Download ! !! ! n nn N P =

Nov 5, 2003
General Chemistry C
Mid-term Examination
(2 hours)
Question 13 need adjustment: B should read CO instead of CH4
Section B, question 4(b) GB0 and GA0 should read GP0 and GR0
You may find the following equations and physical constants useful. However, it does
not mean they are required in answering the questions.
[A] = [A]0 − kat
ln[ A] = ln[ A]0 − akt
1
1
=
+ akt
[A] [A]0
P=
K eq = e
−
∆G 0
RT
a x2 + b x + c = 0
x=
x=
N!
n A ! nB ! n x !
K a2 + 4(K a c ) − K a
2
− b ± b 2 − 4ac
2a
Gas constant R = 8.314 JK-1mol-1
Section A (30 points):
1.
Which one of the following statement is WRONG?
A. The higher the concentration of reactants, the faster the rate of the reaction.
B. The higher the temperature, the faster the rate of the reaction.
C. Reaction rate is higher when the reaction is exothermic.
D. Reaction rate is higher when the surface areas of the reactants are larger.
2.
For the reaction, 2R + S → 3M + N, what is the correct definition of the reaction
rate?
1
A.
3.
d [M ]
dt
B.
d [S ]
dt
C. −
1 d [R ]
2 dt
D. −
1 d [R ] d [S ]
−
2 dt
dt
What is the overall reaction order of the following rate law?
Rate = k [ A] [B ]2
1
A. 1
4.
C. B. 1
D. -1½
If the half life of a reaction with respect to a reactant concentration is
what is the reaction order of the reactant?
A. First order
B. Second order
C. Third order
0.693
,
k
D. Zero order
5.
Which of the following is the Zeroth law of thermodynamics?
A. Energy can never be created or destroyed but it can be changed from one form
to another.
B. Two bodies in thermal contact are at thermal equilibrium with each other if
the two bodies are at the same absolute temperature.
C. Any process carried out in several steps, the overall ∆H is equal to the sum of
the enthalpy changes (signed) for the individual steps.
D. None of the above.
6.
Which of the following is an extensive property?
A. Temperature
B. Volume
C. Molar heat capacity
D. Specific heat capacity
7.
For any spontaneous chemical reactions in a beaker ( ) at constant
temperature, which of the following condition must be satisfied?
8.
A. ∆Ssystem ≥ 0
B.
C. ∆G ≥ 0
D. ∆Suniverse < 0
H − T Ssystem < 0
For the titration of 0.1 M HClO4 against 0.5 M NH3 at 25 C, what is the pH
value at the equivalent point?
A. pH = 7
B. pH < 7
C. pH > 7
2
9.
Which one of the following reactants and product is a Lewis acid?
A. BF3
B. N(CH3)3
C. BF3-N(CH3)3
10. In an exothermic reaction, if we double the amount of reactants, which of the
following statements is correct?
A. The amount of heat evolved will be doubled
B. The amount of heat absorbed will be the same
C. The final temperature will be two times higher
D. The amount of heat evolved will be the same
11. NO2 gas will decompose at 573 K. The concentration of NO2 is measured as a
function of time:
Time, sec
[NO2], M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200
0.00481
300
0.00380
What is the reaction order with respect to NO2?
A. Zero order
B. First order
C. Second order
D. Third order
12. For a [H+] concentration of 0.01410 M, the pH value should read
A. 1.85078
B. 1.8508
C. 1.851
D. 1.85
13. Which of the following has the largest entropy at room temperature?
A. One mole of He atoms
3
B. One mole of CO molecules
C. One mole of CO2 molecules
D. 10 g of water
14. Which of the following reaction is NOT favored by the system entropy?
A. H2 (g) + Br2 (g) → 2 HBr (g)
B. 2H2 (g) + O2 (g) → 2 H2O (g)
C. H2O (l) → H2O (g)
D. HCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq)
15. Given that a saturated NaCl solution has a concentration of 5.4 M, which of the
following condition will cause NaCl to crystallize?
