CHEM 1000 A and V Exam December, 2005
Part A. 60 marks. Answer each question (5 marks each).
1.
Write the net ionic equation for the reaction 2 AgNO3(aq) + PbCl2(aq) → 2 AgCl + Pb(NO3)2
Ag+(aq) + Cl-(aq) → AgCl(s)
2.
3.
State whether or not each of the following processes or reactions is spontaneous as written:
2 Li(s) + Pt+2(aq) ↓ 2 Li+(aq) + Pt(s)
Yes (Li is above Pt in the activity series
Sn+2(aq) + Co(s) ↓ Sn(s) + Co+2(aq)
Yes (Co is above Sn is the activity series)
2 H+(aq) + Hg(l) ↓ H2(g) + Hg+2(aq)
No (Hg is below H2 in the activity series)
CH4(g) + 2 O2(g) ↓ CO2(g) + 2 H2O(g)
Yes (This is a combustion reaction, H<0, S0, thus G<0
H2O(g) ↓ H2O(l) at 95oC
No (The temperature is below the boiling point of water)
What type of hybrid orbitals does the Se atom use in a molecule of SeCl2?
This is of the form AX2E2. There are thus four electron clouds around the Se atom, which must therefore be using
sp3 hybrid orbitals.
4.
Explain the term “effective nuclear charge” and use it to explain why the ionization potential increases going left
to right across a period.
Effective nuclear charge is a measure of how strongly the nucleus attracts the valence electrons. Going left to
right across a period, electrons are being added the same principle shell. They therefore do not shield each other
very effectively. Also, protons are being added to the nucleus, with the result that the effective nuclear charge
increases, and thus the ionization potential increases.
5.
A reaction has a positive value of H and a negative value of S. Will this reaction be spontaneous at high
temperature, low temperature, both or neither? How do you know?
G = H - TS. If H > 0 and S<0, then G will be positive at all temperatures, i.e. the reaction will not be
spontaneous at any temperature.
6.
Why is the O2 molecule paramagnetic?
This is because the molecule has two unpaired electrons in its highest energy molecular orbital.
7.
Why is the ‘V’ in the ideal gas equation replaced with ‘V-nb’ in the van der Waals equation?
The gas molecules themselves actually take up some of the volume of the container, so the volume that any one
molecule has to fly around in is actually less than V. n is the number of moles of gas and b is a measure of the
size of the molecules, both of which reduce the volume from V to a lower value.
8.
Does the position of the equilibrium in the reaction 2 H2(g) + O2(g) ∏ 2 H2O(g) shift left or right when we:
Increase the pressure
Right
Increase the temperature
Left
Add some H2O(g)
Left
Chemistry 1000 A and V
9.
Mid-Year Examination, December, 2005
Page 2 of 8
Remove some O2(g)
Left
Add an inert gas such as N2(g)
Can not tell without knowing if the volume changes or not. Everyone gets one
mark for this one!
Why do deep-sea divers breathe a mixture of oxygen and helium?
This prevents the bends, which is caused by dissolved nitrogen in the blood reverting to small bubbles of gas as
the diver ascends. Helium is not very soluble in blood.
10.
11.
Consider the spontaneous exothermic reaction carried out at atmospheric pressure:
C3H6(g) + 4.5 O2(g) ↓ 3 CO2(g) + 3 H2O(g). State whether the following variables increase or decrease as the
reaction progresses from left to right:
Volume of the system
Increases (5.5 moles → 6 moles)
Temperature of the surroundings
Increases (Exothermic combustion reaction)
Entropy of the system
Increases (5.5 moles → 6 moles)
Entropy of the surroundings
Increases (Exothermic reaction warms the surroundings)
Entropy of the Universe
Increases (Since both system and surroundings entropies increase)
Why is the H-O-H bond angle in H2O less than 109.5o?
The two lone pairs repel the bonding pairs more than the bonding pairs repel each other.
12.
Helium gas is constantly being input into the Earth’s atmosphere as a product of radioactive decay. Why is there
so little helium in the atmosphere?
This is because Helium is a very lightweight gas, and some of the atoms will have sufficient upwards speed to
escape the Earth’s gravitational pull, i.e. some of theh will have a speed higher than the escape velocity of the
Earth.
Part B. Answer both questions. (20 marks each)
B1. For the reaction H2O(g) + Cl2O(g) Ý 2 HOCl(g) Kp = 0.090 at 25oC. 1.0 atm of each gas is placed in a flask at 25o C.
(a) Using the reaction quotient, show whether the reaction moves left or right so as to attain equilibrium.
Q
p2HOCl
12

