5.2 Hess’s Law
From the Data Booklet
• Section 11: Bond enthalpies and average bond
enthalpies at 298K
• Section 12: Selected compounds –
Thermodynamic data
• Section 13: Enthalpies of combustion
5.3
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
ΔHθ = ?
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
ΔHθ = -808 kJ mol-1
5.3 Bond enthalpies
ΔHrxn = ΣE (Bonds broken) – ΣE (Bonds formed)
Combustion reactions: reacting with oxygen O2
Enthalpy of combustion ΔHco (kJ mol-1)
– Energy released when one mole of a particular substance reacts with O2
5.1 Comparing Measured experimental values to Literature value
HL stuff
ΔHfo = Enthalpy of formation (SL)
HL stuff
5.2 Hess’s Law
• Understandings:
– The enthalpy change for a reaction that is carried
out in a series of steps is equal to the sum of the
enthalpy changes for the individual steps.
Law of thermodynamics: Conservation of energy
5.2 Hess’s Law
Applications and skills
• Application of Hess’s Law to calculate
enthalpy changes.
• Calculation of ΔH reactions using ΔHf data.
• Determination of the enthalpy change of a
reaction that is the sum of multiple reactions
with known enthalpy changes.
Using the equations below:
C(s) + O2(g) → CO2(g)
Mn(s) + O2(g) → MnO2(s)
∆Hο = –394 kJ mol–1
∆Hο = –520 kJ mol–1
What is ∆H, in kJ, for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
A. 914
B. 126
C. –126
Determination of the enthalpy change of
D. –914
a reaction that is the sum of multiple
reactions with known enthalpy changes.
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
Equations given:
a. C(s) + O2(g) → CO2(g)
∆Hο = –394 kJ mol–1
b. Mn(s) + O2(g) → MnO2(s) ∆Hο = –520 kJ mol–1
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
Equations given:
a. C(s) + O2(g) → CO2(g)
b. Mn(s) + O2(g) → MnO2(s)
∆Hο = –394 kJ mol–1
∆Hο = –520 kJ mol–1
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
a. C(s) + O2(g) → CO2(g)
∆Hο = –394 kJ mol–1
b. Mn(s) + O2(g) → MnO2(s)
∆Hο = –520 kJ mol–1
• If the direction of b is reversed, the sign of ΔH is also
reversed (exo becomes endo).
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
a. C(s) + O2(g) → CO2(g)
∆Hο = –394 kJ mol–1
b. Mn(s) + O2(g) → MnO2(s)
∆Hο = –520 kJ mol–1
• If the direction of b is reversed, the sign of ΔH is also
reversed (exo becomes endo).
b reversed.
MnO2(s) → Mn(s) + O2(g)
∆Hο = +520 kJ mol–1
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
a. C(s) + O2(g) → CO2(g)
b reversed.
MnO2(s) → Mn(s) + O2(g)
∆Hο = –394 kJ mol–1
∆Hο = +520 kJ mol–1
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
a. C(s) + O2(g) → CO2(g)
b reversed.
MnO2(s) → Mn(s) + O2(g)
∆Hο = –394 kJ mol–1
∆Hο = +520 kJ mol–1
____________________________________________________________________
Combining the two reactions
C(s) + O2(g) + MnO2(s) → CO2(g) + Mn(s) + O2(g)
Main reaction:
MnO2(s) + C(s) → Mn(s) + CO2(g)
a. C(s) + O2(g) → CO2(g)
b reversed.
MnO2(s) → Mn(s) + O2(g)
∆Hο = –394 kJ mol–1
∆Hο = +520 kJ mol–1
____________________________________________________________________
Combining the two reactions
C(s) + O2(g) + MnO2(s) → CO2(g) + Mn(s) + O2(g)
∆Hο = ∆Hοa + ∆Hοb = (-394) + (+520)
= +126kJ mol-1
Using the equations below:
C(s) + O2(g) → CO2(g)
Mn(s) + O2(g) → MnO2(s)
∆Hο = –394 kJ mol–1
∆Hο = –520 kJ mol–1
What is ∆H, in kJ, for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
A. 914
B. 126
C. –126
D. –914
Hess’ Law
• A Swiss-born Russian chemist Germain Henri Hess
(1802 – 1850), proposed a precursor of the 1st Law
of Thermodynamics (the Law of Conservation of
Energy)
• Hess’ 1840 statement referred to a
‘Conservation Law for Heat of Reactions’
(but he did not explicitly relate it to exchanges of heat and work)
Hess’s Law
• The enthalpy change of a reaction depends
only on the identities of the reactants and
products.