[NaCl (molar mass 58.44); NaNO3 (molar mass 85.00)]
A. Addition of 1 M NaNO3 solution to 5.4 M NaCl solution.
B. Addition of 10 g of NaCl crystals to 100 ml of 3 M NaCl solution
C. Addition of 6 M NaNO3 solution to saturated NaCl solution
D. Addition of 1.0 g of NaNO3 crystals to 10.0 ml of 4.4 M NaCl solution
Section B (50 points):
(For the final answer only: deduct 2 marks for wrong unit; deduct 2 marks for
wrong significant figure; accept a deviation of ±1 for the last digit)
1. The compound 2-germaacetic acid (GeH3COOH), is an unstable weak acid. At
25oC, a 0.050 M solution of 2-germaacetic acid has a pH of 2.42.
(a) Calculate the Ka of 2-germaacetic acid. [Hints: Ge is a Group 4 element, same as
carbon]
[H+] = 10-2.42 = 3.80 x 10-3 M
(2 marks)
-3 2
-3
Ka = (3.80 x 10 ) / (0.050 - 3.80 x 10 ) = 3.12 x 10-4
(3 marks)
(b) Given that the pKa value of CH3COOH is 4.75 at 25 oC. Predict which of the
groups, GeH3 and CH3 , has stronger electron donating property. Explain.
pKa for CH3COOH is 4.75 and pKa for GeH3COOH is 3.51. (1 mark)
Since GeH3COOH is a stronger acid, its conjugate base GeH3COO- is
more stable than CH3COO-.
(2 marks)
Since CH3COO is the less stable conjugate base, it means that the CH3
group has stronger electron donating property.
(2 marks)
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2. Cd2+ is one of the very toxic (
) heavy metal ions. Thus, it is very important to
remove Cd2+ from any wastewater ( ) before introducing the water back into our
environment ( ). Cd(OH)2 has a Ksp of 2.3 x 10-14.
(a) Using this solubility-product constant, calculate the concentration of Cd2+ in a
saturated solution of Cd(OH)2.
Cd(OH)2 (s) → Cd2+ (aq) + 2 OH- (aq)
Ksp = 2.3 × 10-14 = [Cd2+][OH-]2
= x (2x)2
[Cd2+] = x = 1.8 × 10-5 M
(5 marks)
(b) Suppose you are given 990 liters ( ) of wastewater containing Cd(OH)2.
Calculate how much of 5.0 M NaOH you have to add so that the final concentration
of [Cd2+] is reduced to 9.2 × 10-12 M.
2.3 × 10-14 = (9.2 × 10-12)[OH-]2
[OH-] = 0.050 M
Let y be the volume we have to add:
(y × 5.0 + 990 × 3.6 × 10-5)/(y+990) = 0.050
y = 10. liters or 1.0 × 101 liters
(5 marks)
3. The organic base aniline (C6H5NH2) forms its conjugate acid, the anilinium ion,
when treated with HCl.
C6H5NH2 (aq) + HCl (aq) → C6H5NH3+ (aq) + Cl- (aq)
(a) If Kb for aniline is 4.0 x 10-10, what is Ka for the anilinium ion?
Ka = Kw / Kb = 2.5 x 10-5
(b) What is the pH of a 0.080 M solution of anilinium chloride?
Using the approximation, x = [H+] = √[(2.5 x 10-5) (0.080)] = 1.41 x 10-3
pH = 2.84
4. Consider the reaction R (aq)
P (aq). Given that the ∆G° of this reaction is
zero.
(a) Calculate the Keq constant for the reaction.
Keq = 1 (1 mark)
5
(b) Sketch the variation of the free energy (G) as a function of the concentration of R.
[Hints: Consider x-axis as the concentration of R, y-axis as the free energy. You must
label the positions of G P0 , G R0 and the equilibrium position.]
G0R
G
G0P
Eqm position
P [%]: 0
100
R [%]: 100
0
0
P
G , G
0
R
and the equilibrium position (8 marks)
5. For the decomposition of carbonyl fluoride, COF2,
2 COF2 (g)
CO2 (g) + CF4 (g)
Keq = 2.00 at 1273 K. If 0.500 mol COF2 (g) is placed in a 3.23-L reaction vessel (
) at 1273 K, how many moles of COF2(g) will remain undissociated ( ) when
equilibrium is reached.
