Chemistry 2300 (Loader)
Winter 2006
Page 1 of 3
Chemistry 2300
Y
TOTAL 50 MARKS
NAME:___________________________
March 22th 2006
MUN STUDENT#:____________________
Answer ALL of the questions in the spaces provided. Show your calculations or
explanations and give final numerical answers to the correct number of significant digits. The
mark that you obtain for this test will be used in calculating your final grade for the course.
1.
(a)
[6]
Some oxidizing agents are very sensitive to pH.
(i) Which of the following oxidizing agents become more or less powerful
and which are unaffected by a decrease in pH?
(A) Cl2(aq)
(B) Fe3+(aq)
(C) MnO −4 (aq) (D) Cr 2 O 2−
7 (aq)
Explain your choice. Write an expression to show how the electrode
potential for (C) is depends on pH.
(A) and (B) are not pH sensitive because H+ ion is not involved in the redox
equilibrium. The redox equilibrium for (C), MnO −4 (aq) is
MnO −4 (aq) +8 H + (aq) +5 e − t Mn 2+ (aq) + 4 H 2 O(l)
The equilibrium constant K=
a Mn 2+ a 4H2 O
a MnO−4 a 8H+
and since the activity of the H+ is raised to
the eighth power is strongly influenced by pH. Decreasing the pH means the
concentration of H+ increases and will lead to a lower value for the expression for
the reaction quotient, Q, and E more positive and the reaction becomes more
spontaneous.
Using the Nernst equation
E = Eo −
RT
5F
ln
a Mn 2+ a 4H
2O
a MnO −4 a 8H +
Increasing the concentration of H+ will Q smaller and ln Q more negative which will
lead to a more positive value for E indicating an increase in the power of the
oxidizing agent. In a similar fashion (D) also involves the hydrogen ion
concentration on the left had side of the equilibrium and thus the oxidizing power
must also increases as the pH decreases.
[10]
(b)
In an experiment to determine the solubility product of thallium chloride,
TlCl(s), the following cell was constructed:
Pt|H2(g, 1 atm)|H+(aq, 1.0 m)||Cl–(aq,0.20 m)|TlCl(s)|Tl(s)
The measured cell potential, E298 K, for the cell was −0.460 V at 298 K.
The standard reduction potentials for Tl+(aq)
E o298 K = −0.336 V
Tl+(aq) + 1 e– t Tl(s);
(i)
Write the equation for the spontaneous cell reaction.
2 Tl(s) + 2 H+(aq) + 2 Cl − (aq) t 2 TlCl(s) + H2(g)
(ii) Calculate the concentration of Tl+(aq) in the right hand electrode.
The concentration of the thallium ion in the right-hand electrode is governed by
the solubility product of TlCl(s). We can use the Nernst equation to find
concentration ot the thallium ion. The spontaneous reaction as far as the Tl+ is
concerned is
2 Tl(s) + 2 H+(aq) t 2 Tl+(s) + H2(g) for which E ocell = 0.336V.
So using the Nernst equation 0.460 V = 0.336 V -
8.314%298
2%96485
so the molality of Tl+ = 7.995 x 10 –3 m
(iii)
Calculate the solubility product, Ksp, for TiCl(s)
TlCl(s) î Tl+(aq) + Cl–(aq)
So the Ksp = 0.20 x 7.885 x 10-3 = 1.6 x 10 –3
Ans: 1.6 x 10 –3
ln
m 2Tl +
1
Chemistry 2300 (Loader)
.
(a)
Winter 2006
Page 2 of 3
The equilibrium
I2(g) + H2(g) î 2 HI(g)
has been very carefully studied. At 300 K the following data is available:
H o300 K = −9.6 kJ mol −1 , S o300 K = 22.18 JK −1 mol −1 .
Given that C p = −9.6 JK −1 mol −1for the reaction between 300 and 500 K,
(i)
calculate the equilibrium constant, Kp, for the reaction at 500 K.
−1
−1
J K mol
H o500 K = −9.6 kJ mol −1 + (− 9.61000
−11.52 kJ mol −1
J kJ −1 % 200 K) =
K
−1
−1
S o500 K = −22.18 JK −1 mol −1 + (−9.6 JK −1 mol −1 ln 500
300 K ) = −22.38 JK mol
−1
−1
JK mol
% 500 K) = −22.71 kJ mol −1
G o500 K = −11.52 kJ mol −1 + ( −22.38
1000 J kJ −1
−1
−1
J K mol %500 K
G o500 K = −22.71 kJ mol −1 = −8.3141000
ln Kp
J kJ −1
Kp = 2.4 x 102
[6]
(ii) Pure HI is introduced into a vessel and heated to 500 K so that the final pressure at
equilibrium was 10.0 atm. Calculate the mole fraction of HI present at equilibrium.
