For kinetics
the reaction is second order in A and first order in B and over all third order.15.4
For kinetics
the reaction is second order in A and first order in B and over all third order.
15.6
For if
the concentration of A is doubled with B concentration fixed the rate will quadrupled.
If the concentration of A is doubled and the concentration of B is halved the rate
will be double the original rate. New rate = 4*0.5*Old rate.
15.9
The first order reaction kinetics is 
; 
.
Second order reaction kinetics is 
; 
(a) A plot of 1/[A] vs time gives a straight line implies second order kinetics.
(b) A plot of Ln[A] vs time giving a straight line implies first order kinetics.
15.18
For the reactions 
the relative rates are:
(a)
(b)
(c)
15.19
From the stoichiometry of the balanced chemical equation we know that for every molecular
nitrogen that reacts three molecules of hydrogen react. Thus, the rate of dissapearance
of hydrogen is three times that nitrogen, or 3 * 0.25 M/min. Similarly, the appearance
of ammonia is twice the rate of dissapearance of molecular nitrogen or 2 * 0.25 M/min.
15.21
Kinetic data for phenyl acetate hydrolysis are given below
| Time (min) | [Phenyl acetate] M |
|
0 |
0.55 |
|
0.25 |
0.42 |
|
0.50 |
0.31 |
|
0.75 |
0.23 |
|
1.00 |
0.17 |
|
1.25 |
0.12 |
|
1.50 |
0.085 |
(a)
The plot of the log of concentration vs. time is more or less linear suggesting first
order kinetics.
(b) Since the time change in the data is constant at 0.25 min it is easy to compute the rate for each pair of concentrations.
| Change in [Phenyl acetate] M | Rate M/min |
|
-0.13 |
-0.50 |
|
-0.11 |
-0.44 |
|
-0.08 |
-0.32 |
|
-0.06 |
-0.24 |
|
-0.05 |
-0.20 |
|
-0.035 |
-0.14 |
(c) See the above table.
(d) If we take the rates from table and divide out concentration we get an estimate for the rate constant (providing first order kinetics).
| [Phenyl acetate] M | Rate M/min | Rate Constant k |
|
0.55 |
-0.50 |
0.91 |
|
0.42 |
-0.44 |
1.05 |
|
0.31 |
-0.32 |
1.03 |
|
0.23 |
-0.24 |
1.04 |
|
0.17 |
-0.20 |
1.18 |
|
0.12 |
-0.14 |
1.17 |
= (1.06)(0.55)(0.77) = 0.45 M/min.
15.24
| [NO] M | [O2] M | Rate of Dissappearance of NO M/s |
|
0.020 |
0.010 |
0.0001 |
|
0.040 |
0.010 |
0.0004 |
|
0.020 |
0.040 |
0.0004 |
(b) k = Rate/[NO]^2[O2] = 25/s as the average k.
(c) When [NO] = 0.045M and [O2] = 0.025M the Rate = 25(0.045)^2(0.025) = 0.00127 M/s.
(d) From the stoichiometry we know 
and so NO reacting = NO_2 forming = 2*0.0005 M/s.
15.26
decomposition
is first order in .
It takes 4.26 min at 55C to decompose 2.56 mg of to 2.50 mg. Find k in minutes and seconds.
; 
; k = 0.00557 min^-1; k = 0.000093
s^-1.
15.28 Cyclopropane transformation is first order with a rate constant k = 0.054 hr^-1. What time is required for the concentration to drop from 0.050M to 0.025M?
; t = (1/0.054)Ln(0.025/0.05)
= 12.8 hr.
15.30 The rate constant for 
with k = 3.40 L/mol-min. The time required to decrease
the concentration from 2.0M to 1.5M is:
t = -(1/3.4)Ln(1.5/2.0) = 0.085 min.
15.34 Phosphine, PH_3, decomposes via first order kinetics with a half
life of 37.9 s. 
How much time is required for 3/4 of the PH_3 to decompose?
One way: 
and n = 2. Two halflives are required = 75.8 s.
Another way: 
.
Solve for t to get the same ans.
Another question: If the half life is 37.9 s what is the rate constant?
= 0.0183
s^-1
15.45 At 2000K ammonia gas decomposes according to the data below.
| time (hr) | [NH_3] M |
|
0 |
8 x 10^-7 |
|
25 |
6.75 x 10^-7 |
|
50 |
5.84 x 10^-7 |
|
75 |
5.15 x 10^-7 |
Plot of Log concentration vs. time
Plot of 1/concentration vs. time
The graph shows second order kinetics.
15.46 The activation energy is 52 kJ/mol for the forward reaction A + B
-> C + D and is 58 kJ/mmol for the reverse reaction.
The reaction is exothermic by 6 kJ/mole.
15.47 The fast reaction between F atoms and hydrogen molecules 
has an activation energy of 8
kJ/mole and has an enthalpy change of -133 kJ/mole. Draw a diagram.
15.48 Calculate the activation energy for the reaction
given the measured rate constants at two different temperatures
Ea = 102 kJ/mole.
15.49 The rate constant triples with a temperature rise from 300K to 310K. What is Ea?
Ea = 84.9 kJ/mole.
15.52 Molecularity means the number of molecules taking part in an elementary
step.
(a) 
;
bimolecular
(b) 
;
bimolecular
(c) 
;
unimolecular
15.53 
1. 
;
fast, steady state, equilibrium.
2. 
Slow, rate determing step, bimolecular
3. 
fast, bimolecular
15.54 Part of the ozone decomposition mechanism in the upperatmosphere
is
The rate is controlled by the slow step 2. Step 1 is fast in both directions and
equilibrium is set up. From step 22
Using equilibrium in step 1
to eliminate [O] in the rate equation
The overall rate is second order in ozone, -1 order in oxygen and overall first order
according to this mechanism.
15.55 The decomposition of hydrogen peroxide by iodine
may go by the following mechanism:
1. 
; slow,
rate determining, bimolecular
2. 
; fast,
bimolecular
3. 
; fast,
bimolecular
(a) The sum of 1 + 2 + 3
(b) See above
(c) 
(d) Intermediates: OI^- and HOI.
15.56 (a) F, (b) F, (c) F, (d) T, (d) T
15.57
(a) A catalyst can significantly lower the activation energy and can affect the "steric
factor."
(b) A catalyst has no effect on the enthalpy change in a reaction.
15.67 (a) T, (b) T, (c) F, (d) F
15.74
(a)
1. enthalpy = -37 kJ/mol
2. enthalpy = -214 kJ/mol
3. enthalpy = -251 kJ/mol
(b) Ea = -17 + 37 = 20 kJ/mol
15.80 Radon has a half-life of 3.82 days and obeys first order kinetics. A basement 12m x 7m x 3m in size contains radon at a partial pressure 1 x 10^-6 mmHg (torr).
(a) Rn atoms/liter of air: n = PV/RT = (10^-6/760 atm)(1L)/(0.082 L-atm/mol-K)(273K) = 5.9 x 10^-11 moles. number of atoms = 5.9 x 10^-11 moles x 6.02 x 10^23 atoms/mole = 3.5 x 10^13 atoms.
(b) In 30 days we have 30 days/3.82 days per half-life = 7.85 half-lives. In this time there is a reduction by 2^7.85 = 231, or the number of atoms left per liter after 30 days = 3.5 x 10^13 atoms/231 = 1.5 x 10^11 atoms/liter