First Order Reaction
Chemical Kinetics Class 12 | Rate Laws, Half-Life & Graphs
1. Definition & Rate Law
A First Order Reaction is a reaction in which the rate depends on the first power of the concentration of the reactant.
Reaction: $R \rightarrow P$
Rate law: $\text{Rate} = -\frac{d[R]}{dt} = k[R]^1$
Rate law: $\text{Rate} = -\frac{d[R]}{dt} = k[R]^1$
Examples: Radioactive decay, Decomposition of $N_2O_5$, Hydrogenation of ethene.
2. Integrated Rate Equation
Integrating the differential rate equation gives us the relation between concentration and time.
$$ \ln[R] = -kt + \ln[R]_0 $$Converting natural log ($\ln$) to base 10 log ($\log$):
$$ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} $$
Exponential Form: $[R] = [R]_0 e^{-kt}$
Where $[R]_0$ is initial concentration and $[R]$ is concentration at time $t$.
3. Half-Life ($t_{1/2}$)
The time required for the concentration of a reactant to reduce to half of its initial value. At $t = t_{1/2}$, $[R] = [R]_0/2$.
$$ k = \frac{2.303}{t_{1/2}} \log \frac{[R]_0}{[R]_0/2} $$ $$ t_{1/2} = \frac{2.303 \log 2}{k} = \frac{2.303 \times 0.301}{k} $$
$$ t_{1/2} = \frac{0.693}{k} $$
Key Point: For a first-order reaction, half-life is independent of the initial concentration of the reactant.
4. Important Graphs
- $\ln[R]$ vs $t$: A straight line with negative slope $= -k$ and intercept $= \ln[R]_0$.
- $\log[R]$ vs $t$: A straight line with negative slope $= \frac{-k}{2.303}$.
5. Useful Relations for JEE/NEET
- Amount left after $n$ half-lives: $[A] = \frac{[A]_0}{2^n}$
- Time for 75% completion: $t_{75\%} = 2 \times t_{50\%}$
- Time for 99.9% completion: $t_{99.9\%} \approx 10 \times t_{50\%}$
- Unit of Rate Constant ($k$): $time^{-1}$ (e.g., $s^{-1}, min^{-1}$)
Practice Quiz
Test your concepts on First Order Kinetics.
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