1.
Iodide ion is oxidized by H2SO3 to triiodide ion, which can be detected by the blue color
of a starch-triiodide compound:
6 I–(aq) + H2SO3(aq) + 4 H+(aq)
a.
S(s) + 2 I3–(aq) + 3 H2O(ℓ)
Estimate the instantaneous rate of formation of I3– at 450 s from the graph.
(4 pts.)
[H 2SO3 ]
(0.445  0.529) M
 
 8.40 x 104 M/s
t
(500  400) s
mole H 2SO3
2 mole I3
8.40 x 104
x
 1.68 x 10–3 M I3–/s
Ls
1 mole H 2SO3
rate  
b.
Initial rate data for the reaction are tabulated below:
[I–]i, M
0.0300
0.0600
0.0300
0.0300
0.0400
Exp#
1
2
3
4
5
[H2SO3]i, M
0.0200
0.0200
0.0400
0.0200
0.0300
[H+]i, M
0.0300
0.0300
0.0300
0.0600
0.0500
Initial rate, M/s
1.05 x 10–4
8.40 x 10–4
2.10 x 10–4
4.20 x 10–4
?
Write the rate law for the reaction. Show work. (4 pts.)
1-2
1-3
1-4
2x [I–]
2x [H2SO3]
2x [H+]
= 8x rate
= 2x rate
= 4x rate
[I–]3
[H2SO3]1
[H+]2
rate = k[I–]3[H2SO3][H+]2
c.
What is the order of the reaction in hydrogen ion? (3 pts.)
2nd order
d.
Calculate the numerical value of the rate constant. (3 pts.)
k 
e.
1.05 x 104 M/s
 2.16 x 105 M–5s–1
(0.0300 M)3 (0.0200 M)(0.0300 M) 2
Calculate the rate of experiment #5. (4 pts.)
rate = 2.16 x 105 M–5s–1 (0.0400 M)3 (0.0300 M) (0.0500 M)2
= 1.04 x 10–3 M/s
f.
What function of [H2SO3], when plotted versus time, would give a straight line?
What information can be obtained from the slope of this line? (3 pts.)
ln[H2SO3] versus t, slope = –k
The rate data of the following reaction were collected at different temperatures and
functions of the data plotted to determine the activation energy.
CH3Br(aq) + OH–(aq)
T (ºC)
17.0
35.7
46.4
57.6
0.0030
-10
k (M–1·s–1)
1.07 x 10–7
1.65 x 10–6
6.71 x 10–6
2.47 x 10–5
CH3OH(aq) + Br–(aq)
0.0031
-11
0.0032
0.0033
0.0034
0.0035
y = (-1.29 x 104)x + 28.4
-12
y = lnk
2.
-13
-14
-15
-16
-17
-1
x = 1/T (K )
a.
Clearly label the x and y axes (above) with the proper function. (2 pts.)
lnk versus 1/T
b.
Determine Ea of the reaction. (4 pts.)
slope = –(Ea/R); Ea = –(–1.29 x 104 K) x 8.314 J/mole·K = 107 kJ/mole
c.
Sketch a reaction profile, to scale, if ΔHreaction = –120 kJ/mole. (3 pts.)
3.
Barium peroxide (BaO2) is a powerful oxidizer that can produce Cl2 from Cl–:
1)
BaO2(ℓ) + 4 HCl(g)
K = 223.7 at 527 oC
a.
2 BaCl2(s) + 4 H2O(g) + 2 Cl2(g)
2
 1 
 1 
K2 = 
 =

 223.7 
 K1 
2 BaO2(ℓ) + 8 HCl(g)
2
= 1.998 x 10–5
Write the K equilibrium expression for reaction 1. (3 pts.)
K =
c.
ΔH = –376 kJ
Calculate the value of K at 527 oC, for reaction 2: (3 pts.)
2)
b.
BaCl2(s) + 2 H2O(g) + Cl2(g)
[H 2O]2 [Cl 2 ]
[HCl]4
Calculate the value of Kp for reaction 1 at 527 oC. (4 pts.)
Kp = K(RT)Δn = 223.7(0.08206 x 800.15)3–4 = 3.41
d.
BaO2, HCl, BaCl2, H2O and Cl2 are placed in a cylinder with a movable piston,
heated to 527 oC and allowed to reach equilibrium. Will the amount of Cl2 in the
cylinder increase, decrease, or not change in response to the following. Explain
briefly. Use reaction 1 as reference. (8 pts.)
i.
H2O(g) is added to the flask.
decrease
adding a product (H2O) shifts to remove product; toward reactant
ii.
The piston is pulled up.
decrease
pressure decreases – reactions shifts in direction of more moles of gas
to increase pressure; 4 moles reactants versus 3 moles product
iii.
BaCl2 is added to the flask.
not change
not part of the equilibrium expression; cannot change the
concentration of a solid
iv.
The temperature is decreased.
increase
goes in the direction to produce heat; to the right in this case
4.
A flask is filled with 0.500 M N2 and 0.800 M H2. The following takes place at 1100 oC:
N2(g) + 3 H2(g)
2 NH3(g)
At equilibrium, the concentration of NH3 is 0.160 M. Calculate K. (6 pts.)
Initial