1
pH2O pCl2O 1(1)
Thus, Q>Kp. Thus the reaction must shift to the left in order that Q decreases until Q=K p
(b) Calculate the equilibrium partial pressures (in atm) of all three gases.
2
Chemistry 1000 A and V
Mid-Year Examination, December, 2005
Initial
Change
Equilibrium
Page 3 of 8
H2O(g) Cl2O(g) HOCl(g)
1
1
1
+x
+x
-2x
1+x
1+x
1-2x
Thus at equilibrium,
p 2HOCl
(1  2x) 2

 K p  0.09
p H2O pCl2O (1  x)(1  x)
1  2x
 0.09  0.30
1 x
0.3(1  x)  1  2x
0.3  0.3x  1  2x
2.3x  0.7
x  0.304
Thus, pHOCl = 1-2x = 0.392 atm
pH2O = pCl2O = 1+x = 1.30 atm
Check:
0.3922
 0.09  K p
1.30(1.30)
(c) Calculate the standard free energy change (in kJ mol-1) for the reaction. Is the reaction spontaneous at 25oC?
Go = RT ln(Kp)
= - 8.314 J K-1mol-1 (25+273)K ln(0.090)
= +5,966 J mol-1
= +5.97 kJ mol-1
The reaction is not spontaneous at 298 K (since Go > 0)
B2. (a) Write the four oxygen-only reactions collectively called the Chapman cycle.
h
O 2 
2O
O  O 2 
 O3
h
O3 
 O  O2
O  O3 
 2 O2
3
Chemistry 1000 A and V
Mid-Year Examination, December, 2005
Page 4 of 8
(b) Beginning with NO(g), show the reactions that produce ozone at ground level, i.e. in smog.
NO  12 O 2 
 NO 2
h
NO 2 
 NO  O
O  O 2 
 O3
(c) Show the reactions in the catalytic cycle by which Cl atoms destroy ozone in the stratosphere.
O3  Cl 
 ClO  O 2
ClO  O 
 Cl  O 2
or
2[Cl  O3 
 O 2  ClO]
2ClO 
 Cl2O 2
h
Cl2O 2 
 2Cl  O 2
(d) Beginning with a sulphide ore such as NiS(s), show the reactions by which acid rain is produced.
NiS(s)  32 O2(g) 
 NiO(s)  SO2(g)
SO2(g)  12 O2(g) 
 SO3(g)
SO3(g)  H 2O(l) 
 H 2SO4(aq)
Part C. Attempt all five questions. The best four will be used to calculate your mark. (20 marks each)
C1. The hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) supplies the energy to drive many
biochemical reactions. Use the data below to answer the following questions about the reaction:
ATP-4(aq) + H2O(l) Ý ADP-3(aq) + HPO4-2(aq) + H+(aq)
Species
ATP-4(aq)
H2O(l)
ADP-3 (aq)
HPO4-2(aq)
H+(aq)
Hof, kJ mol-1
-2981.8
-286.6
-2000.2
-1292.1
0.0
So, J K-1 mol-1
100.0
69.9
237.4
-33.5
0.0
(a) Calculate the standard enthalpy change for the reaction (kJ mol-1).
Ho = Hfo (ADP-3(aq)) + Hfo (HPO4-2(aq)) + Hfo (H+(aq))  Hfo (ATP-4(aq))  Hfo (H2O(l))
= –2000.2 – 1292.1 + 0 – (–2981.8) – (–286.6)
= -23.9 kJ mol-1
(b) Calculate the standard entropy change for the reaction (J K-1 mol-1).
So = So (ADP-3(aq)) + So (HPO4-2(aq)) + So (H+(aq))  So (ATP-4(aq))  So (H2O(l))
= 237.4 – 33.5 + 0 – 100.0 + 69.9
= 34 J K-1 mol-1
(c) Calculate the standard free energy change for the reaction (kJ mol-1).
4
Chemistry 1000 A and V
Mid-Year Examination, December, 2005
Page 5 of 8
Go = Ho – TSo
= –23900 – 298 (34)
= –34037 J mol-1
= –34 kJ mol-1
(d) Calculate Kc for the reaction at 37oC.
 G o 
K c  exp 

