Chem 110 Reci – Week 7 – Kinetics II
More Linear Regression/Integrated Rate Law Examples:
1. The hydrolysis of table sugar (sucrose) occurs by the following overall
reaction:
C12H22O11(s) + H2O(l)  C6H12O6(aq, glucose) + C6H12O6(aq, fructose)
A nutritional biochemist studes the kinetics of the process and obtains the
following data:
[Sucrose] mol/L Time (h)
0.501
0
0.451
0.50
0.404
1.00
0.363
1.50
0.267
3.00
a) use the data to determine the rate constant and the half-life of the
reaction.
Zero order
First Order
Second Order
Slope
-0.077
-0.210
0.590
Corr
-0.994
-0.9998
0.9967
First order, k = 0.210 (-slope)
t1/2 = ln2/k = ln2/0.210 = 3.30 sec.
b) How long does it take to hydrolyze 75% of the sucrose?
75% = 2 half lives = 6.60 sec.
2. In a first order decomposition reaction, 50.0% of a compounds decomposes in
10.5 min.
a) What is the rate constant of the reaction?
t1/2 = 10.5 min = ln2/k  k = 0.0660 min-1
b) How long does it take for 75.0% of the compound to decompose?
75% = 2 half lives = 21 min.
3. A decomposition reaction has a rate constant of 0.0012 yr-1.
a) What is the half-life of the reaction?
First order (yr-1)  t1/2 = ln2/k = ln2/0.0012 yr-1 = 577.6 yr.
b) How long does it take for the reactant to reach 12.5% of it’s original
value?
50%
12.5%

 X  144.4 yr
0.0012
X
4. A biochemist studying the breakdown in soil of the insecticide DDT finds that it
decomposes by a first-order reaction with a half-life of 12 yr. How long
does it take DDT to decompose from 275 ppbm to 10 ppbm in a soil
sample?
t1/2 = 12yr = ln2/k  k = 0.058 yr-1
 [ A] 
 275 ppm 
  (0.058 yr 1 )t  t  57 yr
ln  0   kt  ln 
 10 ppm 
 [ A]t 
5. Aspirin in broken down in the body by a first-order process. The half-life of
aspirin in elderly humans is 3.7 hr.
a) How much aspirin remains in the bloodstream from a 160-mg dose after
24 hours?
t1/2 = 3.7 hr = ln2/k  k = 0.1.87 hr-1
 [ A]0 
 8.88 x10 4 M 
 8.88 x10 4 M 
  kt  ln 
  (0.187hr 1 )( 24hr )  ln 
  4.488
ln 
X
X




 [ A]t 
8.88 x10  4 M
 e 4.488  X  9.99 x10 6
X
b) In young humans, the half-life is 2.4 hr. How much aspirin remains in a
young person 24 hours after the same dose?
t1/2 = 2.4 hr = ln2/k  k = 0.289 hr-1
 [ A]0 
 8.88 x10 4 M 
 8.88 x10 4 M 
  kt  ln 
  (0.289hr 1 )( 24hr )  ln 
  6.936
ln 
X
X




 [ A]t 

8.88 x10  4 M

 e 6.936  X  8.63x10 7 M
X
More Initial Rate Examples:
1. a) Give the individual reaction order for each substance and the overall
reaction order for the following rate laws:
[O3 ]2
+
2
i) rate = k[BrO3 ][Br ][H ] , ii) rate  k
[O2 ]
b) By what factor does rate law i) change if:
i) the [Br-] is halved, ii) the [H+] is quadrupled
i) the rate is halved. ii) increase by a factor of 16
c) By what factor does rate law ii) change if:
i) the [O3] is doubled, ii) the [O2] is doubled
i) increase by a factor of 4; ii) decrease by a factor of 2
2. For the reaction
4A(g) + 3B(g)  2C(g)
the following data were obtained at constant temperature:
Exp’t [A] mol/L [B] mol/L Rate mol/L*min
1
0.100
0.100
5.00
2
0.300
0.100
45.0
3
0.100
0.200
10.0
4
0.300
0.200
90.0
a) What is the order with respect to each reactant?
rate  k[ A] x [ B] y
x
rate2 45.0 [0.300] x [0.100] y


 9  3x  x  2
rate1 5.00 [0.100] x [0.100] y
rate3 10.0 [0.100] x [0.200] y


 2  2y  y 1
rate1 5.00 [0.100] x [0.100] y
b) Write the rate law.
y
rate = k[A]2[B]
c) Calculate the rate constant.
90.0 = k[0.300]2[0.200]  k = 5000 L2/mol2*s
3. For the reaction
A(g) + B(g) + C(g)  D(g)
the following data were obtained at constant temperature:
Exp’t [A] mol/L [B] mol/L [C] mol/L Initial Rate (M/L*s)
1
0.0500
0.0500
0.0100
6.25x10-3
2
0.1000
0.0500
0.0100
1.25x10-2
3
0.1000
0.1000
0.0100
5.00x10-2
4
0.0500
0.0500
0.0200
6.25x10-3
a) What is the order with respect to each reactant?
rate  k[ A] x [ B ] y [C ] z
x
rate2 1.25x10 -2 [0.1000] x [0.0500] y [0.0100] z


