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Data page
Equations
Wave equation: c =
Einstein equation: E = h
Rydberg equation:
1
𝜆
= 𝑅(
1
𝑛 𝑎
2
−
1
𝑛 𝑏
2
)
Balmer equation: 𝜈 = 3. 29 × 10
15 (
1
𝑛 𝑎
2
−
1
𝑛 𝑏
2
)
Bond order = (# bonding electrons - # anti-bonding electrons)
Physical constants
c = 2.998 10
8 m s
h = 6.626 10
R = 1.097 10
7 m
NA = 6.022 10
23 mol
R = 8.314 J K
= 0.08206 atm L mol
1 atm = 1.013 10
5 Pa
1 bar = 1.0 10
5 Pa
Kw at 25 C = 1.0 10
0 C = 273.15 K
Gases
Ideal Gas Equation: pV = nRT
Non-Ideal Gas Equation: (𝑝 +
𝑛
2 𝑎
𝑉
2
)(𝑉 − 𝑛𝑏) = nRT
Total Pressure = Partial Pressures of Component Gases
Thermodynamics
U = q + w
w = - p V
q = mc T
G = H – T S
G
o = - RTlnK
G = G + RT ln Q
Equilibria
Henderson-Hasselbach: pH = 𝑝𝐾 𝑎
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
)
Van’t Hoff equation: ln
𝐾 𝑇 2
𝐾𝑇 1
=
∆𝐻
0
𝑅𝑇
[
1
𝑇 1
−
1
𝑇 2
]
E°cell = E°red - E°ox
Kinetics
Zero-order reaction: [ A ]t = [ A ]o - kt
First-order reaction: [ A ]t = [ A ]o exp(- kt )
Second-order reaction (only one reactant A ):
1
[𝐴] 𝑡
−
1
[𝐴] 0
= 𝑘𝑡
Half-life: t 1/2 = 0.693/ k
Arrhenius equation: k = Ae
ln (
𝑘 2
𝑘 1
) =
−𝐸 𝑎
𝑅
(
1
𝑇 2
−
1
𝑇 1
)
1
2
- A hydrogen atom emits a photon as its electron changes from n = 5 to n = 1.
a) What is the wavelength of the photon? (95.0 nm)
𝑎
2
𝑏
2
7 (
1
1
2
1
5
2
b) In what region of the electromagnetic spectrum is this photon found?
These photons are in the ultraviolet region.
c) What are the frequency and energy in of the photon? ( 3.16 x 10
15 Hz, 2.09 x 10
- 18 J)
c =
8
(95 x 10
- 9 )
= 3.16 x 10
15 Hz
E = h =
15 ) = 2.09 x 10
- 18 J
Week 3
- Which has the most unpaired electrons, S
, S or S
2 - ? Use electron configurations to
support your answer.
S: [Ne] 3 s
2 3 p
4 S
: [Ne] 3 s
2 3 p
3 S
2 – : [Ne] 3 s
2 3 p
6
S
has the most unpaired electrons.
S
2 – , with no unpaired electrons, has fewer unpaired electrons than S or S
.
- Write the full and condensed electron configuration for the Mn
2 + ground state.
Give a set of quantum numbers for all the electron in the least stable orbital.
Mn
2+ , 2 3 electrons, [Ar] 3 d
5 :
n l m l
m s
- The first four ionisation energies of aluminium are as follows:
st IE = 577 kJ mol
nd IE = 1817 kJ mol
rd IE = 2745 kJ mol
th IE 11578 kJ mol
a) Explain the trend of ionisation energies
The electron configuration of aluminium is: 1s
2 2s
2 2p
6 3s
2 3p
1
Succesive ionisation energies are increasing due to higher effective nuclear charge.
The second and third ionisation energies of Al are considerably higher than the first
ionisation as the second and third electrons are removed from a filled 2s orbital. The
fourth ionisation energy is far greater, the electron is from inner shell number 2.
- Russian scientists who developed the nerve agents Novichok claim they are the
deadliest ever made. In the 21st century, Novichok agents came to public attention after
they were used to poison opponents of the Russian government, including the Skripals
and two others in Amesbury, UK (2018) and Alexei Navalny (2020).
a) How many π bonds does Novichok have?
There are two bonds in this molecule.
b) What are the hybridized states and bond angles of each of the labelled atoms?
A: sp
3 , 109.5°, B: sp
2 , 1 20 °, C: sp
3 109.5°, D: sp
3 , 109.5°,
c) Redraw the structure of Novichok adding the lone pairs.
- When an oxalate anion C 2 O 4
2 - adds two protons to form oxalic acid, two C-O bonds
become longer and two become shorter than the bonds in oxalate anions. Which bonds
get longer and which shorter. Use bonding principles to explain the changes.
Visualise what happens when protons add to oxalate anions using Lewis structures:
All the C—O bonds in the oxalate anion, with extended orbitals, have equal
lengths and strengths, with a C—O bond order of 1.5. Adding two protons
removes two oxygen atoms from the extended π system and converts the C—O
bonds into single bonds. The other two C—O bonds become double bonds.
