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ChemistryChemical Kinetics
For a first order reaction, the rate constant $k = 0.03\ \text{s}^{-1}$. How much time is required for the concentration to decrease from $7.2$ mol/L to $0.9$ mol/L?
Options
1
$46.2$ s
2
$69.3$ s
3
$92.4$ s
4
$23.1$ s
Correct Answer
69.3 s
Solution
1

First order: $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$

$$t = \frac{2.303}{0.03}\log\frac{7.2}{0.9} = 76.77 \times \log 8 = 76.77 \times 0.903 \approx 69.3 \text{ s}$$
2

Shortcut: $7.2/0.9 = 8 = 2^3$ (3 half-lives)

$t_{1/2} = 0.693/0.03 = 23.1$ s → $t = 3 \times 23.1 = \boxed{69.3}$ s

$7.2 \to 3.6 \to 1.8 \to 0.9$ (3 half-lives of 23.1 s each = 69.3 s)
Theory: Chemical Kinetics
1. Integrated Rate Laws

Zero order: $[A] = [A]_0 - kt$. Plot $[A]$ vs $t$ linear. $t_{1/2} = [A]_0/2k$ (depends on $[A]_0$). First order: $[A] = [A]_0 e^{-kt}$ or $\ln[A] = \ln[A]_0 - kt$. Plot $\ln[A]$ vs $t$ linear. $t_{1/2} = 0.693/k$ (CONSTANT). Second order: $1/[A] = 1/[A]_0 + kt$. Plot $1/[A]$ vs $t$ linear. $t_{1/2} = 1/(k[A]_0)$ (depends on $[A]_0$). Quick recognition: constant half-life = first order. Units of $k$: first order = s$^{-1}$; second order = L mol$^{-1}$ s$^{-1}$.

2. Half-Life Applications

After $n$ half-lives: fraction remaining $= (1/2)^n$. Examples: $^{14}C$ ($t_{1/2} = 5730$ yr) for carbon dating. $^{131}I$ ($t_{1/2} = 8$ days) for thyroid treatment. $^{238}U$ ($t_{1/2} = 4.5\times10^9$ yr) for geological dating. Drug pharmacokinetics: most drugs cleared by first order kinetics. After 5 half-lives: ~97% eliminated. Radioactive decay is ALWAYS first order (nuclear stability is independent of chemical environment).

3. Arrhenius Equation

$k = Ae^{-E_a/RT}$. $A$ = frequency factor. $E_a$ = activation energy. $\ln k = \ln A - E_a/RT$. Two temperatures: $\log(k_2/k_1) = \frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$. Plot $\log k$ vs $1/T$: slope $= -E_a/2.303R$. Catalyst lowers $E_a$ without changing $\Delta H$ or $K_{eq}$. Doubling rule: rate approximately doubles per 10°C rise (valid for $E_a \approx 50$ kJ/mol near 300 K).

4. Rate Law and Order

Rate law: rate $= k[A]^m[B]^n$ (determined by experiment only). Order = $m+n$. Molecularity (for elementary steps only): number of molecules colliding. For elementary reactions: order = molecularity. For complex reactions: rate-determining step (slowest step) determines observed order. Pseudo-first order: one reactant in large excess appears as constant. Example: hydrolysis of ester in water (water $\approx$ 55 M = constant). Effective $k^\prime = k[H_2O]$.

5. Collision Theory

Rate = collision frequency $\times$ fraction with $E \geq E_a$ $\times$ steric factor. $k = Zp\,e^{-E_a/RT}$. Steric factor $p \ll 1$ for complex molecules. Transition State Theory: reactants form activated complex (transition state) at top of energy barrier. $\Delta G^\ddagger$ determines rate. Catalyst stabilises transition state (lowers $E_a$), not reactants or products.

6. Catalysis

Catalyst: lowers $E_a$, provides alternative pathway, not consumed, does not change $K_{eq}$ (both forward and reverse $k$ increased equally). Homogeneous: H$^+$, enzymes. Heterogeneous: Fe (Haber), V$_2$O$_5$ (Contact), Ni (hydrogenation), Pt (catalytic converter: CO + NO $\to$ N$_2$ + CO$_2$). Enzyme: lock-and-key model. Michaelis-Menten: rate $= V_{max}[S]/(K_m + [S])$. At $[S] \gg K_m$: rate $= V_{max}$ (zero order). At $[S] \ll K_m$: rate $= (V_{max}/K_m)[S]$ (first order).

7. Temperature and Equilibrium

Reaction rate: always increases with $T$ (Arrhenius). Equilibrium constant $K$: exothermic reaction ($\Delta H < 0$): $K$ decreases with $T$ (Le Chatelier). Endothermic: $K$ increases with $T$. Van't Hoff: $d(\ln K)/dT = \Delta H/RT^2$. Important distinction: $E_a$ governs $k$ (kinetics); $\Delta H$ governs $K$ (thermodynamics). Catalyst: changes $k$ but not $K$.

8. Methods to Determine Order

Method of initial rates: vary one concentration at a time, measure initial rates. If doubling $[A]$ doubles rate: first order in A. Integrated rate method: try plotting $[A]$, $\ln[A]$, $1/[A]$ vs $t$; the linear graph gives order. Half-life method: if $t_{1/2}$ constant = first order; if $t_{1/2}$ decreases = zero order; if $t_{1/2}$ increases = second order. Isolation method: use large excess of all reactants except one to isolate each order.

Frequently Asked Questions
1. How does the half-life shortcut work here?
$7.2/0.9 = 8 = 2^3$ means 3 half-lives. $t_{1/2} = 0.693/k = 0.693/0.03 = 23.1$ s. $t = 3 \times 23.1 = 69.3$ s. Works whenever concentration ratio is a power of 2.
2. What is the integrated first order rate law?
$-d[A]/dt = k[A]$. Integrating: $\ln([A]/[A]_0) = -kt$. Rearranging: $t = (2.303/k)\log([A]_0/[A])$. Half-life ($[A] = [A]_0/2$): $t_{1/2} = (2.303/k)\log 2 = 0.693/k$.
3. Why is first order half-life constant?
In first order kinetics, rate $\propto [A]$. As $[A]$ decreases, rate decreases proportionally. The FRACTION consumed per unit time is constant ($= 1 - e^{-k\Delta t}$ for any interval $\Delta t$). This is why the same fraction (50%) is consumed in each $t_{1/2}$, regardless of starting concentration. Compare: zero order has constant rate (not fraction) so $t_{1/2}$ decreases.
4. How do units of k confirm the order?
Rate has units mol L$^{-1}$ s$^{-1}$. Rate $= k[A]^n$. Zero order ($n=0$): $k$ units = mol L$^{-1}$ s$^{-1}$. First order ($n=1$): $k$ units = s$^{-1}$. Second order ($n=2$): $k$ units = L mol$^{-1}$ s$^{-1}$. In this problem: $k = 0.03$ s$^{-1}$ confirms first order.
5. What is pseudo-first order reaction?
A second order reaction behaves as first order when one reactant is in large excess (its concentration is approximately constant). Example: acid hydrolysis of an ester, $k$ apparent $= k[H_2O]$ since water is the solvent ($\approx$ 55 M, essentially constant). This simplification allows easier mathematical analysis and is very common in biochemistry (enzyme reactions at low substrate concentration) and pharmacology.
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