Chemistry 2300 (Loader)
Winter 2004
Page 1 of 3
Chemistry 2300
W
TOTAL 50 MARKS
NAME:___________________________
February 9th 2004
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)
For the dehydrogenation reaction of ethane to give ethene and hydrogen
gas,
C2H6(g) → C2H4(g) + H2(g),
the following data are avail able:
o
H 298 K, reaction = 136.98 kJ mol—1
o
S 298 K for C2H4(g) = 219.5; H2(g) = 130.6; C2H6(g) = 229.5 J K—1 mol—1.
Calculate:
[3]
o
Calculate U 298 K for the reaction.
(i)
In this reaction 1 mole of gas gives 2 mole of gas so work is done on the gas rather than by
the gas in the reaction at constant pressure. n = + 1.
U = H − nRT = 136.98 kJ mol −1 − (1 mol % 8.314 % 10 −3 J K −1 mol −1 % 298 K)
U = 134.50 kJ mol −1
Ans: U = 134.50 kJ mol −1
[4]
o
Calculate S 298 K for the reaction.
(ii)
o
S 298 K = (219.5 + 130.6) − (229.5)J K −1 mol −1
o
S 298 K = 120.6 J K −1 mol −1
o
Ans : S 298 K = 120.6 J K −1 mol −1
[3]
o
Calculate G 298 K for the reaction.
(iii)
G = H − TS
o
−1
−1
J K mol
G 298 K = 136.98 kJ mol −1 − 298 K( 120.6
)
1000 J kJ −1
o
Ans : G 298 K = 101.0 kJ mol −1
[5]
(b)
When a 0.2973 g sample of glucose, C6H12O6(s) (molar mass = 180.2 g
mol—1), was burned in a bomb (constant volume) calorimeter of heat capacity
1215 J K—1 the temperature increased by 3.837 K. Use these data to
calculate the internal energy, U, change of combustion for glucose at the
temperature of the reaction.
1215 J K −1
q calorimeter = 1000
J kJ −1 % 3.837 K
q calorimeter = 4.661 96 kJ = −q glucose
moles of glucose burned =
U =
−4.66196 kJ
0.0016498 mol
0.2973 g
180.3 g mol −1
= −2825. 7 kJ mol −1
Ans : U = −2826 kJ mol −1
= 0.001649 8 mol
Chemistry 2300 (Loader)
(c)
[5]
Winter 2004
Page 2 of 3
One mole of an ideal gas is contained in a cylindrical container fitted with a
movable piston at a pressure of 5.00 atm. The temperature of the apparatus
is maintained constant at 77.0 ºC. Using the piston, the pressure on the gas
was reduced in two steps so that the gas expanded.
Initial pressure,
P1 = 5.00 atm
after the first expansion, P2 = 2.54 atm
final pressure,
P3 = 1.00 atm
(i) Calculate the total work, w, done on the gas in going from the initial to
The work done on the gas, w = -PV. We need to know the volume after each expansion.
−1
−1
1 mol % 0.0821 L atm K mol % 350.15 K
PV = nRT so V = nRT
= 5.74 9 L; after the
P . Initial volume =
5.00 atm
1 mol % 0.0821 L atm K −1 mol −1 % 350.15 K
first expansion volume =
= 11.3 2 L;after the third
2.54 atm
1 mol % 0.0821 L atm K −1 mol −1 % 350.15 K
expansion;
= 28.7 4 L.
1.00 atm
Work in first expansion = -2.54 atm x (11.32 - 5.749) L = -14.15 L atm
Work in second expansion = -1.00 atm x (28.74 - 12.83) L = -17.42 L atm
The total work = (-14.15 - 17.42) atm = -31.57 L atm
In joules this is = -31.57 L atm x 101.3 J L-1 atm-1 = -3198 J
Ans: -3.20 kJ
[4]
(ii)
Calculate the total work, w, done on the gas in going from the initial to
the final state using a reversible path.
