Ch. 14 Kinetics (23 Questions)
14.1) (a) Reaction rate is the change in concentration of reactants or products in a given
amount of time.
(b) Rate can be affected by reactant concentration, surface area, temperature, and
the presence of a catalyst.
(c) The stoichiometry of the reaction (mole relationships between reactants and
products) is needed to relate the disappearance of reactants to the appearance of
products.
14.42) (a) Ea = 7 kJ, E = -66 kJ
(b) Ea(reverse) = 66 kJ + 7 kJ = 73 kJ
14.4) (a) Since A and B are present in a 1:1 ratio, as A disappears, a corresponding
amount of B appears.
Moles A
0.100 mol
0.067 mol
0.045 mol
0.030 mol
0.020 mol
Moles B
0 mol
0.033 mol
0.055 mol
0.070 mol
0.080 mol
(b)
Time Interval
0-40 s
40-80 s
80-120 s
120-160 s
-A/t
-(0.067 mol – 0.100 mol)/40 s
-(0.045 mol – 0.067 mol)/40 s
-(0.030 mol – 0.045 mol)/40 s
-(0.020 mol – 0.030 mol)/40 s
Rate (mol/s)
8.3 x 10-4
5.5 x 10-4
3.8 x 10-4
2.5 x 10-4
(c) We would need the liters of solution to calculate the rate in M/s.
14.10) (a) -[HBr]/2t = +[H2]/t = +[Br2]/t
(b) -[SO2]/2t = -[O2]/t = +[SO3]/2t
(c) -[C4H8]/t = +[C2H4]/2t
14.12) (a) Rate = -[C2H4]/t = +[CO2]/2t = +[H2O]/2t
[CO2]/ t = [H2O]/ t = -2[C2H4]/t = -2(-0.18 M/s) = 0.36 M/s
(b) Rate = -PN2H4/t = +PNH3/2t
PNH3/t = -2PN2H4/t = -2(-60 torr/hr) = 120 torr/hr
PT = (-60 torr/hr + -60 torr/hr + 120 torr/hr) = 0 torr/hr
14.14) (a) A rate law is an algebraic expression that relates the concentration of reactants to
reaction rate. A rate constant is the proportionality constant in the rate law and depends
on the chemical reaction and the temperature.
(b) The reaction order for a specific reaction is the exponent of the concentration of that
reactant in the rate law. The overall reaction order is the sum of the exponents in the rate
law.
(c) If the overall reaction order is 2, the units of k are determined as follows:
Rate = k [A]2
Ms-1 = k M2
k = M-1s-1
14.16) (a) Rate = k [N2O5]
(b) Rate = (4.82 x 10-3 s-1)(0.0240 M) = 1.16 x 10-4 M/s
(c) Rate doubles (2.32 x 10-4 M/s)
14.18) (a) & (b)
Rate = k [C2H5Br] [OH-]
1.7 x 10-7 M/s = k (0.0477 M) (0.100 M)
k = 3.6 x 10-5 M-1s-1
(c) adding an equal volume of ethyl alcohol dilutes the concentration of both species by
half which decreases the rate by a factor by ¼.
Rate = k ([C2H5Br]/2) ([OH-]/2) = k [C2H5Br][OH-]/4
14.20) (a) Rate = k [I-] [OCl-] [OH-]-1 or Rate = k [I-] [OCl-]/[OH-]
(b) If I- is tripled, the rate is tripled since it is first order.
(c) If OH- is doubled, the rate is halved since it is negative first order.
14.22) (a) From experiments 1 and 2 we can see that doubling BF3 while holding NH3 constant
results in a doubling of the reaction rate. Thus, the reaction is first order with regards to
BF3. From experiments 4 and 5 we can see that doubling NH3 while holding BF3
constant results in a doubling of the reaction rate. Thus, the reaction is first order with
regards to NH3.
Rate = k [BF3] [NH3]
(b) overall order = 2nd
(c) 0.2130 Ms-1 = k (0.250 M) (0.250 M)
k = 3.41 M-1s-1
14.24) (a) From experiments 1 and 2 we can see that increasing S2O82- by a factor of 1.5 while
holding I- constant results in an increase in the reaction rate by a factor of 1.5. Thus, the
reaction is first order with regards to S2O82-. From experiments 1 and 3 we can see that
doubling S2O82- and increasing I- by a factor of 1.5 results in an increase in the reaction
rate by 3. Since we are overall tripling the concentration (2 x 1.5 = 3) and the rate is
tripling, the order with respect to I- must be first order.