 RT 


(34000 J mol1
 exp 

1
1
 8.314 J K mol (273  37)K) 
 exp(13.7)
 5.36 105
(e) If at 37oC, [H+(aq)] = 1.0 x 10-7 and [HPO4-2(aq)] = 1.0 x 10-6 M, calculate the ratio [ADP-3(aq)]/[ATP-4(aq)].
Kc 
thus,
3


[ADP(aq)
][HPO 4(aq)
][H (aq)
]
4
[ATP(aq)
]
3
[ADP(aq)
]
4
(aq)
[ATP
]

K
[HPO
c

4(aq)

][H (aq)
]
5.36 105

1107 (1 106 )
 5.36 1018
C2. (a) Two structures for Cl2O are possible, Cl-O-Cl and O-Cl-Cl. Draw the Lewis structures for these and use formal
charge to predict which structure is more likely.
..
: ..
Cl
Cl
7
7
0
Valence
Electrons assigned
Formal Charge
..
: O..
Valence
Electrons assigned
Formal Charge
..
Cl :
..
..
O
..
O
6
6
0
Cl
7
7
0
..
Cl :
..
..
..
Cl
O
6
7
-1
Cl
7
6
1
Cl
6
7
-1
The Cl-O-Cl structure is therefore more likely since all formal charges are zero.
(b) Name the type of hybrid orbitals used by the central atom in each of the following three structures:
5
Chemistry 1000 A and V
Mid-Year Examination, December, 2005
PCl5
H2Se
5+(5x7) = 40 e¯
Thus, AX5
Thus, sp3d hybrids
(2x1)+6 = 8 e¯
Thus, AX2E2
Thus, sp3 hybrids
Page 6 of 8
I3¯
3x7+1 = 22 e¯
Thus, AX2E3
Thus, sp3d hybrids
(c) Use VSEPR to predict the shapes of the following three species:
AsF5
AsF4
SeO2
5+(5x7) = 40 e¯
Thus, AX5
Thus, trigonal bipyramidal
5+(4x7) = 33 e¯
Thus, AX4E
Thus, seesaw
6+(2x6) = 18 e¯
Thus, AX2E
Thus, bent
C3. (a) Use the ideal gas law to calculate the pressure that results from putting 3.0 mol He(g) in a 100.0 L container at
50oC.
nRT
V
3mol(0.082 Latm K 1mol 1 )(273  50)K

100L
p
 0.79atm
(b) Does the ideal gas law give an accurate pressure in part (a)? How do you know?
Yes, the pressure is quite low, and the temperature is quite high, so the ideal gas law should give an accurate pressure
under these conditions.
(c) Use the van der Waals equation to calculate the pressure that results from putting 3.0 mol He (g) in a 1.0 L container at
50oC. For He(g), a = 0.0341 atm L2 mol-2 and b = 0.0237 L mol-1.
nRT
n
p
 a 
V  nb
v
2
 3.0mol 
3.0mol(0.082Latm K 1mol1 )(273  50)K

 0.0341atm L2 mol2 

1
1.0L  3.0mol(0.0237 L mol )
 1.0L 
2
 85.5 atm  0.31 atm
 85.2 atm
(d) Calculate the average speed (m s-1) of He(g) atoms at 50oC.
v
3RT
3(8.314 J K 1mol1 )(273  50)K

