5/7/2014
Enthalpy & Hess’s Law
Quantifying Heats of Reactions with Sequential Steps
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Today:
◦ Enthalpy: Heats of Reactions
◦ Hess’s Law
◦ Bond Dissociation Energies
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See Canvas QUIZ 4 Study
Guide & Lab Practical
Review
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Challenge Problem Set 14 DUE
this Monday, May 12th @ 11 pm
Please Read Chp. 8: Sections 8.88.11, pp. 290-299
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“The visible world is the invisible
organization of energy.”
—Heinz Pagels
iClicker Participation Question:
Hess’s Law & Heats of Vaporization
Which reaction below is the MOST EXOTHERMIC?.
A. C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)
B. C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)
C. C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
D. C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
E. They all will release the SAME AMOUNT OF ENERGY
C5H12(g)
Enthalpy Diagram:
8 O2(g) + C5H12(l)
Fuel VAPORS are
more flammable
than liquid fuels
5 CO2(g) + 6 H2O(g)
6 H2O(l)
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5/7/2014
Hess’s Law:
If a reaction is the sum of 2 or more reactions, then its overall H is
equal to the sum of the H values of the other reactions.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(l) → C5H12(g) ΔHo2 = +26.2 kJ
H2O(l) → H2O(g) ΔHo3 = +44.0 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
Hess’s Law:
Adjust the reactions to eliminate the undesired substances.
THEN add the reactions up to achieve the OVERALL desired reaction.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(l) → C5H12(g) ΔHo2 = +26.2 kJ
H2O(l) → H2O(g) ΔHo3 = +44.0 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
Hess’s Law:
REVERSE the second reaction to eliminate C5H12(l) from the equation.
The reverse of an endothermic reaction is an exothermic one.
Change the sign of ΔH when reversing a reaction.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(g) → C5H12(l) ΔHo2’ = -26.2 kJ
H2O(l) → H2O(g) ΔHo3 = +44.0 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
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5/7/2014
Hess’s Law:
Reverse the 3rd reaction AND MULTIPLE BY 6 to eliminate 6 H2O(g).
Remember to change the sign of ΔH when reversing a reaction.
If a reaction is multiplied by a factor, so is the corresponding ΔH.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(g) → C5H12(l) ΔHo2’ = -26.2 kJ
6 H2O(g) → 6 H2O(l) ΔHo3’ = (-44.0 kJ)·6 = -264 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
Hess’s Law:
The OVERALL enthalpy change for a reaction is equal to the sum of
enthalpy changes for the individual steps in the reaction
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(g) → C5H12(l) ΔHo2’ = -26.2 kJ
6 H2O(g) → 6 H2O(l) ΔHo3’ = (-44.0 kJ)·6 = -264 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
ΔHoOVERALL = ΔHo1 + ΔHo2’ + ΔHo3’ = -3799 kJ
Bond Dissociation Energies:
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5/7/2014
Calculate ΔH using Bond Energies of Reactants & Products
ΔH = Σ(Bond Broken)–Σ(Bond Formed)
Calculate ΔH using Bond Energies of Reactants & Products
H2CO3  H2O + CO2
ΔH = ?
ΔH = Σ(Bond Broken)–Σ(Bond Formed)
ΔH = ?
ΔH = Σ(Bond Broken)–Σ(Bond Formed)
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5/7/2014
ΔH =?
A. ΔH = - (Exothermic)
B. ΔH = + (Endothermic)
C. ΔH = 0 (Thermoneutral)
D. Cannot determine from information
provided: bond energies needed.
Thermochemistry Practice Problems
• Consider the following endothermic reaction:
4 NO(g) + 6 H2O(g)  4 NH3(g) + 5 O2(g)
ΔHo = 906.22 kJ
What quantity of heat will be consumed when 58.0
g of NH3 is produced?
HINT: Enthalpy changes refer to heat
associated with the MOLE QUANTITIES of the
stoichiometric coefficients reacting:
906.22 kJ = 4 mol NO = 6 H2O = 4 NH3 = 5 O2
Thermochemistry Practice Problems
• How much heat is released when a mixture
containing 10.0 g CS2 and 10.0 g Cl2 react by the
equation:
CS2(g) + 3Cl2(g)  S2Cl2(g) + CCl4(g) H = 232 kJ
HINT: The limiting reagent dictates the MAXIMUM
amount of heat that can be absorbed or released.
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5/7/2014
Thermochemistry Practice Problems
• Consider the reaction for which H is unknown:
H = ?
2 S(s) + 3 O2(g)  2 SO3(g)
However, the heats of reaction of the following two
reactions are known:
S(s) + O2(g)  SO2(g)
H1 = 297 kJ
H2 = 198 kJ
2 SO3(g)  2 SO2(g) + O2(g)
Use these two reactions to calculate the change in
enthalpy for the top reaction.
Thermochemistry Practice Problems

Consider the gas-phase reaction:
2 N2O5(g)  4 NO2(g) + O2(g)
H = ?
Based on the gas-phase reactions below with the
enthalpies given, determine the H value for the above
reaction.
4 NO + 3 O2  2 N2O5
H1 = – 445 kJ
2 NO + O2  2 NO2
H2 = – 112 kJ
Thermochemistry Practice Problems
•
Consider the reaction:
W(s) + C(s,graphite)  WC (s)
Hrxn = ?
for which the heat of reaction is not known.
Calculate Hrxn using the following information:
2 W(s) + 3 O2(g)  2 WO3(s)
H1 = 1680.6 kJ
C(graphite) + O2(g)  CO2(g)
H2 =  393.5 kJ
2 WC(s) + 5O2(g)  2 WO3(s) + 2 CO2(g) H3 = 2391.6 kJ
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