1. Representing Energy Changes
UNIT 3
Representing Energy Changes
How can energy changes be represented in chemical reactions?
Thermochemical equations with energy term as product or reactants
e.g N2(g) + 2O2(g) kJ → 2NO2(g)
Thermochemical equations with energy term beside the equation
e.g N2(g) + 2O2(g) → 2NO2(g) ΔHr = kJ
Enthalpy diagrams
Image source: MHR, Chemistry 12 © ISBN ; page 294

2. What to use when UNIT 3 When the problem is about:
Chapter 5: Energy Changes
Section 5.2
What to use when
When the problem is about:
A solid dropped into water (like the copper penny expt):
Use: Qreleased = Q gained where Q= mc∆T
If it is as above but the container c and m is given use:
Q released by reaction = Q gained by water + Q gained by container
If the question is asking for molar enthalpy of reaction, and gives c and m, then calculate Q first (which is ∆H) and then use ∆H = n∆H
If it is about enthalpy in aqueous solutions, use C= n/V and if as above it is a solid and molar enthalpy is needed, then you need: n=m/M

3. UNIT 3
Section 5.3
Hess’s Law
- calculating the enthalpy change of reactions using existing data
- When using calorimetry is impractical
The enthalpy change of any reaction can be determined if:
the enthalpy changes of a set of reactions “add up to” the overall reaction of interest
standard enthalpy change, ΔH°, values are used
Image source: MHR, Chemistry 12 © ISBN ; page

4. UNIT 3
Chapter 5: Energy Changes
Section 5.3
For example
To find the enthalpy change for formation of SO3 from O2 and S8, you can use
Because only 1 mol of sulfur trioxide is in the final step:
Divide the first equation by 8 so 1 mol of sulfur dioxide is in the first step
Divide the second equation by 2
Image source: MHR, Chemistry 12 © ISBN ; page

5. Techniques for Manipulating Equations
UNIT 3
Section 5.3
Techniques for Manipulating Equations
You can reverse an equation
the products become the reactants, and reactants become the products
the sign of the ΔH value must be changed
You can multiply each coefficient
all coefficients in an equation are multiplied by the same integer or fraction
the value of ΔH must also be multiplied by the same number
Image source: MHR, Chemistry 12 © ISBN ; page

6. Standard Molar Enthalpies of Formation
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Standard Molar Enthalpies of Formation
Often used for Hess’ Law is
standard molar enthalpy of formation, ΔH˚f
the change in enthalpy when 1 mol of a compound is synthesized from its elements in their most stable form at SATP conditions
enthalpies of formation for elements in their most stable state under SATP conditions are set at zero
since formation equations are for 1 mol of compound, many equations include fractions (for a balanced eq’n)

7. Formation Reactions and Thermal Stability
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Formation Reactions and Thermal Stability
The thermal stability of a substance is the ability of the substance to resist decomposition when heated.
decomposition is the reverse of formation
the opposite sign of an enthalpy change of formation for a compound is the enthalpy change for its decomposition
the greater the enthalpy change for the decomposition of a substance, the greater the thermal stability of the substance
Image source: MHR, Chemistry 12 © ISBN ; page

8. Using Enthalpies of Formation and Hess’s Law
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Using Enthalpies of Formation and Hess’s Law
So the enthalpy of a reaction = the sum of all the products – the sum of all the reactants. For example:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Image source: MHR, Chemistry 12 © ISBN ; page

9. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
UNIT 3
Section 5.3
Try this
Determine ∆H˚r for the following reaction using the enthalpies of formation that are provided.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
∆H˚f of C2H5OH(l): –277.6 kJ/mol
∆H˚f of CO2(g): –393.5 kJ/mol
∆H˚f of H2O(l): –285.8 kJ/mol

10. UNIT 3 ∆H˚r = [(2 mol)(∆H˚f CO2(g)) + (3 mol)(∆H˚f H2O(l))] –
Chapter 5: Energy Changes
Section 5.3
∆H˚r = [(2 mol)(∆H˚f CO2(g)) + (3 mol)(∆H˚f H2O(l))] –
[(1 mol)(∆H˚f C2H5OH(l)) + (3 mol)(∆H˚fO2(g)]
∆H˚r = [(2 mol)(–393.5 kJ/mol) + (3 mol)(–285.8 kJ/mol)] –
[(1 mol)(–277.6 kJ/mol) + (3 mol)(0 kJ/mol)]
Image source: MHR, Chemistry 12 © ISBN ; page 98
∆H˚r = (– kJ) – (–277.6 kJ)
∆H˚r = – kJ