INTEGRATED RATE LAWS02/13 /13

AP Chemistry

Integrated vs. Differential Rate Laws

Differential Rate Law: A rate law that expresses how the rate depends on the concentration of the reactants. (Often simply called “a rate law.”)

Integrated Rate Law: A rate law that expresses how the concentrations depend on time.

First Order Rate Laws

Consider the reaction: aA productsThe rate is -∆[A]/ ∆t = k[A]n

Since it is often useful to consider the change in concentration in terms of time, the following law is derived from the previous equation by some sort of miracle.

ln[A] = -kt + ln[A]0

ln = natural log, k = rate constant, t = time, [A] is at time t, [A]0 is at time 0 (the start of the experiment).

ln[A] = -kt + ln[A]0

This equation shows how the concentration of A depends on time. If the initial concentration and k are known, the concentration of A at any time can be calculated.

ln[A] = -kt + ln[A]0

Note that the form of this equation is y = mx + b.

y = ln[A] x = t m = -k b = ln[A]0

Therefore, plotting the natural log of concentration vs. time will always produce a straight line for a first order reaction.

If the plot is a straight line, the reaction is first order; if the plot is not a straight line, the reaction is not first order.

ln[A] = -kt + ln[A]0

The integrated rate law can also be expressed in terms of a ratio of [A] and [A]0.

ln([A]0/A]) = kt

First Order Rate Laws

Consider 2N2O5(g) 4 NO2(g) + O2(g)

Using this data, verify that the rate law is first order and calculate k.

[N2O5] (mol/L) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

-6

-4

-2

100 200 300 400

First Order Rate Laws

Since the reaction is first order, then the slope of the line is equal to -k = ∆y/ ∆x = ∆(ln[N2O5])/ ∆t

Since the first and last points are exactly on the line, we will use them to calculate slope.

k = -(slope) = 6.93 x 10-3 s-1

First Order Rate Law

Using the previously obtained data, calculate [N2O5] at 150 seconds after the start of the reaction.

Half Life of a First Order Reaction

Half Life: the time required for a reactant to reach half its original concentration.

ln([A]0/A]) = ktBy definition at t =½: [A] = [A]0/2

ln(2) = kt½

t½ = 0.693/k

Half Life for First Order Reaction

A certain first order reaction has a half-life of 20.0 minutes.

Calculate the rate constant for this reaction.How much time is required for this reaction

to be 75% complete?

ln([A]0/A]) = kt

The decomposition of A is first order, and [A] is monitored. The following data are recorded:

Calculate the rate constant. Calculate the half life. Calculate [A] when t = 5 min.

How long will it take for 90% of A to decompose?

T (min) 0 1 2 4

[A] 0.100 0.0905 0.0819 0.0670

Reaction Rate: ConcentrationNH4

+(aq) + NO2-(aq) N2(g) + 2H2O(l)

The reaction 2ClO2 (aq) + 2OH- (aq) ClO3- (aq) +

ClO2- (aq) + H2O (l) was studied with the

following results.Exp [ClO2] [OH-] Rate (M/s)1 0.060 0.030 0.02482 0.020 0.030 0.002763 0.020 0.090 0.00828

(a) Determine the rate law(b) Calculate the value of the rate law constant