Question

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

Answer 1

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 `["A"]_t`. This gives the zeroth-order integrated rate law:

`["A"]_t = ["A"]_0 -kt`

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["A"]_t`. This gives the first-order integrated rate law:

`ln["A"]_t = ln["A"]_0 -kt`

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`

`1/(["A"]_t )color(white)(l) – 1/(["A"]_0) = kt`

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

`1/(["A"]_t ) = 1/(["A"]_0) + kt`