# Reaction Rates and the Rate Law

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```					                                      KINETICS I
Reaction Rates and the Rate Law

By the end of this lecture, you will be able to:
(1)     Understand the difference between kinetics and thermodynamics.
(2)     Understand rate, average rate, initial rate and instantaneous rate.
(3)     Write a generic expression for the rate of a chemical reaction.
(4)     Understand order of reaction.
(5)     Know what differential and integrated rate laws are and how they differ.
(6)     Know how to find integrated rate laws from differential ones for 1st and 2nd
order reactions.

Kinetics vs. Thermodynamics

Thermodynamics

“Will the reaction proceed?”

(i.e., Is it spontaneous under the given conditions?)

Kinetics

“How quickly will the reaction proceed, and by what mechanism?”

e.g.,

C(diamond) → C(graphite)

Spontaneous (∆G° = − 3 kJ mol-1), but very slow (millions of years).

Kinetics is the study of the rates and mechanisms of chemical reactions.
• We cannot determine rate and mechanism by inspection of chemical equations or
from tabulated data.
• Kinetics are path-dependent.

Rate

Rate is change in concentration per unit time (mol L−1 s−1).

Consider the reaction:

A→B

The rate of loss of A (i.e., decrease in [A] over time) is equal to the rate of buildup of B
(i.e., increase in [B] over time).
That is:

d [ A] d [ B ]
−         =
dt     dt

d [ A]    d [ B]
•     −          or        is the instantaneous rate of the reaction at a given time.
dt        dt
•   Rate is always positive.
•   The instantaneous rate very often (but not always) varies with time.
•   Therefore, chemists often talk about initial rate. This is the instantaneous rate at
the beginning of a reaction (t = 0).

Stoichiometry and Rate

Consider the reaction:

2 NO2(g) → 2 NO(g) + O2(g)

Measuring the concentration of each species with time gives the curves given in Fig.
12.1.

The slope of the tangent to a curve at a specific time is the instantaneous rate of loss or
gain of that component at that time.
From the stoichiometry of the reaction, the instantaneous rates of loss or gain of the
individual components are related by:

d [ NO2 ] d [ NO ]    d [O2 ]
−            =         =2
dt       dt          dt
The rate of a general reaction

aA+bB→cC+dD

is

1 d [ A]    1 d [ B ] 1 d [C ] 1 d [ D ]
−            =−          =        =
a dt        b dt       c dt     d dt
Note:

•   Zumdahl describes these rates using ∆ notation.
•   Strictly speaking, this gives the average rate over a specified time span (∆t) and
not the instantaneous rate.
•   Recall from calculus that d means an infinitesimally small increment.
•   By differentiation of concentration with respect to time we get the slope of the
tangent at a point (which is the true instantaneous rate).
•   ∆/∆t approaches d/dt as ∆t → 0.
•   Often, ∆/∆t ≈ d/dt at the beginning of a reaction where the rate is essentially
constant.

The Rate Law

All chemical reactions are reversible. (They’re all equilibria.)

e.g.,
2 NO2(g) = 2 NO(g) + O2(g)

At equilibrium, the forward and reverse reactions have equal rates.

However, at the very beginning of a reaction, the rate may be described in terms of the
concentrations of the reactants only. (The forward reaction dominates − Le Chatelier.)

1 d [ NO2 ]
Rate = −               = k[NO2]n
2 dt
where
k is the rate constant
n is the order with respect to NO2

•   This expression holds only at the beginning of the reaction
•   Both k and n have to be determined experimentally
•   n is most often a small, positive integer (including 0) but may also be negative or
a fraction
•   This is the differential rate law. It describes how the rate depends on
concentration.
Determining k and n

This must be done my experiment!

Two methods:

(1)     Ask “How does the rate depend on the concentration of a reagent during the
course of a single experiment?”

Consider the reaction:

2 N2O5(soln) → 4 NO2(soln) + O2(g)

1 d [ N 2O5 ] d [O2 ]
Rate = −                =        = k[N2O5]n
2     dt         dt

Variation of [N2O5] with time is given in Fig. 12.3.

If ∆t is chosen so that the curve is nearly linear over the range, then

1 d [ N 2O5 ]    1 ∆[ N 2O5 ]
−                 ≈−
2     dt         2 ∆t

Use this equation to calculate rate and then plot rate vs. [N2O5].

Rate of N2O5 loss                                 [N2O5] (mol L−1)
(mol L−1 s−1)
5.4 × 10−4                                          0.90
2.7 × 10−4                                          0.45

The slope of this line is 2k.

