Integrated Rate Equations — Revision Notes

⚡ 30-Second Revision

Zero-Order: — $[A]_t = [A]_0 - kt$; $t_{1/2} = \frac{[A]_0}{2k}$; Units of $k$: $ext{mol L}^{-1} \text{s}^{-1}$. Plot: $[A]$ vs $t$ (linear, slope $-k$).
First-Order: — $ln[A]_t = ln[A]_0 - kt$ or k = \frac{2.303}{t} logleft(\frac{[A]_0}{[A]_t}\right); $t_{1/2} = \frac{0.693}{k}$; Units of $k$: $ext{s}^{-1}$. Plot: $ln[A]$ vs $t$ (linear, slope $-k$).
Second-Order ($2A o P$): — $rac{1}{[A]_t} = \frac{1}{[A]_0} + kt$; $t_{1/2} = \frac{1}{k[A]_0}$; Units of $k$: $ext{L mol}^{-1} \text{s}^{-1}$. Plot: $1/[A]$ vs $t$ (linear, slope $k$).

2-Minute Revision

Integrated rate equations are crucial for relating reactant concentrations to time. For zero-order reactions, the rate is constant, and concentration decreases linearly with time ($[A]_t = [A]_0 - kt$).

Its half-life ($t_{1/2} = [A]_0/2k$) is directly proportional to initial concentration. The rate constant $k$ has units of $ext{mol L}^{-1} \text{s}^{-1}$. For first-order reactions, the rate is proportional to concentration, and $ln[A]$ decreases linearly with time ($ln[A]_t = ln[A]_0 - kt$).

Its half-life ($t_{1/2} = 0.693/k$) is constant and independent of initial concentration, a key characteristic. The rate constant $k$ has units of $ext{s}^{-1}$. For second-order reactions (type $2A \to P$), the rate is proportional to $[A]^2$, and $1/[A]$ increases linearly with time ($rac{1}{[A]_t} = \frac{1}{[A]_0} + kt$).

Its half-life ($t_{1/2} = 1/(k[A]_0)$) is inversely proportional to initial concentration. The rate constant $k$ has units of $ext{L mol}^{-1} \text{s}^{-1}$. Remember these formulas, their graphical representations, and half-life dependencies for NEET.

5-Minute Revision

Integrated rate equations are derived from differential rate laws to express reactant concentrations as a function of time. They are essential for determining reaction order, rate constants, and predicting concentrations.

1
Zero-Order Reactions:

* Rate is independent of concentration: $-\frac{d[A]}{dt} = k$. * Integrated Rate Law: $[A]_t = [A]_0 - kt$. This shows a linear decrease in concentration over time. * Half-life ($t_{1/2}$): $t_{1/2} = \frac{[A]_0}{2k}$.

Half-life is proportional to initial concentration. * Units of $k$: $ext{mol L}^{-1} \text{s}^{-1}$. * Graphical Plot: $[A]$ vs. $t$ is a straight line with slope $-k$. * *Example:* If $k=0.1,\text{M s}^{-1}$ and $[A]_0=1.

0, ext{M}$, then$t_{1/2} = rac{1.0}{2 imes 0.1} = 5, ext{s}$.

1
First-Order Reactions:

* Rate is proportional to concentration: $-\frac{d[A]}{dt} = k[A]$. * Integrated Rate Law: $ln[A]_t = ln[A]_0 - kt$ or k = \frac{2.303}{t} logleft(\frac{[A]_0}{[A]_t}\right). * Half-life ($t_{1/2}$): $t_{1/2} = rac{0.

693}{k}$. Half-life is constant and independent of initial concentration – a crucial point. * Units of$k$:$ ext{s}^{-1}$. * Graphical Plot:$ln[A]$vs.$t$is a straight line with slope$-k$. * *Example:* If a first-order reaction has$t_{1/2}=10, ext{min}$, then$k = rac{0.

693}{10} = 0.0693, ext{min}^{-1}$. To find time for$87.5%$completion (i.e.,$12.5%$remaining, which is$1/8$th of initial), it takes 3 half-lives ($[A]_0 o [A]_0/2 o [A]_0/4 o [A]_0/8$). So, time$= 3 imes 10 = 30, ext{min}$.

