DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY
CHEMISTRY OF SOLUTIONS
202-NYB-05 15/16
TEST 3
MAY 2, 2012
INSTRUCTOR: I. DIONNE
Answers
Print your name: _____________________________________________
INSTRUCTIONS:
Answer all questions in the space provided.
Duration of this test is 75 minutes.
No books or extra paper are permitted.
Answer the questions in ink in order to preserve the right to grieve.
In order to obtain full credit for your answers, you must clearly show your work. Answers to
problems involving calculations must be expressed to the correct number of significant figures and
proper units.
5. Calculators may not be shared. Programmable calculators are not permitted.
6. Your attention is drawn to the College policy on cheating. This policy will be enforced.
7. A Periodic Table with constants is provided.
1.
2.
3.
4.
Problem 1:
/3
Problem 6:
/3
Sig. Fig.:
/1
Problem 2:
/3
Problem 7:
/3
Units:
/1
Problem 3:
/4
Problem 8:
/3
Problem 4:
/5
Bonus
/1
Problem 5:
/4
Total:
/30
(3 marks)
PROBLEM 1
A saturated aqueous solution of lead(II) iodate, Pb(IO3)2, contains 3.1 x 10−5 M Pb2+ at 25°C.
Assuming that the ions do not react with water, calculate the Ksp of lead(II) iodate at that
temperature.
Pb(IO3)2 (s) ⇌ Pb2+ (aq) + 2IO3− (aq)
Ksp = [Pb2+][IO3−]2
Ksp = (3.1 x 10−5)(2 x 3.1 x 10−5)2 = (3.1 x 10−5)(6.2 x 10−5)2
Ksp =1.1916 x 10−13
−13
1.2 x 10
Answer: __________________
(3 marks)
PROBLEM 2
(a) Which of these salts has the greatest molar solubility? Circle your choice.
1. Ag2S
2. CaF2
smallest
3. MgF2
greatest
(b) Will the solubility of the following salts depend on pH?
yes
no
1. AgCl
2. LiNO3
3. Fe(OH)2
4. NaOCl
Page |2
(4 marks)
PROBLEM 3
A solution is prepared by mixing 50.0 mL of 0.10 M Pb(NO3)2 with 50.0 mL of 1.0 M KCl.
Calculate the concentrations of Pb2+ and Cl− at equilibrium. Assume precipitation of PbCl2(s) is
complete. [Ksp for PbCl2(s) is 1.6 x 10−5]
Pb(NO3)2 (aq) + 2KCl (aq) ⇌ PbCl2 (s) + 2KNO3 (aq)
Pb(NO3)2: 0.10 M x 0.0500 L = 0.0050 mol
KCl: 1.0 M x 0.0500 L = 0.050 mol
After the reaction, we are left with
Pb(NO3)2: 0.0050 mol − 0.0050 mol = 0
KCl: 0.050 mol − (2)(0.0050) = 0.040 mol which is in 100.0 mL
Ksp = [Pb2+][Cl−]2
Ksp = 1.6 x 10−5 = (s)(0.40 + 2s)2 ≈ (s)(0.40)2
s =1.0 x 10−4
−4
0.40 M
1.0 x 10 M
Answers: [Pb2+] = ________________
and [Cl−] = ________________
Page |3
(5 marks)
PROBLEM 4
(a) If ln[reactant] is plotted versus time, and a straight line of negative slope is observed,
what is the order of the reaction?
First-order
If 1/[ reactant] is plotted versus time =2nd order
Answer: ________________
(b) Indicate whether the following statements are true (T) or false (F).
T
1. ______
A rate law follows directly from the molecularity of an elementary step.
T
2. ______
The value of the rate constant increases as the temperature increases.
3. ______
A rate law could include the concentration of an intermediate.
F
4. ______
In a multi-step reaction, only one step can be fast and rate determining.
F
F
5. ______
The half-life of a first-order reaction depends on reactant concentration.
(c) The reaction CO(g) + NO2(g) → CO2(g) + NO(g) is second order in NO2 and zero order in CO
at temperatures less than 500K.
