# How do you find the integrated rate laws for zeroth, first, and second order reactions?

May 3, 2017

We integrate the differential rate laws.

#### Explanation:

Here's how to do it.

Integrated rate law for zero order reaction

The differential rate law is

"-"(d["A"])/(dt) = k

Rearrange this to give

d["A"] = "-"kdt

Integrate both sides of the equation.

int_(["A"]_0)^(["A"]_t)d["A"] = "-"kint_0^tdt

["A"]_t color(white)(l)– ["A"]_0 = "-"kt

Solve for ${\left[\text{A}\right]}_{t}$. This gives the zeroth-order integrated rate law:

${\left[\text{A"]_t = ["A}\right]}_{0} - k t$

Integrated rate law for first order reaction

The differential rate law is

"-"(d["A"])/(dt) = k[A]

Rearrange this to give

1/(["A"])d["A"] = "-"kdt

Integrate both sides of the equation.

int_(["A"]_0)^(["A"]_t)1/(["A"])d["A"] = "-"kint_0^tdt

ln["A"]_t color(white)(l)– ln["A"]_0 = "-"kt

Solve for $\ln {\left[\text{A}\right]}_{t}$. This gives the first-order integrated rate law:

$\ln {\left[\text{A"]_t = ln["A}\right]}_{0} - k t$

Integrated rate law for second-order reaction

The differential rate law is

"-"(d["A"])/(dt) = k["A"]^2

Rearrange this to give

"-"(d["A"])/(["A"]^2) = kdt

Integrate both sides of the equation.

"-"int_(["A"]_0)^(["A"]_t)(d["A"])/(["A"]^2) = kint_0^tdt

$\frac{1}{{\left[\text{A"]_t )color(white)(l) – 1/(["A}\right]}_{0}} = k t$

The final version of the second-order integrated rate law is:

$\frac{1}{{\left[\text{A"]_t ) = 1/(["A}\right]}_{0}} + k t$