• It is independent of the reaction pathway
(how you get there doesn't matter).
• This means enthalpy is a "state function."
• *Hess' Law is essentially the Law of
Conservation of Energy applied to chemical
reactions.
5.2 Hess’s Law
• Understandings:
– The enthalpy change for a reaction that is carried
out in a series of steps is equal to the sum of the
enthalpy changes for the individual steps.
Law of thermodynamics: Conservation of energy
5.2 Hess’s Law
Applications and skills
• Application of Hess’s Law to calculate
enthalpy changes.
• Calculation of ΔH reactions using ΔHf data.
• Determination of the enthalpy change of a
reaction that is the sum of multiple reactions
with known enthalpy changes.
1. Linear representation
2. Energy diagram representation
3. Energy cycle representation
(All main reactions are going from A to C)
Solving Hess’s Law problems using
“simultaneous equation method”
Solving Hess’s Law problems using
“simultaneous equation method”
Solving Hess’s Law problems using
“simultaneous equation method”
• If the equations can be manipulated to "add
up" to a specific reaction, then the enthalpy
change for that reaction can be easily
calculated by adding up the component
enthalpy changes.
Hess’ Law: Details
• You can always reverse the direction of a
reaction when making a combined reaction.
• When you do this, the sign of DH changes.
N2(g) + 2O2(g)
 2NO2(g)
DH = 68 kJ
2NO2(g)
 N2(g) + 2O2(g)
DH = -68 kJ
Hess’ Law: Details
• The magnitude of DH is directly proportional to the
quantities involved
• E.g. if you need to multiply the number of moles, you
also multiply the magnitude of DH
N2(g) + 2O2(g)  2NO2(g)
2N2(g) + 4O2(g)  4NO2(g)
DH = 68 kJ
DH = 136 kJ
5.2 Hess’s Law
• Understandings:
– The enthalpy change for a reaction that is carried
out in a series of steps is equal to the sum of the
enthalpy changes for the individual steps.
Law of thermodynamics: Conservation of energy
Hess’s Law – What’s the point?
• Allows us to calculate enthalpy changes
theoretically that are impossible, or near
impossible, to carry out experimentally.
• Allows us to confirm/ compare how
experimental enthalpy changes and
theoretically calculated ones compare.
– Provides information about how accurate
experimental methods and theoretical models
are.
Example: The following reaction is difficult for its
enthalpy change to be measured experimentally
C (s) + ½ O2 (g)  CO (g)
ΔH = ?
Meanwhile, these two other reactions are easier
to carry out and have ΔH measured accurately.
C (s) + O2 (g)  CO2 (g)
ΔH1 = -394 kJ mol-1
CO (g) + ½ O2 (g)  CO2 (g) ΔH2 = -283 kJ mol-1
C (s) + ½ O2 (g)  CO (g)
C (s) + O2 (g)  CO2 (g)
ΔH = ?
ΔH1 = -394 kJ mol-1
CO (g) + ½ O2 (g)  CO2 (g) ΔH2 = -283 kJ mol-1
C (s) + ½ O2 (g)  CO (g)
C (s) + O2 (g)  CO2 (g)
ΔH = ?
ΔH1 = -394 kJ mol-1
CO (g) + ½ O2 (g)  CO2 (g) ΔH2 = -283 kJ mol-1
...reverse…
CO2 (g)  CO (g) + ½ O2 (g) ΔH2 = +283 kJ mol-1
C (s) + ½ O2 (g)  CO (g)
C (s) + O2 (g)  CO2 (g)
ΔH = ?
ΔH1 = -394 kJ mol-1
CO2 (g)  CO (g) + ½ O2 (g) ΔH2 = +283 kJ mol-1
_______________________________________
Combine, subtract CO2 and ½ O2 from both sides
C (s) + ½ O2 (g)  CO (g)
C (s) + ½ O2 (g)  CO (g)
ΔH = ?