1. Write the rate law for the reaction.
Rate = k[NO2]2
______________________________________
original value
2. How will the reaction rate change if the NO2 concentration is halved? 1/4
____________
(d) For the hypothetical reaction A + B → C + D, the activation energy is 32 kJ/mol. For the
reverse reaction, the activation energy is 58 kJ/mol. Is the reaction A + B → C + D
exothermic or endothermic?
exothermic
Answer: ________________
(e) Assume that an A molecule reacts with two B molecules in a one-step process to give AB 2.
That is, A + 2B → AB2
If the initial rate of appearance of AB2 is 2.0 x 10−5 mol/L·s when the initial concentrations
of A and B are 0.30 M, calculate the rate constant for the reaction.
Rate = k[A][B]2
2.0 x 10−5 M/s = k[0.30 M][0.30 M]2
k = 7.4074 x 10−4 M−2s−1
7.4 x 10−4 M−2s−1
Answer: _______________________
Page |4
(4 marks)
PROBLEM 5
HOF decomposes by a first order reaction to HF and O2 with a half-life of 30.0 minutes at
25°C.
HOF(g) →HF(g) + ½O2(g).
If the partial pressure of HOF in a 1.0 L flask is initially 100. mmHg at 25°C, what is the partial
pressure of HOF and the total pressure in the flask after 45.0 minutes?
1st order:
t1/2 = 0.693/k
k = 0.693 / 30 min
k = 0.0231 min−1
ln[x] = −(0.0231)(45) + ln100
x = 35
HOF (g) → HF (g) + ½O2 (g).
After 45 min
PHOF = 100 – 65 = 35 mmHg
PHOF = 35 mmHg, PHF = 65 mmHg, PO2 = 33 mmHg
133 mmHg
Answers: PHOF = ______________
and Ptotal = ______________
35 mmHg
PROBLEM 6
(3 marks)
Calculate the activation energy (Ea) for the reaction N2O5(g) → 2NO2(g) + ½O2(g), from the
observed rate constants at the following two temperatures:
(a) at 25°C, k = 3.46 x 10−5 s−1 and (b) at 55°C, k = 1.5 x 10−3 s−1.
Ln (k2/k1) = Ea/R(1/T1 – 1/T2)
ln(3.16 x 10−5/1.5 x 10−3) = Ea/8.314(1/328 – 1/298)
Ea = 1.0215 x 105 J/mol
1.0 x 105 J/mol
Answer: ____________________
Page |5
(3 marks)
PROBLEM 7
A study of the reaction 2A + B →C + D gave the following results:
Experiment
[A]0, M
[B]0, M
R0, M/s
1
0.100
0.050
6.0 x 10−3
2
0.200
0.050
1.2 x 10−2
3
0.300
0.050
1.8 x 10−2
4
0.200
0.150
1.1 x 10−1
What is the rate law for this reaction and the overall order of the reaction?
Exp 1 / exp 2
0.50 = 0.50x
6.0 x 10−3 / 1.2 x 10−2 = k(0.100)x(0.050)y / k(0.200)x(0.050)y
x=1
Exp 2 / exp 4
1.2 x 10−2 / 1.1 x 10−1 = k(0.200)x(0.050)y / k(0.200)x(0.150)y
0.109 = 0.333y
y=2
2
Rate = k[A][B]
Third order
Answers: Rate law = ________________________________
Overall order = _____________
Page |6
(3 marks)
PROBLEM 8
The ozone decomposes according to the equation 2O3(g) → 3O2(g)
The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a
slow second step.
1. O3(g) ⇌ O2(g) + O(g)
2. O3(g) + O(g) → 2O2(g)
(fast)
(slow)
(a) Which of the steps is the rate-determining step?
Step 2
_______________________
(b) Write the rate law for the rate-determining step.
Rate = k[O3]2/[O2]
_______________________
(c) What is the molecularity of the first step?
unimolecular
_______________________
Rate = k[O3][O]
but O is an intermediate
k'[O3] = k”[O2][O]
[O] = k'[O3]/k”[O2]
Rate = k[O3][O] = k[O3]{k'[O3]/k”[O2]}
Bonus (no partial mark – max of 30 for the test)
(1 mark)
After five half-life periods for a first order reaction, what fraction of reactant remains?
t1/2
1
2
3
4
5
1/2
1/4
1/8
1/16
1/32
Answer: ____________
Page |7