Rate = k [S2O82-] [I-]
(b)
Rate1 = k1 [S2O82-] [I-]
2.6 x 10-6 Ms-1 = k1 (0.018 M)(0.036 M)
k1 = 4.0123 x 10-3 M-1s-1
Repeat for all trials: k2 = 4.0123 x 10-3 M-1s-1, k3 = 4.0123 x 10-3 M-1s-1,
k4 = 3.8889 x 10-3 x 10-3 M-1s-1
Avg. k = 3.981 x 10-3 M-1s-1 = 4.0 x 10-3 M-1s-1 (2 SF)
(c) -[ S2O82-]/t = +[I-]/3t
(d)
Rate = k [S2O82-] [I-]
Rate = (4.0 x 10-3 M-1s-1)(0.075 M)(0.050 M)
Rate = 1.5 x 10-5 Ms-1
-[ S2O82-]/t = +[I-]/3t
[I-]/t = -3[ S2O82-]/t = 3(1.5 x 10-5 Ms-1) = 4.5 x 10-5 Ms-1
14.25) (a) A first order reaction depends on the concentration raised to the first power of one
reactant.
(b) A graph of ln [A] vs. time yields a straight line for a first order reaction.
(c) The half-life of a first order reaction does NOT depend on the initial concentration,
only the value of k.
14.26) (a) A second order reaction depends on the concentration of a single reactant raised to the
second power or the concentrations of two reactants raised to the first power.
(b) A graph of 1/[A] vs. time yields a straight line for a second order reaction.
(c) The half-life of a second order reaction DOES depend on the initial concentration.
14.28) Since first order:
t1/2 = 0.693/k
2.3 x 105 s = 0.693/k
k = 3.0 x 10-6 s-1
14.30) (a) Since first order:
ln Pt – ln P0 = -kt
ln Pt = -(4.5 x 10-2 s-1)(75 s) + ln(340 torr)
ln Pt = 2.4539
Pt = 12 torr
(b)
ln Pt – ln P0 = -kt
ln (0.1 x 340 torr) – ln(340 torr) = -(4.5 x 10-2 s-1)t
t = 51 s
14.32) Since first order:
ln Pt – ln P0 = -kt
ln (335 torr) – ln (502 torr) = -k(2,000 s)
k = 2 x 10-4 s-1
t1/2 = 0.693/k = 0.693/(2 x 10-4 s-1) = 3 x 103 s
14.38) (a) Although the process is unimolecular, collisions between molecules provides energy
for the methyl isonitrile to isomerizes to acetonitrile. Thus, the more collisions and the
greater the energy of the collisions, the faster the reaction rate.
(b) All collisions with sufficient energy are not necessarily effective, as the collisions
may not be properly oriented.
(c) Kinetic molecular theory predicts that at a high temperature there will be a higher
average kinetic energy. At higher temperatures, there will also be a greater distribution
of molecular energies, thus increasing the probability of an effective collision.
14.40) (a) f = e-Ea/RT = e-(80,000 J/mol/(8.31 J/mol•K x 450 K)) = 5.12 x 10-10
(b) f = e-(80,000 J/mol/(8.31 J/mol•K x 460 K)) = 8.15 x 10-10
8.15 x 10-10/5.12 x 10-10 = 1.6
14.54) (a) molecularity = the number of molecules that participate in a process like an
elementary step
(b) Termolecular processes are extremely rare because it is difficult and unlikely
for 3 molecules to simultaneously collide with the correct energy and orientation.
(c) An intermediate is a temporary product that is produced then quickly
consumed (make it, then take it). It does not appear anywhere in the net reaction,
only the mechanism.
14.56) (a) bimolecular; Rate = k [CH2] [Cl2]
(b) unimolecular; Rate = k [(CH2)3]
(c) unimolecular; Rate = k [SO3]
14.60) The slow (rate determining) step must be a bimolecular collision between H2 and
ICl. This leaves choices (b) and (d). Since they both cancel out to the correct
final equation, they are both possible.
14.82) (a) (CH3)3AuPH3  C2H6 + (CH3)AuPH3
(b) (CH3)3Au, PH3, (CH3)Au are intermediates (make it, then take it!).
(c) Steps 1 and 2 are both unimolecular steps. Step 3 is bimolecular.
(d) Step 2 is the rate-determining step since it’s the slow step.
(e) Rate = k [(CH3)3Au]
Since the above involves an intermediate, we need to substitute in for (CH3)3Au
from Step 1.
Rate = k [(CH3)3AuPH3]/[PH3]
(f) Adding PH3 would actually decrease the rate of the reaction since the reaction
is inversely proportional to the concentration of PH3.
14.64) (a) 2N2O → 2N2 + O2
(b) NO is a catalyst because it is regenerated as a product.
(c) No, the mechanism cannot be ruled out. NO2 is an intermediate and is quickly
consumed in the reaction, thus not accumulating in measurable quantities.