Thus, the C—OH bonds in oxalic acid are longer than the C—O bonds in oxalate
anions, whereas the C=O bonds in oxalic acid are shorter.
Week 6
- What is the density (g L
- 1 ) of SF 6 gas at 1.013 x 10
5 Pa and 25°C? (5.97 gL
First calculate the amount of gas in one litre (1 × 10
3 ).
( 1. 013 × 10
5 Pa)( 1 × 10
− 3 m
3 )
( 8. 314 J mol
− 1 K
− 1 )( 298 K)
= 0. 04089 mol in one litre
𝑚 = 𝑛 × 𝑀 = 0. 040 89 mol × 146 g mol
− 1 = 5. 96 g in one litre
density = 5. 96 g L
− 1
- Humans consumed glucose to produce energy. The products of glucose
consumptions are CO 2 and H 2 O.
C 6 H 12 O 6 (s) + O 2 (g) → CO 2 (g) + H 2 O(l) (unbalanced)
What volume of CO 2 produced under body conditions (37 °C and 1.013 x 10
5 Pa) when
5.00 g of glucose is consumed? (0.00424 m
3 )
First balance the equation:
C 6 H 12 O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(l)
Then calculate the amount of glucose:
𝑛 glucose
=
00 g
16 g mol
− 1
= 0. 0278 mol
Using mole ratios:
𝑛 CO 2
6
=
𝑛 glucose
1
𝑛 CO 2
= 6 × 𝑛 glucose
= 6 × 0. 0278 mol = 0. 167 mol
Volume can be calculated from the ideal gas equation:
𝑉 =
𝑛𝑅𝑇
𝑝
=
( 0. 167 mol) × ( 8. 314 J mol
− 1
K
− 1
) × ( 310 K)
( 1. 013 × 10
5 Pa)
= 0. 004 24 m
3
- List ethanol CH 3 CH 2 OH, propane, CH 3 CH 2 CH 3 and pentane, CH 3 CH 2 CH 2 CH 2 CH 3 in
order of increasing boiling point. Explain your response.
Boiling point depends on the magnitude of intermolecular attractions, the larger
the attractions, the higher the boiling point. The size of intermolecular attractions
depends on the extent of dispersion forces (size of molecule and size of orbitals),
the presence of polar bonds and hydrogen bonding. Among these substances,
ethanol forms hydrogen bonds, so it has the highest boiling point. Propane, being
smaller than n - pentane, has the lowest boiling point: Propane (lowest bp) < n -
pentane < ethanol (highest bp).
Week 7
- The standard enthalpy combustion, ΔH°c of glucose is – 2.80 x 10
3 kJ mol
C 6 H 12 O 6 (s) + O 2 (g) → CO 2 (g) + H 2 O(l) (unbalanced)
Calculate enthalpy change formation of this compound. – 1276 kJ mol
)
Species ∆H˚f (kJ mol
- 1 )
CO 2 ( g ) - 393.
H 2 O (l) - 285. 9
O 2 ( g ) 0.
- A sample of copper was heated to 120 °C and then plunged into an insulated vessel
containing 200 g of water 25.00 °C. The final temperature of the mixture 26. 50 °C.
Specific heat of water = 4.18 J g
a) How much heat was absorbed by the water? (1. 25 kJ)
∆ T water = T f – T i = 2 6. 50 – 25.00 = 1.5°C
q gained by water = mc∆T
= 20 0 g × 4.18 J g
= 1 254 J
b) How much was lost by the copper sample? (– 1.25 kJ)
q lost by copper = – q gained by water
= – 1.25 kJ
c) What was the mass of the copper sample? (34.5 g) Specific heat of copper = 0. 387 J
g
Specific heat of copper = 0. 387 J g
∆ T copper = T f - T i = 2 6. 50 – 12 0 = – 93.5 °C
silver
− 1250 J
( 0. 387 J g
− 1 K
− 1 )(− 93. 5 K)
= 34. 5 g
Week 8
1) Calculate the Standard Gibbs Free Energy ( G r) at 25°C. ( – 88. 18 kJ mol
a) SO 3 (g) + H 2 O (g) → H 2 SO 4 (l)
Species S˚ (J mol
- 1 K - 1 ) ∆H˚f (kJ mol - 1 )
H 2 SO 4 (l) 157 .0 - 811.
SO 2 ( g ) 256.2 - 395.
H 2 O ( g ) 188. 7 - 241.
Gº = Hº – TSº
Hrº = (ΔHfº (298K)products) - (ΔHfº (298K)reactants)
= [(ΔHfº (H 2 SO 4 (l)– ( ΔHfº (SO 3 (g)]+ ΔHfº (H 2 O (g)]
= ( - 811.32 ) – ( - 395.2 + (-241.8)]
= - 174. 02 kJ mol
Srº = (ΔSº (298K)products) - (ΔSº (298K)reactants)
= ([(ΔSfº (H 2 SO 4 (l)– ( ΔSfº (SO 3 (g)]+ ΔSfº (H 2 O (g)]
= ( 157) – ( 2 56.2 + (188.7)]
= - 287. 9 J K
Gº = Hº – TSº
3 J mol
= - 88. 18 kJ mol
b) What is the effect of temperature on reaction spontaneity?