Combine, subtract CO2 and ½ O2 from both sides
C (s) + O2 (g) + CO2 (g)  CO2 (g) + ½ O2 (g) + CO (g)
ΔH1 = -394 kJ mol-1
ΔH2 = +283 kJ mol-1
ΔH = ΔH1 + ΔH2 = (-394) + (+283) = -111 kJ
5.2 Hess’s Law
Applications and skills
• Application of Hess’s Law to calculate
enthalpy changes.
• Calculation of ΔH reactions using ΔHf data.
• Determination of the enthalpy change of a
reaction that is the sum of multiple reactions
with known enthalpy changes.
• Two classics reactions to determine enthalpy
changes for reaction are formation and
combustion
• Formation – forming compounds from their
elements
• Combustion – reacting substances completely
with oxygen
Combustion reactions: reacting with oxygen O2
Enthalpy of combustion ΔHco (kJ mol-1)
– Energy released when one mole of a particular substance reacts with O2
HL stuff
ΔHfo = Enthalpy of formation (SL)
HL stuff
Standard enthalpy of formation
• ΔHfo, the enthalpy change when 1 mole of a
compound is formed from its elements in their
standard states at 298K and 100kPa.
• Elements have a ΔHfo value of zero!
Standard enthalpy of formation
• C (s) + 2H2 (g)  CH4 (g)
ΔHfo = -74 kJ mol-1
• 2C(s) + 3H2(g) + ½ O2(g)  C2H5OH
ΔHfo = -278 kJ mol-1
Enthalpy of Formation and Hess’s Law
• The enthalpy change for a reaction can be
calculated using Hess’s Law and enthalpy of
formations.
ΔHo = Σ(ΔHfo products) - Σ(ΔHfo reactants)
Hess’ Law: Details
• You can always reverse the direction of a
reaction when making a combined reaction.
• When you do this, the sign of DH changes.
N2(g) + 2O2(g)
 2NO2(g)
DH = 68 kJ
2NO2(g)
 N2(g) + 2O2(g)
DH = -68 kJ
Hess’ Law: Details
• The magnitude of DH is directly proportional to the
quantities involved
• E.g. if you need to multiply the number of moles, you
also multiply the magnitude of DH
N2(g) + 2O2(g)  2NO2(g)
2N2(g) + 4O2(g)  4NO2(g)
DH = 68 kJ
DH = 136 kJ
- ΔHf CH4
ΔHf CO2
- ΔHf O2
2 x ΔHf H2O
ΔHo = Σ(ΔHfo products) - Σ(ΔHfo reactants)
CH4 (g) + O2 (g) CO2 (g) + 2 H2O (l)
ΔHo = Σ(ΔHfo products) - Σ(ΔHfo reactants)
ΔHo
= (ΔHfo CO2 + 2 x ΔHfo H2O) - (ΔHfo CH4 + ΔHfo O2)
= [(-394) + 2 x (-286)] – [+75 + 0]
= -966 – 75
= -1041 kJ
Do not mix these two up!!
• ΔHo through formation (Using Table 12)
ΔHo = Σ(ΔHfo products) - Σ(ΔHfo reactants)
• ΔHo through bond enthalpy (Using Table 13)
ΔHo = Σ(BE reactants) – Σ(BE products)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
ΔHθ = ?
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
ΔHθ = -808 kJ mol-1
Discrepancies in the two methods
CH4 (g) + O2 (g) CO2 (g) + 2 H2O (l)
• Through enthalpy of formation
– ΔHθ = -1041 kJ mol-1
• Through bond enthalpy
– ΔHθ = -808 kJ mol-1
• Bond enthalpy assumes all substance in gaseous
state, does not take into account energy needed
to change state.
Do not mix these two up!!
• ΔHo through formation (Using Table 12)
ΔHo = Σ(ΔHfo products) - Σ(ΔHfo reactants)
• ΔHo through bond enthalpy (Using Table 13)
ΔHo = Σ(BE reactants) – Σ(BE products)
Read and practice: page 179 - 184