 1419 m s 1
1
MW
0.004 kg mol
(e) Calculate the partial pressures (atm) of each gas in a mixture of 30.0 g He (g) and 200 g Ne(g) in a 250.0 L container at
25oC.
6
Chemistry 1000 A and V
Mid-Year Examination, December, 2005
Page 7 of 8
30 g He
 7.5 mol He
4 g mol1
200 g Ne
 9.9 mol Ne
20.2 g mol1
n tot  7.5 mol He  9.9 mol Ne  17.4 mol
p tot
n tot RT 17.4 mol(0.082 L atm K 1mol 1 )(273  25)K


 1.7 atm
V
250 L
X He 
7.5 mol
 0.43
7.5 mol  9.9 mol
Thus, p He  X He (p tot )  0.43(1.7 atm)  0.73 atm
and p Ne  X Ne (p tot )  0.57(1.7 atm)  0.97 atm
C4. (a) The reaction Ba+2(aq) + SO4-2(aq) → BaSO4(s) is carried out in a calorimeter. When 2.00 mol Ba+2(aq) are mixed with
2.00 mol SO4-2(aq) in 4.00 L of water at 25oC, the temperature of the solution rises to 28.1oC. Assuming that the
calorimeter itself absorbs a negligible amount of heat, that the final solution has a density of 1.00 g mL -1, and that the
specific heat capacity of the solution is 4.18 J oC-1 g-1, calculate the enthalpy of reaction (H) per mole of BaSO4(s)
formed.
m = 4,000 mL x 1.0 g/mL = 4,000 g
cp = 4.18 J oC-1 g-1
T = 28.1 – 25.0 = 3.1oC
q = m cp T
= 4,000 g (4.18 J oC-1 g-1)(3.1oC)
= 51832 J
H 
51832 J
 25916 J mol1  25.9 kJ mol1
2 mol BaSO4(s)
(note that H is negative because the reaction is exothermic.)
(b) Calculate the standard enthalpy of formation of B2H6(g), i.e. the enthalpy of the reaction 2 B(s) + 3 H2(g) → B2H6(g),
given the following data:
2 B(s) + 3/2 O2(g) → B2O3(s)
B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g)
H2(g) + ½ O2(g) → H2O(l)
H2O(l) → H2O(g)
H = 1273 kJ mol-1
H = 2035 kJ mol-1
H = 286 kJ mol-1
H = +44 kJ mol-1
2 B(s) + 3/2 O2(g) → B2O3(s)
H = 1273 kJ mol-1
7
Chemistry 1000 A and V
Mid-Year Examination, December, 2005
3 (H2(g) + ½ O2(g) → H2O(l))
H = 3(286 kJ mol-1)
B2O3(s) + 3 H2O(g) → B2H6(g) + 3 O2(g)
H = (2035 kJ mol-1)
Page 8 of 8
3(H2O(l) → H2O(g))
H = 3(44 kJ mol-1)
_____________________________________________________
2 B(s) + 3 H2(g) → B2H6(g)
H = +35 kJ mol-1
C5. (a) Use the Balmer-Rydberg equation to calculate the ionization energy (in kJ mol-1) of a hydrogen atom whose
electron is initially in the ground state.
“Ground state” means m = 1.
“Ionization” means removal of the electron from the atom, i.e. n = .
Thus,
1
1
1 
 1
1
 R  2  2   R  2  2   R  0.01097 nm 1

n 
m
1  
1
Thus,  
 91.2 nm
0.01097 nm 1
hc 6.63  1034 J s (3.00  108 m s 1 )

 2.18  1019 J (per photon)
9

91.2  10 m
23
1
6.02  10 mol  1.31  106 J mol1
E
 1310 kJ mol1
(b) Calculate the wavelength of light required (nm) to dissociate H2, which has a bond energy of 432 kJ mol-1.

hc 6.63  1034 J s (3.00  108 m s 1 )

 2.77  107 m  277 nm
432000 J mol1
E
6.02  1023 mol1
8