High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Chemistry Textbook > Kinetics

Integrated rate Laws

The Rate LawPrevious Section
Half-lives for Various OrdersNext Section
Popular Courses
Find My Course
0:00 / 0:00

Integrated rate Laws

  • gives [] of reactants and products at any time after the start of the reaction
  • By integrating general rate equations we can obtain integrated rate laws. These laws allow us to predict the concentration of a species over the entire course of a reaction.
  • The general form of the 0th, 1st, and 2nd order integrated rate laws are on your formula sheet and are also shown below. These equations are applied to a system which depends on the concentration of one species [A].
  • Note your formula sheet has assumed that a = 1. Be very careful that if a does not equal 1 you remember to include it.
aA→bB+cCaA \to bB+cC
[A]=[A]0−kt0th[A]=[A]_0-kt \hspace{30pt} 0^{th}
ln⁡[A]=ln⁡[A0]−kt   1st (linear form)\ln\left[A\right]=\ln\left[A_0\right]-kt\ \ \ 1st\ \left(linear\ form\right)
[A]=[A0]e−kt   1st(exp⁡onential form)\left[A\right]=\left[A_0\right]e^{-kt}\ \ \ 1st \left(\exp onential\ form\right)
1[A]=1[A0]+kt   2nd\frac{1}{\left[A\right]}=\frac{1}{\left[A_0\right]}+kt\ \ \ 2nd


[A] is the concentration of a species at time, t.
[A0] is the initial concentration.
t is time in seconds!
  • In order to determine if a reaction is 0th 1st or 2nd order in a substrate, scientists can plot [A] vs t
  • in 0 order: [A] decreases in a linear fashion as t increases, slope=-ak (show how!)



  • in 1st order: [A] decays exponentially with t, larger k will result in
    faster
    (slower/faster) decay, and
    shorter
    (shorter/longer) half life (see half life equation for 1st order to help: 1st order: t1/2=ln2/ak )
  • but if you plot ln[A] vs t you will get a linear decline in [A] as t increases (this is how you can confirm it is 1st order!)
  • this linear plot also has slope=-ak
  • for 2nd order: if we plot 1/[A] vs t we get a straight line with slope=ak

For each of the above integrated rate laws determine what will need to be plotted in order to obtain a straight line graph. What is the significance of the slope and y intercept? Remember that a straight line has the formula y = mx + b

0th order:
[A]vs t[A]vs\ t
m=−akm = -ak
b=[A]ob=[A]_o

1st order:
ln⁡[A]vs t\ln[A]vs\ t
m=−akm=-ak
b=ln⁡[A]ob=\ln[A]_o

2nd order:
1/[A] vs t1/[A]\ vs\ t
m=akm=ak
b=1/[A]ob=1/[A]_o

N2O4 decomposes to NO2 in a first order process with a rate constant of 0.0007 s-1 . How long will it take for 75% of a sample of N2O4 to decompose?



0:00 / 0:00

First Order Reactions

  • For first-order reactions, the differential rate law is:
Rate=k[A]1=k[A]Rate= k[A]^1 = k [A]
  • The integrated rate law for a first-order reaction is:
[A]=[A]0e−kt[A] = [A]_0 e^{-kt}
  • The integrated rate law can be rearranged to a standard linear equation format:
ln[A]=−kt+ln⁡[A]0y=mx+bln[A]=-kt+ln⁡[A]_0\\ y=mx+b
  • A plot of ln[A] versus t for a first-order reaction is a straight line with a slope of −k and an intercept of ln[A]0.


0:00 / 0:00

Second Order Reactions

  • For second-order reactions, the differential rate law is:
Rate=k[A]2Rate= k[A]^2
  • The integrated rate law for a second-order reaction has the form of the equation of a straight line:
1[A]=kt+1[A]0y=mx+b\begin{gathered}\cfrac{1}{\left[A\right]}=kt+\cfrac{1}{\left[A\right]_0} \\ y=mx+b\end{gathered}
  • A plot of 1/[A] versus t for a second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.

Extra Practice
In an industrial plant producing ammonia from hydrogen and nitrogen a chemist decides to run the reaction at 1000oC instead of the normal 850oC and observes an increase in the reaction rate by a factor of 10!
What is the activation energy for this process?
Determine an empirical rate law for the following reaction using the kinetic data collected at 298 K and evaluate the rate constant k.
A+B+C⟶Pr⁡oductsA+B+C⟶\Pr oducts
[A]0(M)[B]0(M)[C]0(M)Initial Rate(M/s)0.1510.2130.3980.4800.2510.1050.3250.3560.1510.2130.5251.1020.1510.2500.4800.988\def\arraystretch{1.5} \begin{array}{c} \hline [A]_0 (M) &[B]_0 (M) &[C]_0 (M)&Initial \ Rate (M/s) \\ \hline 0.151 & 0.213 & 0.398 &0.480 \\ 0.251& 0.105& 0.325 &0.356 \\ 0.151&0.213 & 0.525 &1.102 \\ 0.151&0.250 &0.480 &0.988 \\ \hline \end{array}
A slightly bruised apple will rot extensively in about 4 days at room temperature (20 °C). If it is kept in the refrigerator at 0 °C the same extent of rotting days about 16 days. What is the activation energy for the rotting reaction?