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TOPIC 5.2: Hess’s Law and Enthalpy of Formation
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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?
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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.
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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!
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Reaction Routes- Example #1
If the reaction of A→ B cannot be
measured, use the data from A→ C→ B
instead!
ΔH1 = ΔH2 + ΔH3
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Enthalpy Graph for Example #1
ΔH1 = ΔH2 + ΔH3
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Practice: #2
Write the
equation for ΔH of the
A → D reaction
ΔH = ΔH1 + ΔH2+ ΔH3
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Practice: #3
Write 2 different equations to
solve for ΔH1
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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
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Draw an energy cycle
for the 3 reactions in the graph.
Write an expression for ΔHc and
solve.
More Practice!
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• 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 :)
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• 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 :)
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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.
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• 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
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• 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
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ΔH reaction = Σ(ΔHf ◦products - Σ(ΔHf
◦reactants)
Standard Enthalpy of Formation ΔHfº
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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