Chemistry 1B Spring 2006
Examination #1 ANSWER KEY
1.
(23 pts. total) Questions A – E below involve the citric acid (H3 C6 H5 O7 )
molecule shown, which possesses the following acid-dissociation constants:
H
O
H
O
C
O
O
C
H
O
Ka1 = 7.4 x 10-4
Ka2 = 1.7 x 10-5
Ka3 = 4.0 x 10-7
C
C
H2
CH2
C
O
O
H
A.
(6 pts.) Write three equations describing the complete ionization of this
polyprotic acid. Why do the acid-dissociation constants become
progressively smaller? Circle the three most ionizable H atoms in the
structure above.
H3 C6 H5 O7 ? H2 C6 H5 O7- + H+
H2 C6 H5 O7 - ? HC6 H5 O7-2 + H+
HC6 H5 O7 -2 ? C6 H5 O7 -3 + H+
***On the basis of electrostatic attractions, one would expect a
positively charged proton to be lost more readily from a neutral
molecule than from a negatively charged ion. Therefore, it is always
easier to remove the first proton from a polyprotic acid than to
remove the second and third, respectively (three most acidic
hydrogen atoms circled above).
B.
(3 pts.) The H2 C6 H5 O7- ion is considered amphiprotic. What is an
amphiprotic substance? Write two balanced chemical equations with
ONLY the H2 C6 H5O7- ion and water which illustrate this behavior.
***An amphiprotic substance is capable of acting as either an acid or
a base in solution (i.e. can either gain or lose a proton).
H2 C6 H5 O7 - ? HC6 H5 O7-2 + H+
H2 C6 H5 O7 - + H2 O ? H3 C6 H5 O7 + OH-
C.
(4 pts.) For the equilibrium reaction below, label the following: acid, base,
conjugate acid, and conjugate base.
Base
HC6H5O72- + H2O
Acid
D.
Conjugate Acid
H2C6H5O7- + OHConjugate Base
(7 pts.) Calculate the pH of a 0.35 M H3 C6 H5 O7 solution and the
equilibrium concentrations of the species H3 C6 H5 O7 , H2C6 H5O7-,
HC6 H5 O7 2-, and C6 H5 O7 3-.
H3 C6 H5 O7
H2 C6 H5 O7 H+
Initial
0.35
0
0
Change
-x
+x
+x
Equilibrium
0.35 – x
x
X
2
-4
-2
x /(0.35 – x) = 7.4 x 10 ; approx. to obtain x = 1.61 x 10 M; pH = 1.79
H2 C6 H5 O7 HC6 H5 O7 -2
H+
Initial
1.61 x 10-2
0
1.61 x 10-2
Change
-y
+y
+y
-2
Equilibrium
1.61 x 10 - y
y
1.61 x 10-2 + y
y[1.61 x 10-2 + y]/[ 1.61 x 10-2 – y] = 1.7 x 10-5 ; approx. to obtain
y = 1.7 x 10-5 M
HC6 H5 O7 -2
C6 H5 O7 -3
H+
Initial
1.7 x 10-5
0
1.61 x 10-2
Change
-z
+z
+z
Equilibrium
1.7 x 10-5 - z
z
1.61 x 10-2 + z
z[1.61 x 10-2 + z]/[ 1.7 x 10-5 - z] = 4.0 x 10-7 ; approx. to obtain z = 4.2 x 10-10 M
Therefore, pH = 1.79; [H3 C6 H5 O7 ] = 0.33 M; [H2 C6 H5 O7 -] = 1.61 x 10-2 M;
[HC6 H5 O7 -2 ] = 1.7 x 10-5 M; and [C6 H5 O7 -3 ] = 4.2 x 10-10 M.
E.
(3 pts.) Consider the –CH2 – component of citric acid labeled with the
arrow. What affect does replacing both hydrogen atoms with chlorine
atoms have on the overall acidity of the molecule? Briefly explain.
The overall acidity of this molecule INCREASES when the H atoms
are replaced with Cl atoms due to the inductive effect, where electronwithdrawing, more electronegative heteroatoms pull the electron
density away from the carboxylic acid hydrogen atoms, thereby
making the latter more susceptible to deprotonation by a potential
base.
2.
(8 pts. total) Consider the following proposed reaction mechanism:
k1
FAST EQUILIBRIUM
H3O+ + H2O2
H3O2+ + H2O
k-1
k2
H O + + IH O + HOI
3
2
HOI + H3O+ + II2 + I-
A.
B.
C.
D.
E.
F.
3.
k3
k-3
k4
2
FAST
2H2O + I2
FAST EQUILIBRIUM
I3-
SLOW
(2 pts.) Determine the rate law for this given mechanism.