Spontaneous at low T
Non-spontaneous at high T
Week 9
- The forward reaction in the equilibrium :
C (s) + 2 H 2 (g) ⇌ CH 4 (g)
has G
o
= - 50. 8 kJ mol
- 1 can be used to convert coal to methane.
b) If not, in which direction will a spontaneous change occur to restore equilibrium?
Since the value of the reaction quotient Qc for this system is larger than that
of the equilibrium constant Kc , the system shifts to the left to make Qc
smaller and hence reach equilibrium.
- At 200 °C, Kc 1.4 x 10
N 2 O(g) + NO 2 (g) ⇌ 3 NO(g)
If 0.225 mol of N 2 O and 0.450 mol of NO 2 are placed in a 4.00 L container what is the
NO concentration at equilibrium? (9.63 10
The balanced equilibrium equation is: N 2 O(g) + NO 2 (g) ⇌ 3NO(g), for which the equilibrium constant
expression is:
3
10
2 2
[NO]
[N O][NO ]
c
K
−
= =
2
1
N O
0.225 mol
0.0563 mol L
4.00 L
n
c
V
−
= = =
2
1
NO
0.450 mol
0.113 mol L
4.00 L
n
c
V
−
= = =
The concentration table is:
N 2 O + NO 2 ⇌ 3NO
Initial 0.0563 0.113 0
Change – x – x +3 x
Equilibrium 0.0563 – x 0.113 – x +3 x
Substituting these equilibrium concentrations into the equilibrium constant expression and
assuming that (0.0500 – x ) ≈ 0.0500 and (0.113 – x ) ≈ 0.113 gives:
3 3
10
2 2
[NO] (3 )
[N O][NO ] (0.0563) (0.113)
c
x
K
−
= = =
This approximation is made since Kc is small.
Solving for x = 3.21 10
- 5 . As [NO] = 3 x , we obtain [NO] = 9.63 10 - 5 .
Week 10
- Phenol, C 6 H 5 OH is sometimes used as a disinfectant. Ka of phenol is 1.
× 10
− 10
. What are the concentrations of all the solute species in a 0.0250 M solution of
phenol? (1.8 × 10 ̄
6
)
C 6 H 5 OH + H 2 O ⇌ H 3 O
𝑎
[H
3
O
][C 6
H
5
O
−
]
[C
6
H
5
OH]
= 1.3 × 10
− 10
The concentration table is:
H 2 O + C 6 H 5 OH ⇌ H 3 O
Initial 0.025 0 0
Change – x + x + x
Equilibrium 0.025 – x + x + x
Since (0.025 – x ) ≈ 0.025, then
𝑎
= 1.3 × 10
− 10
𝑥 = 1.8 × 10 ̄
6
= [H 3
O
] (^) = [C 6
H
5
O
−
]
[C
6
H
5
OH]^ = 0. 025 − 1.8 × 10 ̄
6
= 0. 025 M
Weeks 11- 12
- Cyclopropane, C 3 H 6 is a gas used as general anaesthetic. It undergoes a slow
rearrangement to propene.
At a certain temperature, the following rate data were obtained.
Experiment Initial concentration (M) Initial rate of
reaction (M s
1 0. 050 2. 95 X 10
2 0. 100 5 .90 X 10
3 0. 150 8. 85 X 10
a) What is the rate law for the reaction?
First order wrt to cyclopropane.
R = k[cyclopropane]
1
b) What is the value of the rate constant? Include the correct units.
From experiment 1: R = k[cyclopropane]
1
2.95 X 10
k = 5. 9 x 10
- At 35 °C, the rate constant for the reaction:
C 12 H 22 O 11 + H 2 O → C 6 H 12 O 6 + C 6 H 12 O 6
Sucrose glucose fructose
is k = 6.2 x 10
- 5 s - 1 . The activation energy for the reaction is 108 kJ mol - 1 . What is the
rate constant at 45°C? ( 2.34 x 10
ln (
2
1
𝑎
2
1
ln (
𝑘 2
- 2 𝑥 10 − 5
− 108 , 000
- 314
1
318
1
308
k 2 = 2.34 x 10
14 C dating can be used to estimate the age of archaeological samples. If tissue
samples from a mummy contain about 81% of the
14 C expected in living tissue, how old
is the mummy? (1740 years) The half-life for decay of
14 C is 5730 years.
t 1/2 = 0.693/ k
5730 = 0.693 / k
K =1.209 X 10
- 4 yr - 1
[ A ]t = [ A ]o exp(- kt )
ln[A]t = ln [A] 0 - kt
ln(0.81) = ln (1) – ( 1.209 X 10
- 4 ) (t)
t = 1742 years
