Chemistry 2300 (Loader)
Winter 2005
Page 1 of 3
Chemistry 2300
Y
TOTAL 50 MARKS
NAME:___________________________
March 23th 2005
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.
[6]
(a)
Phosphorus pentachloride, PCl5, when heated gives an equilibrium mixture of
PCl5(g) , PCl3(g) and Cl2(g) in the reaction:
PCl5(g) î PCl3(g) + Cl2(g).
When 4.80 g of pure PCl5 were placed in an evacuated 1.00 litre container
and heated at 503 K until equilibrium was reached, the final pressure in the
container was 760 mmHg and the measured partial pressure due to the
chlorine in the mixture was 28.1 mmHg.
(i) Calculate the equilibrium constant, Kp, for the reaction at 503 K.
We know the final pressure and the partial pressure due to Cl2(g). Since p Cl 2 (g) must
= p PCl 3 (g) as demanded by the stoichiometry p PCl 5 (g) can be calculated using
Dalton's Law.
p PCl 3 = p Cl 2 = 28.1 mmHg so p PCl 3 = (760 − 28.1 − 28.1)mmHg = 703.8 mmHg.
Since 760 mmHg = 1 atm = 101.3 kPa, 1 bar = 750.2 kPa.
P PCl
Kp =
Po
P Cl
3
2
Po
PCl 5
Po
28.1
28.1
750.2
750.2
703.8
750.2
=
= 1.49 5 % 10 −3
(using 1 atm as standard Kp = 1.49 5 % 10 −3 ) Ans: 1.50 % 10 −3
[4]
(ii)
)
and G503K for the reaction.
)
G 503 K = −RT ln K p
)
G503K = −8.314 J K−1mol−1 % 503 Kln1.495 % 10−3
)
[6]
Ans: 27.2 kJ mol −1
G503 K = 27.21 kJ mol −1
(b) Barium oxide can be made by the thermal decomposition of barium
carbonate at high temperature.
(A) BaCO3(s) t BaO(s) + CO2(g),
)
r G 298 K = 219.5 kJ mol-1; r H )298 K = 272 kJ mol-1; r S )298 K = 172 J K-1 mol-1
The temperature required can be lowered by coupling the reaction with the
reaction that is the reduction of carbon dioxide by carbon:
(B) C(s) + CO2(s) t 2 CO(g),
)
r G 298 K = 119.8 kJ mol-1 ; r H )298 K = 172 kJ mol-1; r S )298 K = 182.2 J K-1 mol-1
The overall reaction equilibrium becomes
(C) BaCO3(s) + C(s) î BaO(s) + 2 CO(g)
Estimate the temperatures at which reactions (A) and (C) become
spontaneous and explain why reaction (C) becomes more favorable at higher
temperatures.
Equation (C) can be obtained from equations (A) and (B)
H o
At equilibrium Go = 0 so 0 = H o − TS o and T = S o . Of course the equation
Go = Ho − TS o is only true at constant temperature. If Ho and So are reasonably
constant over a temperature range then the formula 0 = H o − TS o can be used to estimate the
% 103 kJ mol −1
temperature at which Go = 0. So for reaction (A) T = 272
172 J K −1 mol −1 = 158 1 K
Ans: 1.58 x 103 K
and for reaction (C) Ho = 272 + 172 = 444 kJ mol −1
and So = 172 + 182.2 = 354.2 kJ mol −1
3
−1
10 kJ mol
to T = 444%
Ans: 1.25 x 103 K
354.2 J K −1 mol −1 = 125 4 K
Although reaction (C) has a very +ve Go = 339.3 kJ mol −1 and reaction (A) is less
positive Go = 219.5 kJ mol −1 , because r S ) for (C) is so much larger (2 moles of gas
are formed) Go become less positive much more rapidly than for reaction (A) and the
reaction becomes spontaneous at a lower temperature.
Chemistry 2300 (Loader)
2.