Equil
[]
[N2]
0.500 M
–x
0.500 – x
0.500 –0.080 = 0.420 M
[H2]
0.800 M
–3x
0.800 – 3x
0.800 – 3(0.080) = 0.560 M
[NH3]
0M
+2x
2x = 0.160 M
0.160 M
 0.160 
[NH3 ]
= 0.347
=
3
3
[N 2 ][H 2 ]
 0.420  0.560 
2
K=
5.
Cyanogen, C2N2, is known as a pseudohalogen because it undergoes reactions similar to
the halogen, such as forming “inter”halogen compounds..
2 ICN(solv)
I2(solv) + C2N2(solv)
K = 2.08 (at 50.0 °C)
(solv) means dissolved in CHCl3
0.300 mole ICN and 0.450 mole of both I2 and C2N2 are placed in 1.00 L of solution.
a.
Which direction does the reaction go (to the left or to the right) to reach
equilibrium? Show a calculation. (4 pts.)
Q =
[I2 ]i [C2 N 2 ]i
(0.450)(0.450)
=
= 2.25; Q > K, reaction goes to left.
2
[ICN]i
(0.300)2
b.
Calculate the molarity of all substances at equilibrium. (8 pts.)
initial

equilibrium
[ICN]
0.300
+2x
0.300 + 2x
[I2]
0.450
–x
0.450 – x
[C2N2]
0.450
–x
0.450 – x
[I2 ][C2 N 2 ]
(0.450  x)(0.450  x)
(0.450  x) 2
K =
=
=
= 2.08
[ICN]2
(0.300 + 2x)2
(0.300 + 2x) 2
(0.450  x)
=
(0.300 + 2x)
2.08 ; 0.4327 + 2.8844x = 0.450 – x
3.8844x = 0.01733; x = 0.00446 M
[ICN] = 0.300 + 2(0.00446) = 0.309 M
[I2]
= 0.450 – 0.00446 = 0.446 M
[C2N2] = 0.450 – 0.00446 = 0.446 M
Check: K =
[I2 ][C2 N 2 ]
(0.446)(0.446)
=
= 2.08
2
[ICN]
(0.309) 2
6.
Calculate the pH of the following aqueous solutions. To receive full credit, you must
show an appropriate chemical equation for each. Constants in the handout. (30 pts.)
a.
0.20 M HC8H8O2 (phenylacetic acid)
C8H8O2–(aq) + H3O+(aq)
HC8H8O2(aq) + H2O(ℓ)
initial

equilbrium
[HC8H8O2]
0.20 M
–x
0.20 – x
[H3O+]
~0
+x
x
H2O
–
–
–
[C8H8O2–]
0
+x
x
[H3O+ ][C8 H8O2  ]
x2
x2
5
Ka =
= 5.2 x 10 =

[HC8 H8O2 ]
0.20  x 0.20
x = 3.2 x 10–3 M = [H3O+]; pH = –log(3.2 x 10–3) = 2.49
b.
0.10 M NaC7H14O2 (sodium heptanoate)
NaC7H14O2(aq)
Na+(aq) + C7H14O2–(aq)
C7H14O2–(aq) + H2O(ℓ)
initial

equilbrium
Kb =
[C7H14O2–]
0.10 M
–x
0.10 – x
HC7H14O2(aq) + OH–(aq)
H2O
–
–
–
[OH–]
~0
+x
x
[HC7H14O2]
0
+x
x
[OH  ][HC7 H14O2 ]
Kw
1.0 x 1014
x2
10
=
=
=
7.7
x
10

[C7 H14O2  ]
K a  HC7 H14O2 
1.3 x 105
0.10
x = 8.8 x 10–6 M = [OH–]; pOH = –log(8.8 x 10–6) = 5.06;
pH = 14.00 – pOH = 8.94
c.
0.0015 M HClO4 (perchloric acid)
HClO4(aq) + H2O(ℓ)
[H3O+] = [HClO4]i = 0.0015 M
pH = –log(0.0015) = 2.82
H3O+(aq) + ClO4–(aq)
d.
0.25 M Li2O (lithium oxide)
2 Li+(aq) + O2–(aq)
Li2O(aq)
2 OH–(aq)
O2–(aq) + H2O(ℓ)
[OH–] = 2 [O2–] = 0.50 M; pOH = 0.30; pH = 14.00 – pOH = 13.70
e.
0.34 M KBr (potassium bromide)
K+(aq) + Br–(aq)
KBr(aq)
Br–(aq) + H2O(ℓ)
HBr(aq) + OH–(aq)
Br– is the conjugate base of a strong acid, therefore a VERY weak base.
K+ is not an acid or a base.
The only species remaining that affects the pH is H2O. The pH of pure water
is 7.00.
f.
0.15 M (C5H5NH)NO3 (pyridinium nitrate)
(C5H5NH)NO3(aq)
C5H5NH+(aq) + NO3–(aq)
C5H5NH+(aq) + H2O(ℓ)
initial

equilbrium
[C5H5NH+]
0.15 M
–x
0.15 – x
C5H5N(aq) + H3O+(aq)
H2O
–
–
–
[H3O+]
~0
+x
x
[C5H5N]
0
+x
x
[H3O+ ][C5 H5 N]
Kw
1.0 x 1014
Ka =
=
=
= 5.9 x 106
+
9
[C5 H5 NH ]
K b  C5H5 N 
1.7 x 10
5.9 x 106 
x2
0.15
x = 9.4 x 10–4 M = [H3O+]; pH = –log(9.4 x 10–4) = 3.03