1
Chapter 14 - Homework Solutions
14.3)
For a reaction A v B+ C the rate law has the form
r = rate = k [A]x
Determine the value of x (order of the reaction) for the following cases, a-c, below.
This can be done by using two methods, (1) by inspection or (2) and
mathematically (see pages 585-586 in the textbook, especially sample exercise 14.6
on page 586). The simple cases given in this question can be done fairly easily by
inspection. More difficult questions need to be done using the mathematical
method. This method is shown below for two reactants, A and B:
a) There is no rate change when [A] is tripled.
If the rate doesn’t change when the concentration changes then x = 0. This
means the reaction does not depend on A and the reaction is zero-order in A.
2
14.3) (cont.)
b) The rate increases by a factor of 9 when [A] is tripled.
If x = 1, the rate and [A] would be directly proportional and the rate would
triple when [A] triples.
If x = 2, the rate will increase by a factor of 9 if [A] is tripled. That’s because
32 = 9.
r = k [A]2
r1 = k [A]12
[A]2 = 3 [A]1
r2 = k [A]22
substitute this into the eqn for r2
r2 = k [A]22 = k {3 [A]1}2 = k {3}2 [A]12 = 9 {k [A]12} = 9 r1
This shows that the reaction is 2nd order in A.
c) The rate increases by a factor of 8 when [A] is doubled.
If x = 3, the rate will increase by a factor of 8 if [A] is doubled. That’s
because 23 = 8.
r1 = k [A]13
[A]2 = 2 [A]1
r2 = k [A]23
substitute this into the eqn for r2
r2 = k [A]23 = k {2 [A]1}3 = k {2}3 [A]13 = 8 {k [A]13} = 8 r1
This shows that the reaction is 3rd order in A.
3
14.5)
The following three graphs for experiments done at two different temperatures are
obtained for a first-order reaction.
(Copyright by Prentice and Hall)
For a first-order reaction, a plot of ln[A] vs. time is linear (ln[A]t = -kt + ln[A]0).
The slope is -k and the y-intercept is ln[A]0.
The rate constant depends on the frequency factor, A, (this is not concentration),
activation energy (Ea) and temperature, T (in Kelvin).
k = A e&Ea/RT
(rate is directly proportional to the rate constant, k)
This equation shows that as activation energy, Ea, decreases the rate constant (and
therefore rate) increases and as T increases the rate constant (and rate) increases.
a)
Which two lines represent exp done at the same temperature? What accounts
for the difference in the two lines? In what way are they the same?
The rate constant depends on temperature and is related to the slope of the
lines. If the temperature is the same for two experiments then the slopes
should be the same. Exp 1 and 2 have the same slopes (rate constants) and
are thus at the same temperature.
The y-intercepts (ln[A]0) of the two lines (exp 1 and 2) are different. Thus
the two experiments had different initial concentrations of A.
4
14.5) (cont.)
b)
Which two lines represent experiments done with the same starting
concentration but at different temperatures? Which line likely represents the
lower temperature and how do you know?
The lines for exp 2 and 3 have the same y-intercepts (ln[A]0) and thus the
same starting concentrations of A.
The slopes of the two lines for exp 2 and 3 are different so their rate
constants must be different. This means they occur at different
temperatures (Ea will be the same for the same reaction). The line for exp 2
has a larger slope (rate constant) which implies exp 2 occurs at the higher
temperature. You know this because rate increases with increasing
temperature. Since the rate is directly proportional to the rate constant this
means the rate constant increases with increasing temperature (you can
also see this using the Arrhenius equation given above and in section 14.5)
14.8)
The effect of temperature on the rate of two reactions is studied. A graph of the
natural logarithm of the rate constant (ln k) as a function of 1/T (T in Kelvin) is
made. This is based on the Arrhenius equation:
k = A e&Ea/RT
where,
A = frequency factor
Ea = activation energy
T = temperature of Kelvin
If the natural log (ln) is taken of both sides you obtain an eqn for a straight line:
Ea 1
ln(k) = & ----- (----) + ln(A)
R T
Slope = & Ea/R
y-intercept = ln(A)
Larger slope means larger Ea
(slope will always be negative, Ea > 0)
Ea is the difference in energy between reactants and the transition state on a
potential energy diagram (reactants and the top of the hill).
5
14.8) (cont.)
How do graphs of ln(k) vs. 1/T compare under the following conditions?
a)
The two reactions have the same frequency factor but the activation energy of
the second reaction is higher than activation energy of the first reaction.