For the reaction above, the constant of equilibrium expression becomes:
Keq = [CO2][CF4]/[COF2]2 (2 marks)
Keq = (nCO2/V)(nCF4/V)/(nCOF2/V)2
Since the number of moles of gas for both sides of the reaction is the same, there
is no volume dependence:
Keq = (nCO2)(nCF4)/(nCOF2)2
One can work with moles directly. Applying stoichiometry, at equilibrium:
2.00 = (nCO2)2/(0.5 - 2nCO2)2
let x = nCO2,
2.00 = x2/(0.500-2x)2
(2 marks)
Take the square root of both sides:
6
1.414 = x/(0.500-2x)
x=0.185
Remaining COF2 = 0.500 - 2(0.185) = 0.131 moles
(6 marks)
Section C (30 points)
(Wrong unit for the final answer, deduct 1 mark; wrong significant figure for the
final answer, deduct 1 mark)
1. The stoichiometry for the reaction between Cl2 and ClO2- is shown in equations 1
and 2. Acidic conditions and high concentrations of chlorite favor the formation of
chlorine dioxide (ClO2), while basic conditions and low concentrations of chlorite
(ClO2-) favor the formation of the chlorate ion (ClO3-) :
Cl2 (aq) + 2ClO2- (aq) → 2ClO2 (aq) + 2 Cl- (aq)
Cl2 (aq) + ClO2- (aq) + 2OH- (aq) → ClO3- (aq) + 2Cl- (aq) + H2O (l)
(1)
(2)
[J. S. Nicoson and D. W. Margerum, Inorganic Chemistry, vol. 41 (2): 342-347, 2002.]
(a) The proposed mechanism for the reaction of chlorine and chlorite ion is as
follows:
Cl2 (aq) + ClO2- (aq) → ClOClO (aq) + Cl- (aq)
slow
ClOClO (aq) + ClO2 (aq) → 2ClO2 (aq) + Cl (aq)
fast
Based on the above mechanism, what should be the rate law for this reaction?
Do you think reaction 1 is an elementary process? Explain.
rate = k1 [Cl2][ClO2-]
(1 mark)
Reaction 1 is not an elementary process (1 mark)
because the rate law does not correspond to the stoichiometry of the reaction.
(4 marks)
(b) While keeping the total concentration of chlorine constant, the following data have
been obtained:
[ClO2-], mM
observed pseudo rate, s-1
0.211
132
0.421
228
0.632
334
0.842
455
1.050
550
7
What is the order of the reaction with respect to chlorite? Does the proposed
mechanism agree with these results?
The reaction is first order with respect to chlorite.
The proposed mechanism agrees with these results.
(2 mark)
(2 marks)
2. Normal rain is usually a bit acidic due to dissolved carbon dioxide (CO2). Carbon
dioxide dissolving in water makes the solution slightly acidic:
CO2(aq) + H2O(l)
H+(aq) + HCO3-(aq)
Ka = 4.3 x 10-7
Acid rain, on the other hand, is a major pollution problem in many parts of this planet
( ). Acid rain contains a fair amount ( ) of strong acids such as H2SO4
and HNO3. It results largely from the burning or combustion of fossil fuels ( ) that contain sulfur. Sulfur dioxide (SO2) is formed which can further
react with oxygen in the atmosphere (!
" ) forming sulfur trioxide (SO3). Sulfur
trioxide dissolves in water to form sulfuric acid:
S(in fossil fuel) + O2 (g) → SO2 (g)
SO2 (g) + (1/2) O2(g) → SO3 (g)
SO3 (g) + H2O (rain) → H2SO4 (aq)
The net result is to increase the acidity of the rain, which damages (# $ ) trees,
aquatic life (
% & ' ) and corrodes away (( ) ) stones and metal surfaces.
(a) The pH in acid rain can drop to 3 or even a bit lower in heavily polluted areas.
Calculate the concentrations of [H+] and [OH-] at pH 3.30 at 25oC.
pH = -log[H+]
[H+] = 10-pH = 10-3.30 = 5.0 x 10-4 M (1 mark)
in aqueous solution at 25oC, [H+][OH-] = 1.0 x 10-14
[OH-] = (1.0 x 10-14)/ (5.0 x 10-4) = 2.0 x 10-11 M
(1 mark)
(b) When sulfur dioxide (SO2) dissolves in water, sulfurous acid (H2SO3, Ka1=1.7 ×
10-2, Ka2=6.4 × 10-8) is formed. Sulfurous acid can donate a proton to a water
molecule. Write a balanced chemical equation for this reaction, and identify the
stronger Br* nsted-Lowry acid and base in the equation. (Consider only the first proton
8
of H2SO3)
H2SO3(aq) + H2O(l) <==> HSO3-(aq) + H3O+(aq)
(1 mark)
+
H3O is the stronger acid (as indicated by Ka1 < 1) (1 mark)
HSO3- is the stronger base.