− ¶V
V final
w=
initial
PdV = − nRT ¶ V
V final
initial
1
V dV
= − 1 mol % 8.314 J K −1 mol −1 % 350.1 K ln( 51 )
= -4684.6 J
Ans: -4.69 kJ
(iii) Calculate the total entropy change for the expansion of the gas.
S = −1 mol x 8.314 J K −1 mol −1 % ln( 5.00
1.00 )
[3]
S = 13.4 J K−1
Ans:
2.
[4]
Trichloromethane is the liquid commonly known at chloroform that was once used
as an anesthetic. The normal boiling point of trichloromethane is 61.88 ºC and the
o
enthalpy of vaporization under these conditions, H vap = 29.4 kJ mol—1.
(a)
o
o
Calculate the entropy of vaporization S vap and G vap at the normal boiling
point of trichloromethane.
o
S vap =
29.4 kJ mol −1
(61.88+273.15) K
= 0.08775 kJ mol −1 K −1 . Ans:
8.78 J mol −1 K −1
Since this process is taking place at the normal boiling point and the temperature at
o
o
which the value of G vap is measured the G vap = 0
o
Ans:G vap = 0
[4]
(b)
In an industrial process it is desired to keep trichloromethane in the liquid
state at the process temperature of 120 ºC by applying pressure to the liquid.
Calculate the minimum pressure that must be applied so that the
trichloromethane remains a liquid at 120 ºC.
p
vap H
T −T
p
−1
2
s
kJ mol
1
ln P21 = R
= ln 1atm
= 8.314 29.4
T1 T2
% 10 −3 kJ K −1 mol −1
ln p2 = 1.56 03 , so p2 = 4.76 atm
Ans: 4.76 atm
(393.15−335.03) K
(393.15)(335.03) K 2
Chemistry 2300 (Loader)
[4]
3.
Winter 2004
Page 3 of 3
Consider the reaction
N2(g) + 3 H2(g) → 2 NH3(g).
o
Given that G 298 K = -16.6 kJ mol—1 for the reaction at 1.00 bar, calculate G298 K
o
p2
It can be shown that G 298 K = G 298 K − 2RT ln P 1
= −16.6 kJ mol −1 − 2 % 8.314 % 10 −3 kJ K−1 mol −1 % 298 K ln 200.0
1.00
= -42.9 kJ mol-1
Ans: -42.9 kJ mol-1
4:
A well insulated calorimeter contains 200.00 g of water initially at 86.0 ºC. 50.0
grams of pure ice was added to the water in the calorimeter and when equilibrium
had been reestablished all of the ice had melted and the final temperature of the
melted ice/water was 52.0 ºC. The specific heat of water in this temperature range
is 4.18 J g—1 K—1. You may ignore the heat capacity of the calorimeter.
Calculate the enthalpy of fusion, H fus for ice (in kJ mol—1) from these data.
[5]
(a)
He
Heat given to ice =
50.0 g
18.01 g
% H fus + (50.0 g % 4.184 J K −1 gl −1 % 52.0 K
Heat taken from water = [200.00 g % 4.18 J K −1 gl −1 % (86.0 − 52.0) K ] = 284 51.2 J
284 51.2 J = Heat given to ice =
50.0 g
18.01 g
% H fus + (50.0 g % 4.18 J K −1 g −1 % 52.0 K)
Solving gives H fus = 6.32 kJ mol-1
Ans: H fus = 6.32 kJ mol-1
[6]
(b)
Calculate S water , S ice and S total for the process.
T
S water = C p(water) ¶ T 12
Ans: −83.3 J K −1
S ice =
S ice =
H
273.15 K
dT
T
= 200.00 g % 4.18 J g −1 K −1 ln
T
+ C p(melted ice) ¶ T 12
6.024 % 10 3 J mol −1
50.0 g
18.01 g
273.15 K
= −83.27 J K −1
dT
T
+ 50.0 g % 4.18 J g −1 K −1 ln 325.1
273.1
= 61.19 J K-1 + 36.42 J K-1
Ans: +97.61 J K-1
S total =−83. 27 J K−1 + 97.61 J K-1 = 14.33 J K-1
Ans:14.3 J K-1
325.1
359.1