Unit 3 - Chemistry in Society
3C Chemical energy
For industrial processes, it is essential that chemists can predict the quantity of heat energy
taken in or given out. If reactions are endothermic, costs will be incurred in supplying heat
energy in order to maintain the reaction rate. If reactions are exothermic, the heat produced
may need to be removed to prevent the temperature rising.
3C1 Enthalpy changes
The specific heat capacity, mass and temperature can be used to calculate the enthalpy
change for a reaction.
Reaction of Magnesium with Silver Nitrate
A small piece of magnesium was reacted with 25ml of 1M silver nitrate solution (ie. an
excess). The initial temperature before reaction and the final temperature after reaction
were measured. The heat of reaction for one mole of magnesium was determined from this
data.
Weight of magnesium
=
Initial temperature
=
Final temperature
=
Mass of water heated
=
0.025kg (volume = 25ml)
The enthalpy changes can be calculated using ΔH = c m ΔT
c = specific heat capacity of water and is 4·18 kJ kg−1 ºC−1
m = mass in grams
ΔT = temperature change
ΔH = 4.2 x 0.025 x __________ = __________ kJ
g of magnesium produce __________ kJ
24·3 g of magnesium produce x kJ
x = 24·3/ __________ x __________ = __________ kJmol-1
Theoretical value =
Enthalpy of combustion
The enthalpy of combustion of a substance is the enthalpy change when one mole of the
substance burns completely in oxygen. These values can often be directly measured using
a calorimeter and values for common compounds are available from data books and online
databases for use in Hess’s law calculations.
All ΔH values for enthalpies of combustion are negative.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
ΔH = -891 kJ
The enthalpy of combustion of methanol (CH3OH) can be determined from measurements
using the apparatus shown.
thermometer
100 cm3 water
copper can
methanol
In an experiment, the following results were obtained.
mass of methanol burned = 0.45 g
temperature rise of water = 10.0 °C
Use these results to calculate the enthalpy of combustion, in kJ mol-1 of methanol.
ΔH
= c m ΔT
= 4.18 x 0.1 x 10
= 4.18 kJ
CH3OH
formula mass = 12 + 3 + 16 + 1 = 32 g is the mass of 1 mole
0.45 g
→
32 g
→
4.18 kJ
32
x 4.18
0.45
= 297 kJ mol-1
3C2 Hess's Law
Hess’s law states that the enthalpy change for a chemical reaction is independent of the
route taken. Enthalpy changes can be calculated by applying Hess’s law.
1.
Methanol can be made by a two-stage process.
In the first stage, methane is reacted with steam to produce a mixture of carbon monoxide
and hydrogen.
CH4(g) + H2O(g) ⇌ CO(g) +3H2(g)
Use the data below to calculate the enthalpy change, in kJ mol–1, for the forward reaction.
CO(g) + ½O2(g) → CO2(g)
ΔH = −283 kJ mol–1
H2(g) + ½O2(g) → H2O(g)
ΔH = −242 kJ mol–1
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔH = −803 kJ mol–1
target equation
CH4(g) + H2O(g) → CO(g) +3H2(g)
1
CO(g) + ½O2(g) → CO2(g)
ΔH = −283 kJ mol–1
2
H2(g) + ½O2(g) → H2O(g)
ΔH = −242 kJ mol–1
3
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔH = −803 kJ mol–1
multiply to fit with target equation
-1x(1) CO2(g) → CO(g) + ½O2(g)
ΔH = +283 kJ mol–1
-3x(2) 3H2O(g) → 3H2(g) + 1½O2(g)
ΔH = +726 kJ mol–1
3
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔH = −803 kJ mol–1
add the equations
CO2(g) + 3H2O(g) + CH4(g) + 2O2(g)
→ CO(g) + ½O2(g) +3H2(g) + 1½O2(g) + CO2(g) + 2H2O(g)
cancel species appearing on each side of the equation
1H2O(g)
CO2(g) + 3H2O(g) + CH4(g) + 2O2(g)
→ CO(g) + ½O2(g) +3H2(g) + 1½O2(g) + CO2(g) + 2H2O(g)
overall
CH4(g) + H2O(g) → CO(g) +3H2(g)
this is the same as the target equation so the enthalpies can be summed
ΔH
= +283 + 726 −803 kJ mol–1
= 206 kJ mol–1
2.
The enthalpy of formation for pentan-1-ol is shown below.
5C(s) + 6H2(g) + O2(g) → C5H11OH(l)
ΔH = –354 kJ mol–1
Using this value, and the enthalpies of combustion of carbon and hydrogen from the data
booklet, calculate the enthalpy of combustion of pentan-1-ol, in kJ mol–1.
target equation
C5H11OH(g) + 15/2O2(g) → 5CO2(g) + 6H2O(g)
data
1
5C(s) + 6H2(g) + O2(g) → C5H11OH(l)
ΔH = –354 kJ mol–1
2
C(s) + O2(g) → CO2(g)
ΔH = −394 kJ mol–1
3
H2(g) + ½O2(g) → H2O(g)
ΔH = −286 kJ mol–1
multiply to fit with target equation
-1x(1) C5H11OH(l) → 5C(s) + 6H2(g) + O2(g)
ΔH = +354 kJ mol–1
5x(2) 5C(s) + 5O2(g) → 5CO2(g)
ΔH = −1970 kJ mol–1
6x(3) 6H2(g) + 3O2(g) → 6H2O(g)
ΔH = −1716 kJ mol–1
add the equations
C5H11OH(l) + 5C(s) + 5O2(g) + 6H2(g) + 3O2(g)
→ 5C(s) + 6H2(g) + O2(g) + 5CO2(g) + 6H2O(g)
cancel species appearing on each side of the equation
15
/2O2(g)
C5H11OH(l) + 5C(s) + 5O2(g) + 6H2(g) + 3O2(g)
→ 5C(s) + 6H2(g) + O2(g) + 5CO2(g) + 6H2O(g)
overall
C5H11OH(g) + 15/2O2(g) → 5CO2(g) + 6H2O(g)
this is the same as the target equation so the enthalpies can be summed
ΔH
= +354 − 1970 − 1716 kJ mol–1
= -3332 kJ mol–1
3C3 Bond enthalpies
Molar bond enthalpies A diatomic molecule contains only two atoms; eg. Br2 or HCl
The molar bond enthalpy is the energy required to break one mole of the bonds to give
separate atoms - everything being in the gas state.
Mean molar bond enthalpies The bond enthalpy of a C-H bond depends on the molecule. It will be different in methane
compared to ethane which will be different again from that in ethanol. Mean molar bond
enthalpies are average values over a range of molecules.
Bond enthalpies can be used to estimate the enthalpy change occurring for a gas phase
reaction by calculating the energy required to break bonds in the reactants and the energy
released when new bonds are formed in the products.
Bond breaking is endothermic
Bond forming is exothermic
Examples
1.
Part of the industrial synthesis for hydrogen is
CO(g) + H2O(g) → CO2(g) + H2(g)
Bonds broken (+)
2 O−H
1 C=O
Bonds formed (-)
2 x (463)
743
1 H−H
2 C=O
+1669
ΔH = -1253 kJ mol-1
436
2 x (743)
-1922
2.
Chlorine reacts with ethane to give chloroethane and hydrogen chloride gases.
+ Cl2
+ HCl
Bonds broken (+)
6 C−H
1 C−C
1 Cl−Cl
Bonds formed (-)
6 x (412)
348
243
+3063
ΔH = -115 kJ mol-1
5 C−H
1 C−C
1 C−Cl
1 H−Cl
5 x (412)
348
338
432
-3178