+ 2 - 3
k1 k2 k3 k4 [H 3O ] [I ] [H2O 2]
rate = k k
[H 2O]3
-1 -3
(1 pt.) What is the overall reaction for this process? 2H3 O+ + H2 O2 + 3I? I3 - + 4H2 O
(1 pt.) List any intermediate(s) present in the mechanism, if any. H3 O2 +,
HOI, I2 .
(1 pt.) List any catalyst(s) present in the mechanism, if any. None .
(2 pts.) Draw the appropriate reaction profile energy diagram for this
mechanism, making sure to label all the appropriate components. (see
course guide for examples; your drawing should contain 4 “humps”,
where the last portion of the curve rises the most since it is the RDS)
(1 pt.) How does a catalyst affect the rate of a reaction? Briefly explain
AND clearly illustrate on your proposed reaction profile energy diagram.
A catalyst speeds up a reaction by lowering the activation energy (for
an illustrative example of this, see your course guide)
(8 pts.) Consider the equilibrium reaction shown below, where Kc = 54.5:
H2(g) + I2(g)
2HI(g)
A mixture containing 4.562 x 10-3 mol H2 (g), 7.384 x 10-4 mol I2 (g), and 1.355 x
10-2 mol HI(g) in a 1.000 L container at 425 ºC is at equilibrium. If 1.000 x 10-3
mol I2 (g) are added at the same temperature, what is the composition of all three
species once equilibrium is re-established?
Equilibrium
Initial
Change
Equilibrium
(again!)
H2 (g)
4.562 x 10-3 M
4.562 x 10-3 M
-x
4.562 x 10-3 - x
I2 (g)
7.384 x 10-4 M
1.738 x 10-3 M
-x
1.738 x 10-3 - x
2HI(g)
1.355 x 10-2 M
1.355 x 10-2 M
+2x
1.355 x 10-2 + 2x
[1.355 x 10-2 + 2x]2 / {[4.562 x 10-3 – x][1.738 x 10-3 – x]} = 54.5
50.5x2 - 0.398x + (2.49 x 10-4 ) = 0; solve quadratic to obtain x = 6.85 x 10-4 M
Therefore, [H2 ] = 3.877 x 10-3 M; [I2 ] = 1.053 x 10-3 M; [HI] = 1.492 x 10-2 M
4.
(9 pts. total) The half- life of a reaction of compound A to give compounds C and
D is 12.3 min when the initial concentration of A is 0.360 M. How long will it
take for the concentration to drop to 0.0182 M if:
A.
B.
C.
(3 pts.) the reaction is ZERO order with respect to A?
(3 pts.) the reaction is FIRST order with respect to A?
(3 pts.) the reaction is SECOND order with respect to A?
A.
t1/2 = Ao /2k; solve for k to obtain k = 0.0146 M min-1
A – Ao = -kt; solve for t to obtain t = 23.4 min
B.
t1/2 = 0.693/k; solve for k to obtain k = 0.0563 min-1
A = Ao e -k t; solve for t to obtain t = 53.0 min
C.
t1/2 = 1/kAo ; solve for k to obtain k = 0.226 M -1 min-1
(1/A) – (1/Ao ) = kt; solve for t to obtain t = 231 min
5.
(8 pts.) Cocaine is a weak organic base whose molecular formula is C17 H21NO4 .
An aqueous solution of cocaine was found to have a pH of 8.53 and an osmotic
pressure of 52.7 torr at 15 ºC. Calculate Kb for cocaine.
p (osmotic pressure in atm) = iMRT; note that i = 1 and p = 0.0693 atm
M = p/iRT = 2.93 x 10-3 M
Initial
Change
Equilibrium
C17 H21 NO4
2.93 x 10-3
-x
2.93 x 10-3 – x
H2 O
-
HC17 H21NO4 +
0
+x
x
OH0
+x
x
pH = 8.53; pOH = 14 – 8.53 = 5.47; therefore, [OH-] = 3.39 x 10-6 = x
Kb = [HC17 H21NO4 +][OH-]/[C17 H21NO4 ]
= [3.39 x 10-6 ]2 / [0.00293 – (3.39 x 10-6 )]
= 3.93 x 10-9
6.
(10 pts.) Consider the reaction 2HCl(g) + I2 (s) ↔ 2HI(g) + Cl2 (g), where
Kc = 1.6 x 10-34 . If 4.00 L of HCl(g) at 1.00 atm and 273 K is mixed with 26.0 g
of I2 (s), transferred to a 12.00 L reaction vessel, and heated to 25 °C, what will
the equilibrium concentrations of HCl, HI, and Cl2 be?
