Ch 13 Rates of Reaction (Kinetics)
Kinetics
- Kinetics is the study of reaction rates. Kinetics studies how reaction rates change
under varying conditions, and what molecular events occur during chemical reactions.
- The reaction rate is how fast reactants are converted to products.
- The rate is measured as an increase in molar concentration per unit time of a reaction product,
or as a decrease in molar concentration per unit time of a reactant.
- Reaction rates are affected by the concentrations of reactants, products, and catalysts,
as well as by the areas of reacting surfaces, and by temperature.
Concentration of Reactants
- The rate is generally proportional to the concentration raised to an exponent:
- Rate may be directly proportional to a reactant concentration:
- Or the rate may be unaffected by a reactant:
- The exponent can be negative or a fraction:
Rate  [A]n
[A]1 = ?
[A]0 = ?
[A]-1/2 = ?
Concentration of a Catalyst
- A catalyst is a substance which increases the reaction rate, without being consumed by the reaction.
- A catalyst is not a reactant or product. It is usually written above the reaction arrow,
not on the left or right side of the equation.
- The reaction rate is proportional to its concentration raised to an exponent: Rate  [catalyst]n
Temperature
- Increasing the temperature generally increases reaction rates by increasing
both the kinetic energy and collision frequencies of molecules.
Surface Area of a Solid Reactant or Catalyst
- The reaction rate is proportional to the surface’s concentration, which is expressed as the amount
surface area (m2) per unit volume (cm3). This results in m2/cm3 for the overall units.
2N2O5(g)  4NO2(g) + O2(g)
Expressing the Reaction Rate
-
The progress of the reaction can be observed by formation of O2, and the average rate is
-
The rate is usually not constant. To express an instantaneous rate (as t approaches 0),
use a calculus derivative, so that rate =
-
.
We can also express the reaction rate by its decomposition of N2O5, so that the
average rate is −
-
d[O2 ]
dt
[O2 ]
.
t
[N2 O5 ]
t
.
Note that there is a sign change. (See Ex 13.2)
The rate depends on substance chosen, because there are two N2O5 consumed for each O2 formed.
[O2 ]
2 molecules N O
) (  molecule O2 5 )
t
2
(
= −
[N2 O5 ]
t
or
[O2 ]
)
t
(
1 molecule O2
2 O5
= − (2 molecules N
)
[N2 O5 ]
t
Measurement of the rate requires an indicator of concentration.
- The gas phase decomposition of N2O5 creates five moles of products from two moles of reactants.
The moles created can be calculated from the pressure using the ideal gas law: n = (PV)/(RT).
- NO2 is reddish-brown, while N2O5 and O2 are colorless. So, intensity of color can be used as well.
Rate Law
- The rate law is an equation which relates reaction rate to concentrations of reactants and catalysts.
- For 2NO2 + F2  2NO2F, the equation is: rate = k[NO2][F2].
- k is the rate constant, a proportionality constant relating rate to concentrations.
- For the general reaction aA + bB  dD + eE with catalyst C, the rate is k[A]m[B]n[C]p.
Exponents in the Rate Law
- The exponents are not usually the same as the stoichiometric coefficients.
- Exponents need to be determined experimentally.
- If the exponent is one for a reactant, the reaction is 1st order with respect to that reactant.
An exponent of 2 is 2nd order, and so forth.
- The overall order of reaction is the sum of all exponents in the rate law.
If m, n, and p are all 1 in the general rate law above, then overall order is 3rd (1+1+1=3).
(See Ex 13.3)
Determining a Rate Law
- The rate law must be determined experimentally, with initial rate experiments. (See Ex 13.4)
- Change concentration of one species, while keeping all others constant, to find its exponent.
(
rate2
)=
k[A]m
2
=(
[A]2 m
)
-
Divide the two rate laws for the two trials:
-
If doubling concentration will double rate, then (2) = (2)m, so m = 1.
If doubling concentration will quadruple rate, then (4) = (2)m, so m = 2.
Once all of the exponents are determined, k can be calculated.
k is a function of T, so the temperature dependence of k can be determined experimentally as well.
See Example 13.4.
-
rate1
k[A]m
1
[A]1
Concentration -Time Equations
- A concentration-time equation gives concentration as a function of time.
- The equation can be derived from the rate law by using calculus.
-
Concentration vs. Time can be found for 1st order where rate =
-
This rearranges to
-
Which integrates to
-
And then rearranges to
dA
[A]
─ d[A]
dt
= k[A]1
= ─ (k)dt
ln[A] ─ ln[A]0 = ─ k(t) and
[A]
ln ([A] ) = ─ kt
and
0
ln[A] = ln[A]0 ─ k(t)
[A] = [A]0(e-kt)
Half-Life
- The half-life is the time when the concentration of a reactant is half of it original value.
- At t1/2, we have [A]1/2 = (½)[A]0 .