1
**Second-Order Reactions (for $2A \to P$):**

* Rate is proportional to square of concentration: $-\frac{d[A]}{dt} = k[A]^2$. * Integrated Rate Law: $rac{1}{[A]_t} = \frac{1}{[A]_0} + kt$. This shows $1/[A]$ increases linearly with time. * Half-life ($t_{1/2}$): $t_{1/2} = \frac{1}{k[A]_0}$.

Half-life is inversely proportional to initial concentration. * Units of $k$: $ext{L mol}^{-1} \text{s}^{-1}$. * Graphical Plot: $1/[A]$ vs. $t$ is a straight line with slope $k$. * *Example:* If $k=0.

05, ext{L mol}^{-1} ext{s}^{-1}$and$[A]_0=0.2, ext{M}$, then$t_{1/2} = rac{1}{0.05 imes 0.2} = rac{1}{0.01} = 100, ext{s}$.

Remember to correctly identify the reaction order from the problem statement or given data, choose the appropriate formula, and be careful with logarithmic calculations and units.

Prelims Revision Notes

Integrated rate equations are derived by integrating differential rate laws and are essential for NEET.

1. Zero-Order Reactions:

Rate Law: — Rate $= k$
Integrated Rate Law: — $[A]_t = [A]_0 - kt$
Half-life ($t_{1/2}$): — $t_{1/2} = \frac{[A]_0}{2k}$. (Proportional to initial concentration)
Units of $k$: — $ext{mol L}^{-1} \text{s}^{-1}$ (or $ext{M s}^{-1}$)
Linear Plot: — $[A]$ vs. $t$ (slope $= -k$, intercept $= [A]_0$)

2. First-Order Reactions:

Rate Law: — Rate $= k[A]$
Integrated Rate Law (ln form): — $ln[A]_t = ln[A]_0 - kt$
Integrated Rate Law (log form): — k = \frac{2.303}{t} logleft(\frac{[A]_0}{[A]_t}\right) or $log[A]_t = log[A]_0 - \frac{kt}{2.303}$
Half-life ($t_{1/2}$): — $t_{1/2} = \frac{0.693}{k}$. (Constant, independent of initial concentration)
Units of $k$: — $ext{s}^{-1}$ (or $ext{min}^{-1}$, $ext{hr}^{-1}$)
Linear Plot: — $ln[A]$ vs. $t$ (slope $= -k$, intercept $= ln[A]_0$) or $log[A]$ vs. $t$ (slope $= -k/2.303$, intercept $= log[A]_0$)
Key property: — After 'n' half-lives, [A]_t = [A]_0 left(\frac{1}{2}\right)^n.

3. Second-Order Reactions (for $2A o P$ type):

Rate Law: — Rate $= k[A]^2$
Integrated Rate Law: — $rac{1}{[A]_t} = \frac{1}{[A]_0} + kt$
Half-life ($t_{1/2}$): — $t_{1/2} = \frac{1}{k[A]_0}$. (Inversely proportional to initial concentration)
Units of $k$: — $ext{L mol}^{-1} \text{s}^{-1}$ (or $ext{M}^{-1} \text{s}^{-1}$)
Linear Plot: — $1/[A]$ vs. $t$ (slope $= k$, intercept $= 1/[A]_0$)

General Tips:

Always identify the reaction order first.
Pay attention to units of $k$ and time.
Remember $ln x = 2.303 log x$.
For first-order, if a reaction is $X$ complete, then $(100-X)$ remains. Use this for $[A]_t$ calculation.

Vyyuha Quick Recall

To remember the linear plots for different orders: Zero-order: Zero change in concentration for A (plot A vs. t). First-order: For Log (plot ln A vs. t). Second-order: Second Inverse (plot 1/A vs. t).

And for half-life dependence: Zero: Zero dependence on $k$, but Always on initial A ($[A]_0$). ($t_{1/2} propto [A]_0$) First: Fixed half-life, Independent of initial Intensity ($[A]_0$). ($t_{1/2}$ constant) Second: Shrinking half-life with Increasing initial Intensity ($[A]_0$). ($t_{1/2} propto 1/[A]_0$)