PV = nRT; n HCl = PV/RT = 0.1786 mol
M HCl = 0.1786 mol/12.00 L = 0.015 M
2HCl(g)
I2 (s)*
2HI(g)
Cl2 (g)
Initial
0.015
0
0
Change
-2x
+2x
+x
Equilibrium
0.015 – 2x
2x
x
*recall that solids are NOT included in the equilibrium equation!
4x3 /[0.015 – 2x]2 = 1.6 x 10-34 ; approximate to obtain x ˜ 2.08 x 10-13 M
***Check to see that the approximation is valid by plugging back into the
original equilibrium equation; approximation is indeed valid!
Therefore, [HCl] = 0.015 M; [HI] = 4.16 x 10-13 M; [Cl2 ] = 2.08 x 10-13 M
7.
(9 pts. total) The rate constant for the hydrolysis of the sugar sucrose to the sugars
glucose and fructose is 2.1 x 10-11 s-1 at 27 ºC and 8.5 x 10-11 s-1 at 37 ºC.
C12 H22O11 + H2O ?
A.
B.
C.
8.
C6 H12O6 + C6 H12O6
(3 pts.) Determine the energy of activation for this hydrolysis reaction in
kJ/mol.
(3 pts.) What is the rate constant at 47 ºC?
(3 pts.) Determine the frequency factor A for this reaction.
A.
ln (k2 / k1 ) = [-Ea /RT2 + Ea /RT1 ] where at 600 K, k = 3.4 x 10-4 M -2s -1
Ea = [R ln (k2 / k1 )]/[(1/T1 - 1/T2 )]
= [0.0083145 ln (8.5 x 10-11 /2.1 x 10-11 )]/[(1/300) – (1/310)]
Ea = 108 kJ/mol
B.
k2 / k1 = e -Ea/RT2 / e -Ea/RT1 = e^[-Ea /RT2 + Ea /RT1 ]
Solve for k2 to obtain k2 = 3.1 x 10-10 s -1
C.
k = Ae -Ea/RT; solve for A to obtain A = 1.3 x 108 s -1
(12 pts. total; 2 pts. each) Briefly define/explain each of the following terms, and
give an example of each.
A.
Lewis acid – an electron pair acceptor (many examples!)
B.
activated complex – the transition state of a reaction profile energy
diagram where bonds are made and bonds are broken simultaneously
(e.g. see reaction profile energy diagrams in course guide)
9.
C.
Brønsted-Lowry base – a molecule or ion that acts as a proton acceptor
(many examples!)
D.
Le Châtelier’s Principle – when a system at equilibrium is disturbed,
the system shifts so as to relieve the imposed stress (many examples!)
E.
heterogeneous catalyst – a catalyst that is in a different phase from that
of the reactant substances (many examples!)
F.
Arrhenius equation – an equation that relates the rate constant for a
reaction to the frequency factor A, the activation energy EA, and the
temperature T according to the form k = Ae -Ea/RT (many examples!)
(8 pts. total) Experiments were conducted to study the rate of the reaction
represented by the equation below:
2NO(g) + 2H2 (g) → N2 (g) + 2H2 O(g)
10.
Experiment
[NO], M
[H2 ], M
1
2
3
0.0060
0.0060
0.0010
0.0010
0.0020
0.0060
Initial Rate of
Formation of N2 .
M/min
1.8 x 10-4
3.6 x 10-4
0.30 x 10-4
A.
B.
(5 pts.) Determine the overall rate law for the reaction.
(3 pts.) Calculate the value of the rate constant, k, for experiment #1
(including units).
A.
Take the ratio of exp 2: exp 1 to obtain y = 1
Take the ratio of exp 2: exp 3 to obtain 6x (1/3) = 12; x = 2
Therefore, rate = k[NO]2 [H2 ]
B.
1.8 x 10-4 M/min = k[0.0060 M]2 [0.0010 M]; k = 5.0 x 103 M -2 min-1
(5 pts.) Calculate the pH of a 0.25 M solution of NaC7 H5 O2 (K a for HC7 H5 O2 is
equal to 6.3 x 10-5 ).
Initial
Change
Equilibrium
C7 H5 O2 0.25 M
-x
0.25 – x
H2 O
-
HC7 H5 O2
0
+x
x
x2 /(0.25 – x) = (1.0 x 10-14 )/(6.3 x 10-5 ) = 1.59 x 10-10
***Approximate to obtain x ˜ 6.3 x 10 -6 M = [OH-]
pOH = -log [OH-] = 5.20; therefore, pH = 14 – pOH = 8.80
OH0
+x
x