TOPIC 5.2: Hess’s Law and Enthalpy of Formation

Introduction Discussion

How can you calculate the heat of a reaction when it cannot be directly

measured?

Why might a reaction not be able to be directly measured?

Introduction Discussion

How can you calculate the heat of a reaction when it cannot be directly measured?

Use Hess’s Law!Why might a reaction not be able to be directly measured? • Reaction too slow to measure enthalpy• The reaction is an intermediate in a

series• Difficult to attain specific conditions req.

Hess’s Law

• The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes for the individual steps in the process

KEY IDEA: Route doesn’t matter!

Reaction Routes- Example #1

If the reaction of A→ B cannot be

measured, use the data from A→ C→ B

instead!

ΔH1 = ΔH2 + ΔH3

Enthalpy Graph for Example #1

ΔH1 = ΔH2 + ΔH3

Practice: #2

Write the

equation for ΔH of the

A → D reaction

ΔH = ΔH1 + ΔH2+ ΔH3

Practice: #3

Write 2 different equations to

solve for ΔH1

Effect of Direction Change

If the arrows don’t match the direction you need to go, flip the arrow and the

sign on the ΔH value!

ΔH1 = ΔH2 - ΔH3

Draw an energy cycle

for the 3 reactions in the graph.

Write an expression for ΔHc and

solve.

More Practice!

• Manipulate a series of reactions (called step equations) to reach the required reaction which is most always given.

Rules 1. If you reverse a reaction, reverse the

sign of ΔH. 2. If you multiply or divide a step equation

by an integer, you must also multiply or divide your ΔH by the same factor.

Hess’s Law without Visuals :)

• Manipulate a series of reactions (called step equations) to reach the required reaction which is most always given.

Rules 1. If you reverse a reaction, reverse the

sign of ΔH. 2. If you multiply or divide a step equation

by an integer, you must also multiply or divide your ΔH by the same factor.

Hess’s Law without Visuals :)

C(s) + O2(g) CO2(g) ΔHcº = -393 kJ mol-1

H2(g) + 1/2 O2(g) H2O(l) ΔHcº = -286 kJ mol-1

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔHcº = -890 kJ mol-1

Main Equation: C(s) + 2H2(g) CH4(g)

• Need CH4 on the right and 2 – H2’s on the left.• Reverse the 3rd eq. ∴ change the sign of ΔH• Double 2nd equation & multiply the ΔH by 2.

• Now cross out anything that is the same on both sides.

C(s) + O2(g) CO2(g) ΔHcº = -393 kJ mol-1

2H2(g) + 2/2 O2(g) 2H2O(l) ΔHcº = 2(-286) kJ mol-1

CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔHcº = +890 kJ mol-1

Adding the equations will now give us the main equation.

• C(s) + 2H2(g) CH4(g)

ΔHfº = (+890 kJ mol-1 ) + 2(-286) kJ mol-1 + (-393 kJ mol-1)

ΔHfº = -75 kJ mol-1

• The change in enthalpy for the formation of one mole of a compound from its elements with all substances in standard states

• Degree symbol = measured at standard conditions

• Can be calculated from lit. values.

Standard Enthalpy of Formation ΔHfº

ΔH reaction = Σ(ΔHf ◦products - Σ(ΔHf

◦reactants)

Standard Enthalpy of Formation ΔHfº

What is the standard heat of formation of CO2?Hint: Elements in their standard state ΔHfº = 0

2CO (g) + O2 (g) → 2CO2 (g)

Practice: ΔHfº

Substance ΔHfº (kJ mol-1)

CO -110.5

O2 0

CO2 -393.5

ΔH reaction = (2 x -393.5) - [(2 x -110.5) + o]ΔH reaction = -566.0 kJ mol-1