[5]
Winter 2005
Page 2 of 3
(a)
In an experiment 1.0 mL of 0.10 mol L–1 hydrochloric acid was first diluted to
1.0 L with pure water and then 1.0 mL of the solution formed was diluted a
second time to 1.0 L. Calculate the pH of the solutions formed after the first
and after the second dilutions.
HCl present in 1.0 mL = (1.0 x 10-3 L) x 0.10 mol L–1 = 1.0 x 10-4 mol. Since HCl is a
strong acid the [H+] from the acid = 1.0 x 10-4 mol. The contribution from the
K
ionization of the water = [H + ]wtotal would give a value of no more than 1.0 x 10-10 mol
so can be ignored in this solution. So, pH = -log 1.0 x 10-4 = 4.00. :ANS
HCl present in 1.0 mL = (1.0 x 10-3 L) x 1.0 x 10-4 mol L–1 = 1.0 x 10-7 mol. Since HCl is a
strong acid the [H+] from the acid = 1.0 x 10-7 mol. The contribution from the
K
ionization of the water = [H + ]wtotal and this would give a value of close to 1.0 x 10-7 mol
K
so must not be ignored in this solution. So, [H+] = 1.0 x 10-7 + [H + ]wtotal.
−14
So [H+] = 1.0 x 10-7 + 1.0%10
[H + ] . Solving the quadratic or using successive
approximations gives [H+] = 1.62 x 10-7 giving pH = 6.79 :ANS
(b) The freezing point of a 0.400 mol L–1 solution of methanoic acid (HCOOH, a
monoprotic acid) was 0.758 °C lower than that of pure water.
[4]
Calculate the van't Hoff factor i for the methanoic acid solution. (You
may assume that the molarity is equal to the molality of the solution.
The freezing point depression constant, Kf for water = 1.86 K mol–1 kg.)
The freezing point depression is given by T = K f % m
(i)
1.86 K mol–1 kg x m = 0.758 K so, m = 0.40752 mol kg–1. If this can be taken to be 0.40752 mol
L–1 for this question.
The van't Hoff factor i =
[5]
0.40752
0.400 =
1.0188
Ans: van't Hoff factor i = 1.02
(ii)
Calculate the acidity constant, Ka, for methanoic acid at the temperature
of the experiment. (See Chang Q. 11.70)
HCOOH î HCOO− + H+
Formic acid ionizes in water according to:
If the initial concentration before ionization is 0.400 mol L–1 and x mol L–1 ionizes
then the final number of particles in the solution is
(0.400 mol L–1- x mol L–1) + 2 x mol L–1 = 0.400 mol L–1 + x mol L–1 and we
calculated this in (a). Now, 0.400 mol L–1 + x mol L–1 = 0.40752 mol L–1
x mol L–1 = 7.52 x 10-3 mol L–1 = [H+]=[HCOO-]
Ka =
[6]
[H + ][HCOO − ]
[7.52x10−3mol L –1 ]
[HCOOH] = [0.400 −7.52x10 −3 ] molL–1
=1.44 x 10-4
Ans: Ka =1.44 x 10-4
(c)
List all of the ionic species present and calculate their concentrations in a
0.20 mol L –1 solution of oxalic acid, HOOCCOOH, (a diprotic acid) for which
pKa1 = 1.19 and pKa2 = 4.21.
The ions present are HOOCCOO − , − OOCCOO− , H+ and OH−
and Ka1 = 6.457 x 10-2 and Ka2 = 6.1660 x 10-5.
Since the pKa values are well separated [H+]first ionization >> [H+]second ionization.
We can start by calculating the [H+]first ionization making the assumption that other sources can
be ignored as being insignificant.
The first ionization is HOOCCOO î HOOCCOO− + H+
Ka1 =
[HOOCCOO − ][H + ]
[HOOCCOOH]
+
and since [HOOCCOO − ] l [H+ ] then
[H ] = K a1 % [HOOCCOOH] so [H+ ] = 6.467 % 10 −2 (0.20 − [H + ]) .