Ea(2) > Ea(1)
and A2 = A1
Both lines will have a negative slope (always will). Since the y-intercept of
the plot is ln(A) and the A (frequency factor) values are the same the lines
will have the same y-intercept.
Since Ea(2) > Ea(1) the slope of line 2 should be steeper (since Ea = -slope*R)
(Copyright by Prentice and Hall)
b)
The two reactions have the same activation energies frequency factor but the
frequency factor of the second reaction is higher than frequency factor of the
first reaction.
Ea(2) = Ea(1)
and A2 > A1
Both lines will be parallel with the same negative slopes since Ea(2) = Ea(1).
Since the frequency factors are different the y-intercepts will be different,
with that for reaction 2 higher than for reaction 1 (A2 > A1)
(Copyright by Prentice and Hall)
6
14.9)
The diagram below represents the two steps in a two-step mechanism for a reaction.
(Copyright by Prentice and Hall)
Red = oxygen
a)
Blue = nitrogen Green = fluorine
Write the chemical equation for each step (i.e. the mechanism).
NO2 + F2 v NO2F + F
(1)
NO2 + F v NO2F
b)
(slow)
(2)
Write the balanced equation for the overall reaction (add the steps and cancel
anything that appears on both sides of the reaction arrow).
2 NO2 + F2 + F v 2 NO2F + F
Cancel the F atoms since they appear on both sides:
2 NO2 + F2 v 2 NO2 F
(overall balanced eqn.)
c)
The F atom is an intermediate. Intermediates generally appear for the first
time as a product in one step and used as a reactant in subsequent step. (A
catalyst usually appears for the first time as a reactant and reproduced in a
subsequent step as a product, so it can be used again and again.)
d)
You write the rate law based on the reactants in the slow step (the ratedetermining step). Assume the first step is the slow step.
r = k [NO2] [F2] 1st order w/rsp to NO2 and F2; 2nd order overall
7
14.12)
The diagram below represents a two-step mechanism for a reaction.
(Copyright by Prentice and Hall)
Orange = element A
a)
Green = element B
Write the chemical equation for each step (i.e. the mechanism) and the net
reaction (overall balanced equation).
A2 + AB v A2B + A
A + AC v A2 + C
-------------------------------AB + AC v A2B + C
b)
Blue = element C
(1)
(2)
(overall balanced equation)
What is the intermediate?
A is the intermediate. It appears for the first time as a product in one step
and used as a reactant in subsequent step. (It is produced and consumed.)
c)
What is the catalyst?
A2 is the catalyst. It appears for the first time as a reactant and
reproduced in a subsequent step as a product, so it can be used again and
again. (It is consumed and reproduced.)
8
14.13)
9
14.15)
Given a reaction, A(aq) -> B(aq) and moles of A at 10 minute time intervals.
Calculate moles of B, molarity of A at each 10 min interval ([A]), change in
molarity of A ( )[A]), avg rate of disappearance of A, -()[A]/) t), in M/s and avg
rate of appearance of B (between 10 and 30 min). (100.0 mL soln).
b) Ex. calc. for avg. disap. of A for each 10 min interval.
r = &()[A]/) t) = &{(-0.14 M)/(10-0 min) x (1 min/60 sec)} = 2.3 x 10&4 M/s
10
14.18)
a)
The following table shows the results of calculations for the average rate of
reaction in M/s (avg rate of disappearance of Hcl).
CH3OH (aq) + HCl (aq) v CH3Cl (aq) + H2O (R)
(Copyright by Prentice and Hall)
b)
Graph of [HCl] vs. time. Instantaneous rates in M/min and M/s at t = 75.0
min and t = 250 min (the slope of tangents to the curve at these times).
(Copyright by Prentice and Hall)
Try this in Excel. Put in time and [HCl] and then graph it. If you don’t have
Excel (or a different spreadsheet program or graphing program) try it on your
graphing calc. To see what it looks like.
11
14.20)
a) 2 H2O(g) --------> 2 H2(g) + 1 O2(g)
1
)[H2O]
1 )[H2]
1
)[O2]
rrxn = ---- ( ! ----------- ) = ---- ( --------- ) = ---- ( --------- )
2
)t
2
)t
1
)t
Coeff. in bal. eqn. are in the denominators
ˆ
)[H2O]
)[H2]
)[O2]
( ! ----------- ) = ( --------- ) = 2 ( --------- )
)t
)t
)t
)[H2O]
( ! ----------- ) rate of disappearance H2O
)t
)[H2]
( --------- )
)t
rate of appearance H2
)[O2]
( --------- )
)t
rate of appearance O2
12
14.21)