So, k = 3.0 × 10−4 s−1 in this example.

If there linear relationship between rate and [N2O5] (i.e., doubling [N2O5] doubles the
rate), then the order with respect to N2O5 (n) is 1. This is the case here.

1 d [ N 2O5 ]                                d [O2 ]
−                                    −4    −1 −1
= 3.0 × [N2O5] × 10 mol L s =
2     dt                                        dt

(2)     The method of initial rates
Ask “How does the initial rate of a reaction vary with initial concentration of a reagent
across multiple experiments?”

Recall: the initial rate is the instantaneous rate when t = 0 (at the beginning of the
reaction).

Consider the reaction:

NH4+(aq) + NO2− → N2(g) + 2 H2O(l)

Rate = k[NH4+]n[NO2−]m

Expt.             [NH4+]0                 [NO2−]0                    Initial rate
(M)                     (M)                        (M s−1)
(1)               0.100                   0.005                     1.35 × 10−7
(2)               0.100                   0.010                     2.70 × 10−7
(3)               0.200                   0.010                     5.40 × 10−7

Method:

(a) Hold [NH4+]0 constant, double [NO2−]0, i.e., compare (1) and (2).

Effect:                         initial rate doubles
Conclusion:                     1st order (linear) with respect to NO2− (m = 1)

(b) Hold [NO2−]0 constant, double [NH4+]0, i.e., compare (2) and (3)

Effect:                         initial rate doubles
Conclusion:                     1st order (linear) with respect to NH4+ (n =1)

(c) Get k from the slope of plot of rate vs. [NH4+]0[NO2−]0 product.

Here, k = 2.7 × 10−4 L mol−1 s−1

Be mindful of units!
In this case, slope has units:
mol L−1 s−1 / (mol L−1)2 = L mol−1 s−1

So, overall rate equation is:

Rate = 2.7×[NH4+][NO2−]×10−7 mol L−1 s−1

The overall order of the reaction = n+m = 2.
The Integrated Rate Law (1st order)

Recall: the differential rate law expresses how rate depends on concentration.

The integrated rate law expresses how concentration depends on time.

Consider the reaction:

A→B

Assume the reaction is 1st order in A.

Thus,
d [ A]
−          = k[ A]
dt

1
d [ A] = −kdt
[ A]
[ A]                          t
1
∫] [ A] d[ A] = −k ∫ dt
[A 0                0

ln[A] − ln[A]0 = −kt (1)

 [ A] 
 [ A ]  = −kt
ln       
 0 
[A] = [A]0e−kt             (2)

Important points:
(a)     The integrated rate equation relates concentration to time. If [A]0 and k are
known, [A] can be calculated at any time.
(b)     Equation (1) can also be written as:

ln[A] = −kt + ln[A]0

So, plotting ln[A] vs. t gives a straight line whose slope is −k and whose
intercept is ln[A]0.

If a reaction is 1st order with respect to a reagent, A, then doubling [A] doubles the rate.
Half Life

The time taken for half of the reactant to be consumed, t1/2.

At this point, [A]/[A]0 = 0.5.

Thus, from equation (2)

0.5 = e−kt
−ln(0.5) = kt1/2
ln(2) = kt1/2

t1/2 = ln(2) / k

The half-life is constant over the course of the reaction.

Nuclear decay is a first order process.

Integrated Rate Law (2nd order)

Consider the reaction:

A→B

Assume the reaction is second order in A.

d [ A]
rate =    −            = k[A]2
dt
d [ A]
= − k[A]2
dt

1
d [ A] = −kdt
[ A]2
[ A]                         t
1
∫] [ A]2 d [ A] = −k ∫ dt
[A 0                  0

1    1
−         +      = − kt
[ A] [ A]0
1    1
=      + kt               (3)
[ A] [ A]0
•   A plot of 1/[A] vs. t is linear with an intercept of 1/[A]0 and a slope of k.
•   Doubling [A] quadruples the rate. Tripling [A] raises the rate by a factor of 9.

For a second order reaction, the half-life is not constant.

When one half-life has elapsed, [A] = [A]0/2

From equation (3), we have:

2                 1
= kt1 / 2 +
[ A]0             [ A]0

1
= kt1 / 2
[ A] 0

1
t1 / 2 =
k [ A] 0
•   The 1st order half-life depends only on k.
•   The 2nd order half-life depends on k and [A]0.
•   t1/2 is not constant.

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