If 0.20 >> [H+] then [H+ ] = 6.467 % 10 −2 (0.20) = 0.11373 so 0.20 (([H+]
and the quadratic must be solved. Only the +ve root is meaningful and gives
[H+ ] = 0.085 8 mol L –1. We can now use this to calculate [H+] from the second ionization.
If [H+ ] first >> [H + ] sec ond ionization = [ − OOCCOO − ]. If [H+ ] total l [H+ ] from first ionization then
[ − OOCCOO− ] =
Also, [OH] =
6.1660 % 10 −5 % 0.085 8
=
0.085 8
Kw
[H + ] .
+
6.1660 x 10-5. This is too small to be significant.
So [HOOCCOO − ] = 086 mol L −1 ,
[H ] = 086 mol L −1 , [ − OOCCOO− ] = 6.2 x 10 −5 mol L −1
and [OH− ] = 1.2 % 10−13 mol L−1
Chemistry 2300 (Loader)
3.
(a)
Winter 2005
Page 3 of 3
The following cell was constructed
Ag(s)|AgCl(s)|Cl–(aq, 1.00 mol L–1)||Ag+(aq , 1.00 mol L–1)|Ag(s)
and the cell potential, Ecell = 0.5766 V at 25°C. Use this information to
calculate the solubility product for AgCl(s).
The equation for the cell reaction as written is
Ksp
[Cl − ]
Ag + (aq, 1.00 mol L –1 ) î Ag + (aq,
so E = E o −
0.5766 V =
Ksp
[Cl − ]
RT
o
nF ln 1 and E
K sp
% 298
− 8.314
ln 1
96485
Ksp = 1.77 x 10
), a concentration cell,
=0
V
-10
Ans: Ksp = 1.77 x 10-10
[7]
(b)
[7]
In a study of the redox reaction
Ce4+(aq) + Fe2+(aq) t Ce3+(aq) + Fe3+(aq)
various volumes of a solution containing 0.10 mol L–1 of Fe2+(aq) were added
to 50.0 mL of a solution containing 0.10 mol L–1 Ce4+(aq).
Given that for
Ce4+(aq) + e– t Ce3+(aq), E° = 1.72 V
Fe3+(aq) + e– t Fe2+(aq), E° = 0.771 V
Calculate the half-cell potential for the electrode that initially contained
50.0 mL of 0.10 mol L–1 of Fe2+(aq) after a total of (a) 25.0 mL of 0.10 mol L–1
Ce4+(aq) solution and (b) 30.0 mL of 0.10 mol L–1 Ce4+(aq) solution is added.
Both the Ce4+/Ce3+ and the Fe3+/Fe2+ couples are in the SAME vessel forming one
electrode. The half-cell potential for both couples must be the same or one would
oxidize the other. Either couple can thus be used to calculate the half-cell potential.
In this case it is easier to use the Fe3+/Fe2+ as the concentrations are simple to
calculate.
When 25.0 mL of 0.10 mol L–1 Ce4+(aq) has been added, half of the Fe2+ has been used so
the concentration of Fe2+ = the concentration of Fe3+. Now
EFe3+/Fe2+ = EoFe3+/Fe2+ -
RT
nF
ln
[Fe 2+ ]
[Fe 3+ ]
= 0.77 1 V ANS: (a) =0.77 V
When 30.0 mL of 0.10 mol L–1 Ce4+(aq) has been added and the total volume is V,
[Fe2+] = moles
and [Fe3+] = moles
V
V . Now
moles Fe 2+
V
moles Fe 3+
V
E Fe 3+ /Fe2+ = 0.771 V-
RT
nF
E Fe 3+ /Fe2+ = 0.771 V-
8.314 % 298
1 % 96485
ln
ln
=V
20.0 % 10 −3 % 0.10 mol L −1
30.0 % 10 −3 % 0.10 mol L −1
ANS: (b) 0.78 V
V = 0.78 14 V