Nth-Order Half-Life Derivation

The half-life $t_{1/2}$ is the time required for the concentration of a reactant to fall to half its initial value. For a general reaction $A \longrightarrow \text{Products}$ with rate law:

$$-\frac{d[A]}{dt} = k[A]^n$$

where $n$ is the reaction order (any real number), we derive the general half-life formula.

1. Case 1: $n \neq 1$ (General Case)

Step 1 — Set up the differential equation

$$-\frac{d[A]}{dt} = k[A]^n$$

Step 2 — Separate variables

$$-\frac{d[A]}{[A]^n} = k,dt$$

Using $[A]^{-n},d[A]$:

$$-[A]^{-n},d[A] = k,dt$$

Step 3 — Integrate both sides

Integrate from initial concentration $[A]_0$ at $t = 0$ to concentration $[A]_t$ at time $t$:

$$-\int_{[A]_0}^{[A]_t} [A]^{-n},d[A] = k \int_0^t dt$$

Recall $\int x^{-n} dx = \frac{x^{-n+1}}{-n+1}$ for $n \neq 1$:

$$-\left[ \frac{[A]^{-n+1}}{-n+1} \right]_{[A]_0}^{[A]_t} = kt$$

$$-\frac{[A]_t^{1-n}}{1-n} + \frac{[A]_0^{1-n}}{1-n} = kt$$

Simplify:

$$-\frac{[A]_t^{1-n} - [A]_0^{1-n}}{1-n} = kt$$

Step 4 — Simplify

Multiply numerator and denominator by $-1$:

$$\frac{[A]_t^{1-n} - [A]_0^{1-n}}{n-1} = kt$$

Or more commonly written as:

$$\boxed{[A]_t^{1-n} = [A]_0^{1-n} + (n-1)kt} \quad (n \neq 1)$$

This is the integrated rate law for an $n$th-order reaction.

[!check] Sanity check — $n = 2$ Set $n = 2$: $[A]_t^{1-2} = [A]_0^{1-2} + (2-1)kt$ $$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$ Matches the known second-order integrated rate law. ✓

[!check] Sanity check — $n = 0$ Set $n = 0$: $[A]_t^{1} = [A]_0^{1} + (0-1)kt$ $$[A]_t = [A]_0 - kt$$ Matches the known zero-order integrated rate law. ✓

Step 5 — Apply half-life condition

At $t = t_{1/2}$, the concentration has dropped to half the initial value:

$$[A]_t = \frac{1}{2}[A]_0$$

Substitute into the integrated rate law:

$$\left(\frac{1}{2}[A]_0\right)^{1-n} = [A]0^{1-n} + (n-1)kt{1/2}$$

Step 6 — Solve for $t_{1/2}$

Factor $[A]_0^{1-n}$ out of the left term:

$$[A]_0^{1-n}\left(\frac{1}{2}\right)^{1-n} = [A]0^{1-n} + (n-1)kt{1/2}$$

Bring $[A]_0^{1-n}$ to the left:

$$[A]0^{1-n}\left[\left(\frac{1}{2}\right)^{1-n} - 1\right] = (n-1)kt{1/2}$$

Step 7 — Simplify the exponent

Note that $\left(\frac{1}{2}\right)^{1-n} = 2^{n-1}$ because:

$$\left(\frac{1}{2}\right)^{1-n} = 2^{-(1-n)} = 2^{n-1}$$

Therefore:

$$[A]0^{1-n}\left[2^{n-1} - 1\right] = (n-1)kt{1/2}$$

Step 8 — Final formula

$$\boxed{t_{1/2} = \frac{2^{n-1} - 1}{(n-1),k,[A]_0^{,n-1}}} \quad (n \neq 1)$$

2. Case 2: $n = 1$ (First Order)

The integrated rate law for $n = 1$ is derived separately because the integration of $[A]^{-1}$ gives $\ln[A]$:

$$-\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = k \int_0^t dt$$

$$-\big[\ln[A]\big]_{[A]_0}^{[A]_t} = kt$$

$$\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$$

or

$$[A]_t = [A]_0 e^{-kt}$$

Apply half-life condition $[A]_t = \frac{1}{2}[A]_0$:

$$\ln\left(\frac{[A]_0}{\frac{1}{2}[A]0}\right) = k t{1/2}$$

$$\ln 2 = k t_{1/2}$$

$$\boxed{t_{1/2} = \frac{\ln 2}{k}} \quad (n = 1)$$

3. Summary of All Orders

Order ($n$) Half-Life Formula Dependence on $[A]_0$
0 $\displaystyle t_{1/2} = \frac{[A]_0}{2k}$ Proportional to $[A]_0$
1 $\displaystyle t_{1/2} = \frac{\ln 2}{k}$ Independent of $[A]_0$
2 $\displaystyle t_{1/2} = \frac{1}{k[A]_0}$ Inversely proportional to $[A]_0$
$n$ ($n \neq 1$) $\displaystyle t_{1/2} = \frac{2^{n-1} - 1}{(n-1)k[A]_0^{n-1}}$ Inversely proportional to $[A]_0^{n-1}$

