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A problem set for general chemistry 2 with lab, focusing on chemical kinetics. It includes questions related to calculating average and instantaneous reaction rates, determining reaction orders, writing rate laws, and calculating rate constants. Additionally, it covers topics such as radioactive decay, half-life calculations, and chemical equilibrium. The problem set provides detailed solutions for each question, making it a valuable resource for students studying chemical kinetics and related concepts. It also includes problems related to determining the decay constant and the half-life of a radioactive nucleus, as well as calculating the equilibrium constant kc for a given reaction. Designed to help students practice and reinforce their understanding of these key concepts in chemistry.
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In the reaction of 0.200 M gaseous N 2
5
to yield NO 2
gas and O 2
gas as shown
below:
2
5 (g)
2 (g)
2 (g)
the following data table is obtained:
Time (sec) [N 2 O 5 ] [O 2 ]
(a) The average rate over the measured time interval from 0 to 3000
secs is:
rate = ∆[O 2
] / ∆t = (0.054 - 0) / 3000 - 0 = 1.80 x 10
mol/L•s
(b) The instantaneous rate early in the reaction from 0 to 300 secs is:
rate = ∆[O 2
] / ∆t = (0.010 - 0) / 300 - 0 = 3.33 x 10
mol/L•s
(c) The instantaneous rate late in the reaction from 2400 to 3000secs
is:
rate = ∆[O 2
] / ∆t = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10
mol/L•s
(d) It can be seen that the early instantaneous rate is the largestsince
the concentrations of reactants is highest during the earlieststages of the
reaction and the late instantaneous rate is smallest since the
concentrations of reactants is lowest during the late stages of the
reaction.
In the reaction of gaseous CH 3
CHO to yield CH 4
gas and CO gas as
shown below:
3
(g)
4 (g)
(g)
the following data table is obtained:
Question 2
(a) The average rate over the measured time interval from 0 to20,
secs is:
rate = - ∆[ CH 3
CHO] / ∆t = - (0.0044 - 0.0500) / 20,000 - 0 = 2.30 x
mol/L•s
(b) The instantaneous rate early in the reaction from 0 to 1200 secs is:
rate = - ∆[ CH 3 CHO] / ∆t = - (0.0300 - 0.0500) / 1200 - 0 = 1.67 x
mol/L•s
(c) The instantaneous rate late in the reaction from 15,000 to20,
secs is:
rate = - ∆[ CH 3
CHO] / ∆t = (0.0043 - 0.0056) / 20,000 - 15,000 =
2.60 x 10
mol/L•s
(d) It can be seen that the early instantaneous rate is the largest since
the concentration of reactant is highest during the earliest
stages of the reaction and the late instantaneous rate is smallest since the
concentrations of reactants is lowest during the late stages of the
reaction.
The following rate data was obtained for the reaction which takes place in
a solution of OH
ClO
Question 3
If data from experiments 2 and 3 was substituted into the equationwe would
obtain:
3.6 x 10
= k [0.0020]
x
y
z
7.2 x 10
k [0.0020]
x
y
z
and if we cancel all common terms we would obtain3.6 x
= k [0.0020]
x
y
z
7.2 x 10
k [0.0020]
x
y
z
3.6 x 10
y
y
7.2 x 10
y
(2) which yields y = 1 as the order of reaction with respect to I
If data from experiments 1 and 4 was substituted into the equationwe would
obtain:
1.8 x 10
= k [0.0010]
x
y
z
9.0 x 10
k [0.0010]
x
y
z
and if we cancel all common terms we would obtain1.8 x
= k [0.0010]
x
y
z
9.0 x 10
k [0.0010]
x
y
z
1.8 x 10
z
z
9.0 x 10
z
(3) which yields z = 1 as the order of reaction with respect to OH
The following rate data was obtained for the reaction:
2 ClO 2 + 2 OH
→ ClO 3
Determine the reaction order with respect to (1) ClO 2 and (2) OH
write the rate law and then (4) determine the value of the rate constant, k.
Your Answer:
Rate = k [ClO 2
x
y
If data from experiments 1 and 2 was substituted into the equation
we would obtain
Question 5
Experiment # [ClO 2 ] [OH
] rate
1 0.060 0.030 2.48 x 10
2 0.020 0.030 2.76 x 10
3 0.020 0.090 8.28 x 10
The overall rate law can now be written as follows:
(4) Rate = k [ClO
1
1
1
(5) and using the data from experiment 1 we can determine the
rate constant as follows:
1.8 x 10
= k [0.0010] [0.0030] [1.00]
k = 1.8 x 10
2.48 x 10
= k [0.060]
x
y
2.76 x 10
k [0.020]
x
y
and if we cancel all common terms we would obtain2.48 x
= k [0.060]
x
y
2.76 x 10
k [0.020]
x
y
2.48 x 10
x
x
2.76 x 10
x
(1) which yields x = 2 as the order of reaction with respect to ClO 2
If data from experiments 2 and 3 was substituted into the equationwe would
obtain
2.76 x 10
= k [0.020]
x
y
8.28 x 10
k [0.020]
x
y
and if we cancel all common terms we would obtain2.76 x
= k [0.020]
x
y
8.28 x 10
k [0.020]
x
y
2.76 x 10
y
y
8.28 x 10
y
(2) which yields y = 1 as the order of reaction with respect to OH
the overall rate law can now be written as follows:
Rate = k [ClO 2 ]
2
1
and using the data from experiment 1 we can determine therate
constant as follows:
2.48 x 10
= k [0.060]
2
t 1/
= 0.693 / 3.47 x 10
= 199.7 days
0.693 = (3.47 x 10
) t 1/
0.693 = k t 1/
- 3 - 1.3863 = - k (400) k = - 1.3863 / - 400 = 3.47 x
days) 3.2189 - 4.6052 = - k (400)
ln 25 - ln 100 = - k (
ln [A] - ln [A] 0
= - k t
Your Answer:
Determine the decay constant (k) and the half-life of a radioactive nucleus
if 75% of the material has decayed in 400 days.
Question 6
A sample of wood from an ancient tomb was found to contain 15.7 %
14
content as compared to a present-day sample. The t 1/
for
14
C is 5720
yrs. What is the age of the wood?
Your Answer:
Question 7
Calculate the K c
for the following reaction if an initial reaction mixture of
0.500 mole of CO and 1.500 mole of H 2 in a 5.00 liter container forms an
equilibrium mixture containing 0.198 mole of H 2 O and corresponding
amounts of CO, H 2 , and CH 4.
(g)
2 (g)
4 (g)
2
(g)
Your Answer:
Question 8
If the equilibrium mixture contains 0.198 mole of H 2 O then the
following amounts of materials must be present in the equilibrium
mixture:
H 2 O = 0.198 mole (as stated)
CH 4 = 0.198 mole (1 mole of CH 4 forms for every mole of
H 2 O that is formed)
CO = 0.500 - 0.198 mole (1 mole of CO reacts for every
mole of H 2 O that is formed)
H 2 = 1.500 - 3 x 0.198 mole (3 mole of H 2 reacts for every
mole of H 2 O that is formed)
Change all amounts to moles/L before entering in Kc
expression:
2
O = 0.198 mole / 5.00 L = 0.0396 M
4
= 0.198 mole / 5.00 L = 0.0396 M
CO = 0.302 mole / 5.00 L = 0.0604 M
2
= 0.906 mole / 5.00 L = 0.1812 M
Kc = [CH 4 ] [H 2 O] = [0.0396] [0.0396] = 4.
3
3
