DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY
CHEMISTRY OF SOLUTIONS
202-NYB-05 05
TEST 3
5 DECEMBER 2011
INSTRUCTOR: I. DIONNE
Answers
Print your name: _____________________________________________
INSTRUCTIONS:
Answer all questions in the space provided.
1.
2.
3.
4.
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.
Problem 1:
/2
Problem 6:
/3
Sig. Fig.:
/1
Problem 2:
/3
Problem 7:
/4
Units:
/1
Problem 3:
/4
Problem 8:
/6
Problem 4:
/2
Problem 9:
/4
Problem 5:
/3
Bonus:
/3
Total:
/33
(2 marks)
PROBLEM 1
(a) Write the balanced equation for the dissolution reaction of solid (NH4)3PO4 in water.
+
3−
answer: (NH4)3PO4 (s) ⇌ 3NH4 (aq) + PO4 (aq)
(b) Write the corresponding solubility product expression.
+ 3
3−
answer: Ksp = [NH4 ] [ PO4 ]
PROBLEM 2
(3 marks)
What is the molar solubility of manganese(II) sulfide (MnS) in 0.00010 M sodium sulfide
(Na2S)?
I
C
E
MnS (s) ⇌ Mn2+ (aq) + S2− (aq)
---0
0.00010
−s
+s
+s
---s
0.00010 + s
Ksp = 2.3 x 10−13 ≈ (s)(0.00010)
s = 2.3 x 10−9
and
(2.3 x 10−9 / 0.0010) x 100 = 0.00023%
5% rule answer: 2.3 x 10−9 M
Page |2
(4 marks)
PROBLEM 3
Will Ag2S precipitate when 20.0 mL of 1.3 x 10−4 M silver nitrate (AgNO3) is mixed with 30.0 mL
of 3.5 x 10−5 M sodium sulfide (Na2S)?
Need to compare Qsp and Ksp
Ksp = 1.6 x 10−49
Ag2S (s) ⇌ 2Ag+ (aq) + S2− (aq)
Qsp = [Ag+]o2 [S2−]o
[Ag+]o = (1.3 x 10−4 M)(20.0 mL) / (50.0 mL) = 5.2 x 10−5 M
[S2−]o = (3.5 x 10−5 M)(30.0 mL) / (50.0 mL) = 2.1 x 10−5 M
Qsp = (5.2 x 10−5)2(2.1 x 10−5) = 5.7 x 10−14
Qsp > Ksp therefore Ag2S will precipitate
answer: Ag2S will precipitate
(2 marks)
PROBLEM 4
Consider the following reaction: 4NH3 (g) + 7O2 (g) →4NO2 (g) + 6H2O (g)
(a) Write the rate of reaction in terms of each of the reactants and products.
answer:
Rate = −¼Δ[NH3]/Δt = -1/7Δ[O2]/Δt = ¼Δ[NO2]/Δt = 1/6Δ[H2O]/Δt
(b) Which species in this reaction above either appears or disappears the fastest?
answer:
O2
Page |3
(3 marks)
PROBLEM 5
At 25°C, the rate constant for a first-order decomposition of a pesticide solution is 6.40 x 10 –3
min–1. If the initial concentration is 0.0314 M, what concentration remains after 62.0 min at
25°C?
ln[A]t = -kt + ln[A]o
ln[A]t = -(6.40 x 10−3 min−1)(62.0 min) + ln(0.0314)
ln[A]t = −3.8577474
[A] = 0.0211155
answer: 0.0211 M
PROBLEM 6
(3 marks)
Use the data provided to determine the activation energy for the reaction.
Expt.
Rate constant
Temperature
1
2.5 x 10−3
75°C
2
3.6 x 10−4
35°C
ln(k2/k1) = Ea/R ( 1/T1 – 1/T2)
ln(3.6 x 10−4/2.5 x 10−3) = (Ea/8.314) ( 1/348 – 1/308)
Ln(0.144) = (Ea/8.314) (−0.0003732)
Ea = 4.316 x 104
4
answer: 4.32 x 10 J/mol
Page |4
(4 marks)
PROBLEM 7
Given the following data for the reaction NO2 + 2HBr → NO + H2O + Br2
Expt.
Initial rate, M/s
[HBr]o, M
[NO2]o, M
1
1.20 x 10−4
0.10
0.50
2
2.40 x 10−4
0.20
0.50
3
6.48 x 10−3
0.20
1.50
(a) Write the rate law expression.
expt 2/ expt 3;
(2.40 x 10−4 / 6.48 x 10−3) = (k)(0.20)x(0.50)y / (k)(0.20)x(1.50)y
3.7037 x 10−2 = 0.33y
log(0.037) = y0.33
y=3
expt 1/ expt 2;
(1.20 x 10−4 / 2.40 x 10−4) = (k)(0.10)x(0.50)y / (k)(0.20)x(0.50)y
0.50 = 0.50x
x=1
3
answer: Rate = k[HBr][NO2]
(b) Calculate the value of the rate constant (with units), using the data from experiment 1.
Expt 1
answer:
1.20 x 10−4 M/s = k(0.10 M)(0.50 M)3
K = (1.20 x 10−4 M/s) / (0.10 M)(0.50 M)3 = 9.6 x 10−3 M−3s−1
9.6 x 10−3 M−3s−1
Page |5
(6 marks)
PROBLEM 8
A proposed mechanism for a reaction is
step 1
CH3CH2Cl + AlCl3 → CH3CH2+ + AlCl4−
(fast)
step 2
CH3CH2+ + C6H6 → C6H5CH2CH3 + H+
(slow)
step 3
H+ + AlCl4− → HCl + AlCl3
(fast)
(a) What is the overall balanced equation for the mechanism?
answer: CH3CH2Cl + C6H6 → C6H5CH2CH3 + HCl
(b) Identify the catalyst(s), if any.
answer:
AlCl3
(c) Identify the intermediate(s), if any.
+
+
−
answer: CH3CH2 , H , AlCl4
(d) What is the molecularity of the first step?
answer: bimolecular
(e) What is the rate-determining step?
answer: Step 2
Page |6
(4 marks)
PROBLEM 9
Consider the following potential energy profile for A → D
(a) How many elementary steps are there?
answer:
3
(b) How many intermediates are formed?
answer:
2
(c) Is the overall reaction exothermic or endothermic?
answer:
exothermic
answer:
ii
(d) The rate-determining step is:
(i) A → B
(ii) B → C
(iii) C → D
(iv) A → D
Page |7
(3 marks)
BONUS
Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the
resulting solution after 40.0 mL of KOH have been added.
HClO4 (aq) + KOH (aq) → H2O (l) + KClO4 (aq)
0.200 M
0.100 M
x 0.0400 L
x 0.0400 L
= 0.00800 mol = 0.00400 mol
Total volume = 80.0 mL = 0.0800 L
[HClO4] = 0.00800 mol / 0.0800 L = 0.100 M
[KOH] = 0.00400 mol / 0.0800 L = 0.0500 M
Left with
HClO4 (aq) + KOH (aq) → H2O (l) + KClO4 (aq)
0.100 M
− 0.0500 M
= 0.0500 M
0.0500 M
− 0.0500 M
0
0
0
+ 0.0500 M
+ 0.0500 M
Since HClO4 is a strong acid, pH = −log (0.0500) = 1.30103
answer: 1.301
Page |8