- The half-life can be found from the concentration-time equation.
[A]
½[A]0
)=
[A]0
-
For 1st order, we have
─ kt1/2 = ln ( [A]½ ) = ln (
-
This rearranges to
t1/2 =
0
𝟎.𝟔𝟗𝟑
𝐤
for 1st order.
ln(½) = ─ 0.693
2nd Order Concentration-Time Equation
d[A]
= ─ k[A]2
-
Derive from rate law
rate =
-
This rearranges to
− [A]2 = (k)dt
-
Which integrates to
-
And then rearranges to
dt
dA
1
[A]
𝟏
[𝐀]
−
=
Second Order Half-Life
1
[A]1/2
-
kt1/2 =
Use conc-time equation
1
[A]0
𝟏
[𝐀]𝟎
=
= kt
+ 𝐤𝐭
1
=
1
( )[A]0
2
2
[A]0
𝟏
−
1
[A]0
2
[A]0
1
= [A]
0
nd
-
This rearranges to
-
Note that t1/2 is now dependent on the original concentration (unlike 1st order).
t1/2 = 𝐤[𝐀]
𝟎
for 2 order
Graphing Concentration-Time Equations
- For 1st order, use ln[A] = ln[A]0 ─ k(t) to create a straight line (y = b + mx).
- The plot is ln[A] versus t, where ln[A]0 is the y-intercept (b), and ─ k is the slope (m).
0.5
y = Ln[A]
y = mx + b
m=-k
b = ln[A]0
0.4
0.3
0.2
0.1
0
0
200
400
600
800
1000
x = time (seconds)
-
1
1
+ kt to create the straight line.
[A]0
1
where [A] is the y-intercept (b), and +k is
0
For 2nd order, use [A] =
1
Plot is [A] versus t,
0.6
y = mx + b
m = +k
0.5
y = 1/[A]
the slope (m).
0.4
0.3
0.2
b = 1 / [A]0
0.1
0
0
500
1000
x = time (seconds)
-
See the summary of 1st and 2nd order equations in Table 13.2 of the text.
Effect of Temperature on Rate
Collision Theory
- For a reaction to occur and create products, the reactant molecules must collide.
When they collide, they must possess more than the activation energy (Ea),
and also have the proper orientation. (See Figure 13.12)
- According to Collision Theory, the rate constant (k) is the product of:
1) the collision frequency (Z)
2) the fraction (f) of molecules with more than Ea
3) the fraction (p) of colliding molecules with the proper orientation
- The overall equation is:
k = Zfp
3RT
-
Z is proportional to molecular speed, which is √
-
R is the universal gas constant, but in units of energy per unit temperature, R = 8.314 J/(K·mol)
So, Z is proportional to T1/2, but k actually increases much faster than that with T.
That is, k ≈ doubles with each 10 oC rise in T. (See Figure 13.11)
The dependence of k on T is related primarily to the activation energy (Ea).
f = e─ (Ea/RT) where e ≈ 2.718
or
ln(f) = ─ Ea/RT.
Compared with the equation for Z, the equation for f produces a larger increase in k as T increases.
p is a simple constant (p ≈ 1/2 in Figure 13.12). It relates molecular orientation during collision.
For NO + Cl2  NOCl + ½Cl2, the N must strike Cl2. If O strikes Cl2, there is no reaction.
-
-
MM
where MM is the molar mass in kg/mole.
Transition State Theory
- Colliding reactant molecules form an activated complex.
- The activated complex (‡) is an unstable molecular formation that can become products.
O=N + Cl2  [O=N···Cl···Cl]‡  O=N–Cl + Cl·
- In this complex, the reactant bonds are partially broken, and the product bonds are partially formed.
- The activated complex can exist when the molecules have their ground state E plus Ea.
- The enthalpy (or heat) of a reversible reaction can be found from the difference between
the Ea’s for the forward and reverse reactions:
Hreaction = Ea(fwd) – Ea(rev)
- For an endothermic reaction, H is positive. See Figure 13.13.
- For an exothermic reaction, H is negative. See Figure 13.14.
Arrhenius Equation
k = Ae– (Ea/RT)
- A is called the frequency factor. Nearly constant. Equivalent to Z times p (from Collision Theory)
- Exponential portion is equivalent to f, where f = e – (Ea/RT).
Ea
1
ln(k) = ln(A) – ( R ) (T)
-
Graph equation by taking ln of both sides:
-
Plot ln(k) vs. (1/T) to get straight line (y vs. x)
Ea
R
ln(A) is y-intercept and – ( ) is the slope
-
Also, subtract equations at two different temperatures to solve for Ea (or for k at different T)
k
Ea
1
k1
R
T1
ln(k2) – ln(k1) = ln ( 2 ) = ( ) (
(See Example 13.7)
1
−T )
2
Elementary Reactions
- An elementary reaction is a single molecular event which results in a reaction.
