Kinetics Worksheet
1.
Distinguish between average rate and instantaneous rate. Which of the two
rates gives us an unambiguous measurement of reaction rate? Why?
Average rate is exactly what it sounds like.
Instantaneous rate is the speed at which something is consumed or
produced at any point in the reaction.
2.
Suggest experimental means by which the rates of the following reactions
could be followed:
a)
CaCO3(s)
→
CaO(s) + CO2(g)
Check the pressure created by CO2 over a period of time.
b)
Cl2(g) + 2 Br –(aq)
→
Br2(aq) + 2 Cl –(aq)
Chlorine gas is yellowish-green, bromine is a brownish gas. Use a
colorimeter to check for variations in absorbance.
3.
On which of the following properties does the rate constant of a reaction
depend?
a)
b)
c)
4.
reactant concentration
nature of the reactants
temperature.
For a hypothetical chemical reaction that has the stoichiometry 2 X + Y → Z,
the following initial rate data were obtained. All measurements were made
at the same temperature.
Initial Rate of Formation
of Z, (mol/L-sec)
7.0 x10-4
1.4 x10-3
2.8 x10-3
4.2 x10-3
Initial [X]o,
(mol/L)
0.20
0.40
0.40
0.60
Initial [Y]o,
(mol/L)
0.10
0.20
0.40
0.60
a)
Identify the rate law for this reaction from the data above.
r = k [X]0 [Y]1
b)
Calculate the specific rate constant for this reaction and the units.
k = 7.0 x10-4 mol/L-s/[0.10 mol/L] = 7.0 x10-3/s
c)
How long must the reaction proceed to produce a concentration of Z
equal to 0.20 M, if the initial reaction concentrations are [X]o = 0.80 M,
[Y]o = 0.60 M and [Z]o = 0 M?
2X+Y→Z
Z : Y
[Y] = 0.60 - 0.20 = 0.40
ln [A]0/[A]t = kt
ln [0.60]o/[0.40]t = 7.0 x10-3 t
t = 0.41/7.0 x10-3 = 58 s
5.
The rate constant for the second order reaction
2 NOBr(g)
→
2 NO(g) + Br(g)
is 0.80 L/mol-s at 10° C.
a)
Starting with a concentration of 0.086 M, calculate the concentration of
NOBr after 22 s.
1/[B]t - 1/[B]0 = kt
1/[B]t - 1/[0.086] = 0.80 M-s (22 s)
[B]t = 0.034
b)
Calculate the half-lives
[NOBr]0 = 0.054 M
when
[NOBr]0
=
0.072
M
and
t1/2 = 1/k [B]
t1/2 = 1/0.80 M-s [0.072] = 17. s
t1/2 = 1/0.80 M-s [0.054] = 23. s
6.
The rate constant for the second order reaction
2 NO2(g)
→
2 NO(g) + O2 (g)
is 0.54 L/mol-s at 300° C. How many seconds would it take for the
concentration of NO2 to decrease from 0.62 M to 0.28 M?
1/[0.28] = 1/[0.62]0 + 0.54 t
t = 3.6 s
7.
The rate law for the reaction
2 NO(g) + Cl2(g)
→
2 NOCl(g)
is given by r = k [NO][Cl2].
a)
What is the overall order of the reaction?
2
b)
A mechanism involving the following steps has been proposed for the
reaction
NO(g) + Cl2(g)
NOCl2(g) + NO(g)
→
→
NOCl2(g)
2 NOCl(g)
If this mechanism is correct, what does it imply about the relative rates
of these two steps?
The first step is the rate determining step, the second step is the fast
step.
8.
For the reaction
X2 + Y + Z
→
XY + XZ
it is found that doubling the concentration of X2 doubles the reaction rate,
tripling the concentration of Y triples the rate, and doubling the
concentration of Z has no effect.
a)
What is the rate law for this reaction?
r = k [X2] [Y]
b)
Why is it that the change in the concentration of Z has no effect on the
rate?
Occurs after the slow step.
c)
Suggest a mechanism for the reaction that is consistent with the rate
law.
X2 + Y → XY + X
slow
→
fast
X + Z
9.
XZ
The decomposition of N2O to N2 and O2 is a first order reaction. At 730° C
the half-life of the reaction is 3.58 x103 min. If the initial pressure of N2O is
2.10 atm at 730° C, calculate the total gas pressure after one half-life.
Assume that the volume remains constant.
2 N2O → 2 N2 + O2
moltotal = 1 + 1 + 0.5 = 2.5 mol
2.10 atm/2 mol gas (2.5 mol gas) = 2.63 atm after one half-life