 2  2x  x  1
-3
x
y
z
rate1 6.25x10
[0.0500] [0.0500] [0.0100]
y
rate3 5.00x10 -2 [0.1000] x [0.1000] y [0.0100] z


 4  2y  y  2
-2
x
y
z
rate2 1.25x10
[0.1000] [0.0500] [0.0100]
rate4 6.25x10 -3 [0.0500] x [0.0500] y [0.0200] z


 0  2z  z  0
-3
x
y
z
rate1 6.25x10
[0.0500] [0.0500] [0.0100]
b) Write the rate law.
Rate = k[A][B]2
c) Calculate the rate constant.
6.25x10-3 = k[0.0500][0.0500]2[0.0200]0  k = 50 L2/mol2*s
4. Give the units of the rate constant for the following orders:
a) first
b) fifth
c) third
d) 5/2
-1
4
-4
-1
2
-2
a) time ; b) L *mol *time ; c) L *mol *time-1; d) L3/2*mol-3/2*time-1
5. Give the overall reaction order that corresponds to rate constants with the
following units:
a) mol/L*s b) yr-1
c) (mol/L)1/2*s-1
d)(mol/L)-5/2*min-1
a) zero; b) first; c) 1/2; d) 7/2
order 1
 L 


mol 

kUnits 
time
Mechanism Examples:
1. The proposed mechanism for a reaction is
1) A(g) + B(g)  X(g)
fast
2) X(g) + C(g)  Y(g)
slow
3) Y(g)  D(g)
fast
a) What is the overall equation?
A+B+CD
b) Identify any intermediates.
X and Y
z
c) what is the molecularity for each step?
1) bimolecular; 2) bimolecular; 3) unimolecular
d) If the mechanism instead were A(g) + B(g) + C(g)  D(g), would this be
consistent with the rate law rate = k[A][B][C]?
yes
2. In a study of nitrosyl halides, a chemist proposes the following mechanism for
the synthesis of nitrosyl bromide:
NO(g) + Br2(g)  NOBr2(g)
fast
NOBr2(g) + NO(g)  2NOBr(g)
slow
If the rate law for this reaction is rate = k[NO]2[Br], is the proposed
mechanism valid?
Without substitution, no. With substitution, yes, but you don’t know that
yet.
3. The rate law for 2NO(g) + O2(g)  2NO2(g) is rate = k[NO]2[O2]. The following
mechanisms have been proposed:
I.
2NO(g) + O2(g)  2NO2(g)
II.
2NO(g)  N2O2(g)
N2O2(g) + O2(g)  2NO2(g)
III.
2NO(g)  N2(g) + O2(g)
N2(g) + 2O2(g)  2NO2(g)
a) which of these mechanisms is consistent with the rate law?
Based on what you know now (without substitution); Just I.
Real answer: I and II
b) which is the most reasonable chemically? II
Extra Practice:
1. Consider the following general reaction and data:
2A + 2B + C  D + 3E
Exp’t [A] mol/L [B] mol/L [C] mol/L Initial Rate (M/L*s)
1
0.024
0.085
0.032
6.0x10-6
2
0.096
0.085
0.032
9.6x10-5
3
0.024
0.034
0.080
1.5x10-5
4
0.012
0.170
0.032
1.5x10-6
a) What is the reaction order with respect to each reactant?
rate  k[ A] x [ B ] y [C ] z
rate2 9.6x10 -5 [0.096] x [0.085] y [0.032] z
x


 16  4 x  x  2
-6
x
y
z
rate1 6.0x10
[0.024] [0.085] [0.032]
rate4 1.5x10 -6 [0.012] 2 [0.170] y [0.032] z
y


1  2y  y  0
-5
2
y
z
rate2 9.6x10
[0.096] [0.085] [0.032]
rate3 1.5x10 -5
[0.024] 2 [0.034]0 [0.080] z


 2.5  2.5 z  z  1
-3
2
0
z
rate1 6.25x10
[0.024] [0.085] [0.032]
b) Calculate the rate constant.
1.5x10-6 = k[0.012]2[0.170]0[0.032]  k = 0.326 L2/mol2*s
c) Write the rate law for this reaction.
z
Rate = (0.326 L2/mol2*s)[A]2[C]
d) Express the rate in terms of changes in concentration with time for each
of the components.
1 [ A]
1 [ B]
[C ] [ D] 1 [ E ]





2 t
2 t
t
t
3 t
2. Is each of these statements true or false?
a) At a given temperature, all molecules posses the same kinetic energy.
False
b) Halving the pressure of a gaseous reaction doubles the reaction rate.
False
c) If reactant molecules collide with greater energy than the activation
energy, they change into product molecules.
Not necessarily true (they still need the correct orientation).
d) The rate of a reaction increases as a reaction proceeds.
False
e) Exothermic reactions always have higher rates than endothermic ones.
False
f) The initial rate of a reaction is its maximum rate.
True
g) A bimolecular reaction is generally twice as fast as a Unimolecular
reaction.
False
h) The molecularity of an elementary reaction is proportional to the
molecular complexity of the reactants.
False