It is either a single collision of two molecules or a single molecule breaking into smaller pieces.
- A reaction mechanism is a series of elementary reactions which
combine in a sequence to create the net chemical equation.
- The net reaction NO2(g) + CO(g) NO(g) + CO2(g) has two-steps in its mechanism.
NO2 + NO2 NO3 + NO is followed by NO3 + CO NO2 + CO2 (two elementary reactions)
- NO3 is a reaction intermediate. It is produced in one elementary reaction, but does not appear
in the net chemical equation because it is consumed in a subsequent elementary reaction.
- The sum of the two steps is NO2(g) + NO2(g) + NO3(g) + CO(g)  NO3(g) + NO(g) + NO2(g) + CO2(g)
After cancelling duplicated molecules, the sum reduces to the net chemical equation.
(See Example 13.8)
Molecularity is the number of reactant molecules involved in an elementary reaction.
(1) Unimolecular reaction
O3*  O2 + O
(2) Bimolecular reaction
NO2 + NO2  NO3 + NO
(3) Termolecular reaction (very rare)
Br2  Br + Br, then Br + Br + Ar  Br2 + Ar
Rate Equation for an Elementary Reaction
- The rate is directly proportional to concentration of reactant molecules for an elementary reaction.
- For A  B + C, rate = k[A]
- For A+ B  C + D, rate = k[A][B]
Rate Equation for a Net Chemical Equation
- For the net chemical equation, the rate is determined by the slowest elementary reaction.
- The slowest elementary rxn is called the “rate determining step”
(1)
NO2 + F2 NO2F + F
(slow, w/ rate constant k1)
(2)
F + NO2  NO2F
(fast, w/ k2)
Net: 2NO2 + F2  2NO2F
Rate = k1[NO2][F2]
- The overall rate for the net chemical equation is the same as the rate for equation (1).
(See Example 13.11)
Mechanisms with an Initial Fast (Equilibrium) Step
- Net: 2N2O5(g) 4NO2(g) + O2(g)
Net Rate = k[N2O5(g)]
(1)
N2O5(g) ⇌ NO2(g) + NO3(g)
(fast, stays at eqm, w/ rate constants k1 and k-1)
(2)
NO2(g) + NO3(g)  NO(g) + NO2(g) + O2(g) (slow, w/ rate constant k2)
(3)
NO3(g) + NO(g)  2NO2(g)
(fast, w/ rate constant k3)
- Rate determining step gives us rate = k2[NO2(g][NO3(g)], but NO3(g) is an intermediate,
and intermediates usually have very small concentrations which are difficult to measure.
- Since equation (1) is at equilibrium, we know that its forward rate equals its reverse rate.
(k1)[N2O5(g)] = (k-1)[NO2(g)][NO3(g)]
-
Rate = k2[NO2(g][NO3(g)] =
Rate =
k k
( 2 1 )[N2O5(g)]
k−1
and
k2 [NO2(g) ]× k1 [N2 O5(g) ]
k−1 [NO2(g) ]
(See Example 13.12)
[NO3(g)] =
k k
= ( k2 1 )
−1
k1 [N2 O5(g) ]
k−1 [NO2(g) ]
[NO2(g) ][N2 O5(g) ]
[NO2(g) ]
Catalysis
- A catalyst increases the reaction rate without being consumed.
It is used in one elementary reaction, but is then regenerated in a subsequent step.
- The catalyst provides an alternate pathway for a reaction.
The alternate pathway has a lower Ea than the uncatalyzed reaction.
Industrial H2SO4 Production:
2SO2 + O2  2SO3 (w/ NO as the catalyst)
Alternate (catalyzed) pathway:
2NO + O2  2NO2 followed by NO2 + SO2  NO + SO3
- Decomposition of O3 (ozone) in the stratospheric ozone layer occurs due to
residual chlorine atoms from chlorofluorocarbon refrigerants like CCl2F2 (R-12).
(1)
Cl(g) + O3(g) ClO(g) + O2(g)
(2)
ClO(g) + O(g)  Cl(g) + O2(g)
Net:
O3(g) + O(g)  2O2(g)
There is no Cl in the net equation. Cl is a catalyst. It is regenerated and is not consumed.
- If the catalyst and all reactants are in the same phase (gas or liquid solution), then the reaction
has homogeneous catalysis. The ozone decomposition above has homogeneous catalysis
because all of the molecules involved are in the gas phase.
- Industrial reactions typically use a solid catalyst with fluid (gaseous) reactants, which is
heterogeneous catalysis. Gases bind to surface by chemisorption (chemically bonded to
the catalyst surface) in order to convert to products at that surface.
See the sequence of a catalytic hydrogenation reaction in Figure 13.18.
- Enzymes are large protein molecules, with specific 3D geometries, and are biological catalysts.
They convert Substrate into Products (E+SESE+P), with lock and key mechanism.
(See Figure 13.20)