How do you find the integrated rate law for a first-order reaction?

Given a first order reaction

#aA -> bB#,

its rate law is:

#r(t) = k[A]#

where:

• #r(t)# is the initial rate.
• #k# is the rate constant in the appropriate units.
• #[A]# is the current concentration of #A#.
• #a# and #b# are stoichiometric coefficients.

#A# will disappear at the rate of the reaction, based on the stoichiometric coefficient in front of #A#. Hence, we put a negative sign in front of its rate to equate it with the forward reaction rate.

#r(t) = k[A] = underbrace(overbrace(-1/a)^((-)) overbrace((d[A])/(dt))^((-)))_((+))#

By separation of variables,

#-akdt = 1/([A])d[A]#

Integration of both sides from the initial to the final state gives:

#-akint_(0)^(t)dt = int_([A]_0)^([A]) 1/([A])d[A]#

where #[A]_0# is the initial concentration and #[A]# is the current concentration.

This results in:

#-ak(t-0) = ln[A] - ln[A]_0#

Therefore, the integrated rate law for a general first-order reaction with arbitrary stoichiometric coefficients is:

#color(blue)(barul|stackrel(" ")(" "ln[A] = -akt + ln[A]_0" ")|)#

If the stoichiometric coefficient in front of #A# happens to be #1#, then

#ln[A] = -kt + ln[A]_0#

which may look more familiar.