[Note: Use Kp = 131.9 if you did not answer (i) above].
The number of moles does not change in the reaction, the initial pressure if the gas
was pure HI would be 10.0 atm. Let the final H2 partial pressure = X then
[HI] 2
Kp = [H 2 ][I 2 ]
[10.0−2X ] 2
= 131.9
X2
10.0−2X
= 11.485 and x
X
Kp =
so
a perfect square
= 0.7416 atm
so P H 2 = 10.0 atm - (2 x 0.7416) atm = 8.517 atm so the partial pressure of
8.517
HI = 10 = 0.85
Ans: 0.85
[5]
(iii)
How will the mole fraction of HI present at equilibrium change if the total
pressure is increased?
[2]
Since there are equal moles on each side of the equation, the equilibrium constant
does not change when the total pressure is increases. The mole fraction of each gas
is also constant (The total pressure cancels in the above calculation.
[5]
(b)
Given that
o
2 Hg 2+ (aq) + 2 e − t Hg 2+
2 (aq);E = 0.920 V
2+
−
o
Hg 2 (aq) + 2 e t 2 Hg(l);E = 0.920 V
calculate the value of r G and K at 25 oC for the reaction
2+
Hg 2+
2 (aq) t Hg (aq) + Hg(l)
Using the equations as giver, if we reverse the first equation and add...
−
o
2+
Hg 2+
2 (aq) t 2 Hg (aq) + 2 e ;E = -0.920 V
−
o
Hg 2+
2 (aq) + 2 e t 2 Hg(l);E = 0.920 V gives the required equation
2+
o
Hg 2+
2 (aq) t Hg (aq) + Hg(l); E = 0.000 V
Since E o = 0.000 V we have r G o = 0 and K = 1
Chemistry 2300 (Loader)
[5]
3.
(a)
Winter 2006
Page 3 of 3
The enthalpy of vaporization of trichloromethane, CHCl3(l), is 29.4 kJ mol–1 at
the normal boiling point of 61.7 oC.
(i) Calculate the molar entropy of vaporization trichloromethane at the
normal boiling point.
Hvap
q rev
29.4kJ mol –1
−1
S = T = T = (273.15
= 87.8 J K mol −1
+ 61.7) K x1000 J kJ
(ii) Calculate the entropy change in the surroundings per mole of CHCl3
when the vaporization takes place.
The vaporization takes place at the normal boiling point so is reversible and
Ssurroundings =
[6]
(b)
−q rev
T
=
−Hvap
T
–1
29.4kJ mol
−1
= − (273.15
= −87.8 J K mol −1
+ 61.7) K x1000 J kJ
Monochloroethanoic acid, ClCH2COOH, has pKa = 2.87
(i) Calculate the percentage ionization of the of a 0.010 mol L–1 solution of
ClCH2COOH.
The equilibrium is ClCH 2 COOH î ClCH 2 COO − (aq) + H + (aq) and
[H + ] 2
K A = C o − [H + ] = 10 −2.87 If [H+]<<Co and [H+] from the water is small
[H+ ] 2
compared to the [H+] from the water then KA = Co = 1.35 x10 −3
[H+] = 0.00367 mol L-1 and is not [H+]<<Co so the quadratic must be solved
and give 3.1 x 10-3 mol L-1. So the percentage ionization
= 0.00305
0.010 % 100% = 31% : Ans
(ii) Calculate the pH of the solution formed when 5.00 mL of 0.10 mol L–1
NaOH is mixed with 10.00 mL of 0.10 mol L–1 ClCH2COOH.
This is half-neutralization so pH = KA = 2.87
(c)
For the cell
Pt|H2(g, 1 bar)|HCl(aq, 0.010 m)|AgCl(s)|Ag(s)
the cell reaction is:
½ H2(g) + AgCl(s) t Ag(s) + H+(aq) +Cl–(aq).
The temperature coefficient for the cell,
Calculate r S for the reaction.
o
( ØE
ØT ) P =
–8.665 x 10-5 V K–1.
[5]
ØE
Since r S = νF( ØT )P = 1 mol x 96485 C x –8.665 x 10-5 J C-1 K–1
= -8.360 J K-1 mol-1
o