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

1 Answer

Answer:

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#