(1 mark)
(c) Ignore (+ ,
) the further ionization (since Ka2 << Ka1) of HSO3-, and calculate the
pH of a solution in which the initial concentration of H2SO3 is 4.0 × 10-4M.
[Hints: Consider whether you need to use the quadratic formula]
Ka1 = [H+][HSO3-] / [H2SO3]
[H+]
[HSO3-]
[H2SO3]
initial
0
0
4.0 x 10-4M
shift
+x
+x
-x
equilibrium
+x
+x
4.0 x 10-4M - x
1.7 x 10-2 = x2/(4.0 x 10-4M - x)
x2 + (1.7 x 10-2)x - 6.8 x 10-6 = 0
Using the quadratic formula, we get
x = 0.00039 = [H+] so pH = 3.41
(2 marks)
(d) Now suppose (- . ) that all the dissolved SO2 from part (c) has been oxidized
initially to SO3, so that the initial concentration of H2SO4 is 4.0 × 10-4M. Calculate the
pH in this case. (H2SO4, Ka1 is very large, assume complete ionization, Ka2 = 1.2 x
10-2)
[Hints: Consider whether you need to use the quadratic formula]
H2SO4(aq) + H2O(l) ==> HSO4-(aq) + H3O+(aq) (assume complete)
HSO4-(aq) + H2O(l) <==> SO42-(aq) + H3O+(aq) Ka = 1.2 x 10-2
Thus, we need to deal with the second reaction:
1.2 x 10-2 = [H+][SO42-] / [HSO4-]
[H+]
[SO42-]
[HSO4-]
initial
4.0 × 10-4 (from 1st
rxn)
0
4.0 × 10-4M (from
1st rxn)
shift
+x
+x
-x
equilibrium
4.0 × 10-4 + x
+x
4.0 × 10-4M - x
9
1.2 × 10-2 = (4.0 × 10-4 + x) x / (4.0 × 10-4M - x)
x2 + (1.2 × 10-2) x - (4.8 × 10-6) = 0
Using the quadratic formula, we get
x = 0.00039
[H+] = 4.0 × 10-4 + x = 0.00079, so pH =3.10
(4 marks)
(e) Nature (! / ) also has its own way of fighting (0 1 ) acid rain. Lakes have a
natural buffering capacity, especially in regions ( 2 ) where there is limestone ( 3
) which gives rise to dissolved calcium carbonate. Write an equation for the
reaction of a small amount of acid rain containing sulfuric acid (H2SO4) falling into a
lake containing carbonate (CO32-) ions. Discuss how the lake will resist (1 4 ) further
pH changes. What happens if a large excess of acid rain is deposited in the lake?
CO32- is the conjugate base of the weak acid HCO3- and, thus, can react with the
strong acid H3O+ in the sulfuric acid solution according to
CO32-(aq) + H3O+(aq) <==> HCO3-(aq) + H2O(l)
(1 mark)
with K = 1/Ka2 of H2CO3 (thus, large K)
Likewise, HSO4- will also react with CO32-:
CO32-(aq) + HSO4-(aq) <==> SO42-(aq) + HCO3-(aq)
with K = (Ka2 of H2SO4)/(Ka2 of H2CO3) (likewise, large K)
(1 mark)
Thus, a HCO3-/CO32- buffer will be produced, which will resist further pH
changes.
An excess of acid rain will consume all the CO32- and HCO3-, converting all to
H2CO3 (no more buffer)
(1 mark)
3. Assume we have the following rate law:
−
d [ A]
2
= k [ A]
dt
Determine the half life for the concentration of A.
From the given equations, we have the following integrated equation for the
second order reaction:
1
1
=
+ kt
(1 marks)
[A] [A]0
10
Assume at time t1/2, the concentration [A] becomes [A]0/2:
1
1
=
+ kt
1
[A]0 1 / 2
2 [ A]0
kt1 / 2 =
t1 / 2 =
2
1
−
[A]0 [A]0
1
k [ A]0
(5 marks)
End
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