Key observations

  • First order is the only order with a constant half-life (independent of concentration).
  • For $n > 1$: half-life increases as the reaction proceeds (because $[A]_0$ decreases for successive half-lives).
  • For $n < 1$: half-life decreases as the reaction proceeds.
  • For $n = 0$: plugging $n = 0$ into the general formula gives: $$t_{1/2} = \frac{2^{-1} - 1}{(0-1)k[A]_0^{-1}} = \frac{\frac{1}{2} - 1}{-k/[A]_0} = \frac{-\frac{1}{2}}{-k/[A]_0} = \frac{[A]_0}{2k}$$ which matches. ✓

4. Worked Examples

Example 1: Third-Order Reaction ($n = 3$)

For $A \longrightarrow \text{Products}$ with rate $= k[A]^3$, derive the half-life formula.

Solution:

Using the general formula with $n = 3$:

$$t_{1/2} = \frac{2^{3-1} - 1}{(3-1)k[A]_0^{3-1}} = \frac{2^{2} - 1}{2k[A]_0^{2}} = \frac{4 - 1}{2k[A]_0^{2}} = \frac{3}{2k[A]_0^{2}}$$

So for a third-order reaction:

$$t_{1/2} = \frac{3}{2k[A]_0^2}$$

Example 2: Half-Order Reaction ($n = \frac{1}{2}$)

For $A \longrightarrow \text{Products}$ with rate $= k[A]^{1/2}$, find the half-life.

Solution:

$$t_{1/2} = \frac{2^{1/2 - 1} - 1}{(\frac{1}{2} - 1)k[A]_0^{1/2 - 1}} = \frac{2^{-1/2} - 1}{(-\frac{1}{2})k[A]_0^{-1/2}}$$

$$t_{1/2} = \frac{\frac{1}{\sqrt{2}} - 1}{-\frac{k}{2\sqrt{[A]_0}}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{k}{2\sqrt{[A]_0}}} = \frac{2\sqrt{[A]_0}}{k}\left(1 - \frac{1}{\sqrt{2}}\right)$$

$$t_{1/2} = \frac{2(\sqrt{2} - 1)\sqrt{[A]_0}}{\sqrt{2},k} = \frac{(2 - \sqrt{2})\sqrt{[A]_0}}{k}$$

Example 3: Determining Order from Half-Life Data

A reaction has the following half-lives at different initial concentrations:

$[A]_0$ (M) $t_{1/2}$ (s)
0.10 120
0.050 240
0.025 480

Analysis: When $[A]0$ is halved, $t{1/2}$ doubles. This means $t_{1/2} \propto 1/[A]_0$, which from the table corresponds to $n-1 = 1$ or $n = 2$.

Check: $t_{1/2} = \frac{1}{k[A]_0}$. Using the first row: $k = \frac{1}{120 \times 0.10} = 0.0833\ \text{M}^{-1}\text{s}^{-1}$. Verify with second row: $\frac{1}{0.0833 \times 0.050} = 240\ \text{s}$ ✓

5. Derivation at a Glance (Quick Reference)

Step Operation
1 Start with $-\frac{d[A]}{dt} = k[A]^n$
2 Separate: $-[A]^{-n},d[A] = k,dt$
3 Integrate: $\displaystyle -\int_{[A]_0}^{[A]_t} [A]^{-n},d[A] = k \int_0^t dt$
4 Get: $\displaystyle \frac{[A]_t^{1-n} - [A]_0^{1-n}}{n-1} = kt$
5 Set $[A]_t = \frac{1}{2}[A]_0$: $\displaystyle \frac{(\frac{1}{2}[A]_0)^{1-n} - [A]0^{1-n}}{n-1} = kt{1/2}$
6 Factor: $\displaystyle [A]0^{1-n}\frac{2^{n-1} - 1}{n-1} = kt{1/2}$
7 Solve: $\displaystyle t_{1/2} = \frac{2^{n-1} - 1}{(n-1)k[A]_0^{n-1}}$
8 For $n = 1$, derive separately: $\displaystyle t_{1/2} = \frac{\ln 2}{k}$

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