Measuring instantaneous rates as we have described above is the most direct way of determining the rate law of a reaction, but is not always convenient, and it may not even be possible to do so with any precision.

- If the reaction is very fast, its rate may change more rapidly than the time required to measure it; the reaction may be finished before even an initial rate can be observed.
- In the case of very slow reactions, observable changes in concentratons occur so slowly that the idea of a truly "instantaneous" rate loses its meaning.

The rate law tells us how the rate of a reaction depends on the concentrations of the reactants. But for many practical purposes, it is more important to know how the concentrations of reactants (*and* of *products*) change with time.

For example, if you are carrying out a reaction on an industrial scale, you would want to know how long it will take for, say, 95% of the reactants to be converted into products.

This is the purpose of an integrated rate law.

In the terms of calculus, the rate laws we have been discussing are essentially *derivatives* of concentrations

*d*[A]/*dt* = –*k* [A] (i)

and they can thus be integrated with respect to time to give the concentration of a component after a given interval. Integrating the above expression is denoted by

∫*d*[A] = -*k* ∫[A] *dt* (ii)

or

which, by the rules of calculus, gives

ln [A] = –*kt* + *C* (iii)

To evaluate the integration constant *C*, let [A]_{o} be the concentration of the reactant at time 0. This yields

*C* = ln [A]_{o} (v)

so that integrated rate law is

ln [A] = –*kt* + ln [A]_{o} (i)

which we rearrange to

Solving for [A] at time *t*, we obtain the exponential form of the integrated rate law

[A]_{t} = [A]_{o} e^{–kt}

The integrated rate law

ln [A] = –*kt* + ln [A]_{o}(4-1)

has the form of an equation for a straight line

*y* = *mx* + *b*

in which the slope *m* corresponds to the rate constant *k*. This means that, for a first-order reaction, a plot of ln [A] as a function of time gives a straight line with a slope of –*k*.

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value. It is the single most useful parameter for describing the speed of a reaction. Additionally, it provides a convenient way of determining the reaction order.

If the initial concentration of a reactant is [A]_{o}, then after one half-life t_{½} has passed, [A] = ½[A]_{o}.

Making this substitution into the integrated rate law, we obtain

ln ½[A]_{o} = –*kt*_{½} + ln [A]_{o} (4-2)

Combining the [A]_{o} terms gives the simple result

ln ( [A]_{o} / 0.5[A]_{o} ) = ln 2 = 0.693 (4-3)

so the half-life becomes

*t _{½}* = 0.693 /

which shows that the half-life of a first-order reactant is independent of its concentration — an extremely useful and important relationship.

To integrate the second-order instantaneous rate law

we first rearrange it to

Integration yields

To evaluate the integration constant, let [A]_{o} = [A] at time *t* = 0, so that

After one half-life, [A] = ½[A]_{o}

or

The important thing to notice here is that the half-life depends on the initial concentration.

A bimolecular reaction step always follows a second-order rate law

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