly due to increasing fraction of effective collisions (more molecules exceed $E_a$). At room temperature (300 K) with $E_a = 50$ kJ/mol: fraction of effective collisions $= e^{-50000/(8.314 \times 300)} = e^{-20.05} \approx 2 \times 10^{-9}$. Even though only 1 in $5 \times 10^8$ collisions is effective, billions of collisions occur per second, so reactions proceed.

Enzymes are biological catalysts (proteins). Michaelis-Menten equation: $v = \dfrac{V_{max}[S]}{K_m + [S]}$ where $v$ = reaction velocity, $V_{max}$ = maximum velocity, $[S]$ = substrate concentration, $K_m$ = Michaelis constant. At $[S] << K_m$: $v \approx (V_{max}/K_m)[S]$ — first order in $[S]$. At $[S] >> K_m$: $v \approx V_{max}$ — zero order (enzyme saturated). $K_m$ = $[S]$ at which $v = V_{max}/2$ — measure of enzyme-substrate affinity. Smaller $K_m$ = higher affinity. Turnover number $k_{cat}$ = reactions per enzyme molecule per second. Carbonic anhydrase: $k_{cat} = 10^6$ s$^{-1}$ (fastest known enzyme). Linweaver-Burk plot: $1/v$ vs $1/[S]$, straight line, $y$-intercept $= 1/V_{max}$, $x$-intercept $= -1/K_m$, slope $= K_m/V_{max}$. Inhibitors: competitive (increases apparent $K_m$), non-competitive (decreases $V_{max}$).

Haber process: rate-determining step = N2 dissociation on Fe surface. $E_a$ is high → needs elevated temperature (400-500°C). But high T reduces yield (exothermic equilibrium). Compromise conditions. Contact process: SO2 + 1/2 O2 → SO3 on V2O5. Temperature: 450-550°C. Below 400°C: rate too slow. Above 600°C: V2O5 deactivates. Catalytic cracking (petroleum): zeolite catalysts break long-chain alkanes into shorter chains. Corrosion: iron rusting follows complex kinetics. Rate depends on O2, H2O, electrolytes, pH. Protective coatings prevent O2 access. Food preservation: microbial reactions follow Arrhenius — refrigeration slows reaction rates, preventing spoilage. Nuclear reactors: neutron moderation, criticality control — all governed by nuclear reaction kinetics.