Title: Energy Part III: Calculation of H from a Thermochemical Equations b Heat of Formation Chapter 6 Sec
1Energy Part IIICalculation of ?H froma)
Thermochemical Equationsb) Heat of
FormationChapter 6 Sec 6 Sec 8of Brady
Senese 5TH ed
1
2Standard Heat of Reaction
?H enthalpy of rxn or heat of reaction heat
transferred in a rxn (usu. in kJ, not
kJ/mol) ?Ho standard heat of reaction
?H at standard conditions (1 atm, 25oC,
1M if aq soln) Remember these conditions!
2
3Thermochemical Equations
N2 (g) 3H2 (g) 2NH3 (g) ?Ho
-92.39 kJ This tells us that 1 mole of N2 would
produce 92.39 kJ of heat, that 3 moles of H2
would produce 92.39 kJ of heat, that production
of 2 moles of NH3 would be accompanied by a
release of 92.39 kJ of heat. NOTE that ?Ho is in
kJ and not kJ/mol.
3
4Thermochemical Equations
N2 (g) 3H2 (g) 2NH3 (g) ?Ho
-92.39 kJ The value of ?Ho depends on the
coefficients in the equation. If coefficients are
doubled, ?Ho would be doubled 2N2 (g) 6H2
(g) 4NH3 (g) ?Ho -92.39x2 kJ Note
also that ?Ho is dependent on the physical states
as stated in the equation. CH4 (g) 2O2 (g)
CO2(g) 2H2O (l) ?Ho - 890.5kJ CH4 (g) 2O2
(g) CO2(g) 2H2O (g) ?Ho - 802.3 kJ
4
5Thermochemical Equations
- If we reverse a reaction, the magnitude of ?Ho is
the same but the sign is changed - C (s) O2 (g) ?? CO2 (g) ?H ? 393.5 kJ
- CO2 (g) ?? C (s) O2 (g) ?H 393.5 kJ
6Example 6.6 p.228
- The following thermochemical equation is for the
exothermic reaction of hydrogen and oxygen that
produces water. - 2H2 (g) O2 (g) 2H2O (l) ?Ho
-571.8kJ - What is the thermochemical equation for this rxn
when it is conducted to produce 1.000 mol H2O? - Do Pract Exer 7 8 p.229
7Determination of ?H
- Example 6.8 p. 233
- Fe2O3 (s) 3CO (g) ?? 2Fe(s) 3CO2(g) ?H ?
26.7 kJ - CO(g) ½ O2 (g) ?? CO2 (g)
?H ? 283.0 kJ - Calculate the value of ?H for the following
reaction - 2 Fe (s) ?O2 (g) ?? Fe2O3 (s)
8Determination of ?H
- Ethylene glycol, HOCH2CH2OH, is used as
antifreeze. It is produced from ethylene oxide,
C2H4O, by the reaction - C2H4O (g) H2O (l) ?? HOCH2CH2OH (l)
- What is the heat of reaction of this reaction...
- Given
- 2C2H4O (g) 5O2 (g) ?? 4CO2 (g) 4H2O (l)
- ?H ?2612.2 kJ
- HOCH2CH2OH(l) 5/2 O2(g) ??2CO2(g) 3H2O(l)
- ?H ?1189.8 kJ
- Do Pract Exer 11, 12, 13 p.234
9Hess' Law
- The value of ?H for any reaction that can be
written in steps equals the sum of the values of
?H of each of the individual steps. - This is based on the fact that enthalpy is a
state function. - The implication is that regardless of how many
steps are taken the overall enthalpy change is
the same.
10Enthalpy as a State Function
- Enthalpy (H) depends only on its current state
and not the path taken to get there. - It is not affected by how many steps are used to
get there. - A ????? B
-
- C F
- D E
11Hess' Law
?H1
?H3
?H2
?H1 ?H2 ?H3 This is because
H is a state function. This is extremely useful
because if ?H1 cannot be measured, it can be
calculated from ?H2 and ?H3.
12Hess' Law
- What is ?H1 in terms of ?H2 and ?H3 ?
- F G J
- K L
?H1
?H2
?H3
?H1 ?H2 - ?H3
13Enthalpy Diagrams
- C (s) ½ O2 (g) ?? CO (g) ?H ?110.5 kJ
- Construct an enthalpy diagram for the reaction.
- Learn the terminology, "enthalpy diagram."
- Know what is asked for on an exam.
14Enthalpy Diagrams
- N2 (g) O2 (g) ?? 2 NO (g) ?H 181 kJ
- Draw the enthalpy diagram for this reaction.
- We are skipping Example 6.7, Pract Exer 9 10.
You do not need to know how to draw enthalpy
diagrams of that sort. However, you should know
how to add equations together to determine the
enthlapy change for the overall reaction.
15- Use the two equations below to determine the
standard enthalpy change for the reaction - H2O2 (l) ?? H2O (l) ½ O2 (g)
- H2 (g) O2 (g) ?? H2O2 (l) ?H ?188kJ
- H2 (g) ½ O2 (g) ?? H2O (l) ?H ?286kJ
- Solve the problem by both methods
- By sketching enthalpy diagram
- By manipulating the given eqns.
16- 2Cu (s) O2 (g) ?? 2CuO (s) ?H ?310 kJ
- 2Cu (s) ½ O2 (g) ?? Cu2O (s) ?H ?169 kJ
- Use the two equations above to determine the ?H
of this reaction - Cu2O (s) ½ O2 (g) ?? 2CuO (s)
- Is this exothermic or endothermic?
- Solve the problem by both methods
- By sketching enthalpy diagram
- By manipulating the given eqns. ANS
-141 kJ - Use the equations given in Pract Exer 10 p.232
find the ?H for the reaction - ½ N2 (g) O2 (g) ?? NO2 (g)
- Is this exothermic or endothermic?
- You do not need to construct an enthalpy diagram
to solve this problem.
17Standard Heat of Combustion
- ?Hoc standard heat of combustion
- It is the amount of heat released when one mole
of a fuel substance is completely burned in pure
oxygen gas with all reactants and products
brought to 25oC and 1 bar pressure (1 atm). - Combustion reactions are always exothermic.
Therefore, ?Hoc is always negative.
18- Example 6.9
- How many moles of carbon dioxide gas are produced
by a gas-fired power plant for every 1.00 MJ
(megajoule) of energy it produces? the plant
burns methane, CH4 (g), for which ?Hoc is -890
kJ/mol - Do Pract Exer 14, 15 p.235
19Standard Enthalpy of Formation
- LEARN THIS DEFINITION!
- ?Hfo standard enthalpy of formation
- It is the amount of heat absorbed or evolved when
specifically one mole of substance is formed at
25oC at 1 atm from its elements in their standard
states.
20- ?Hfo for solid potassium sulfate is -1433.7 kJ.
Write the thermochemical equation corresponding
to this value. - ?Hfo for solid ammonium chloride is 315.4 kJ.
Write the thermochemical equation corresponding
to this value. - ?Hfo for solid calcium hydroxide is -986.6 kJ.
Write the thermochemical equation corresponding
to this value. - Do Example 6.10 and Pract Exer 16 17 on p.237
21?Hfo of Elements
- Write the thermochemical equation corresponding
to the ?Hfo of chlorine gas. - What do you think the value of ?Hfo would be for
chlorine gas? - What about the value of ?Hfo of solid silver? of
liquid mercury? - Remember this! You will not be provided ?Hfo
elements in their standard states. They are
always ZERO kJ.
22Applying Hess' Law to Heats of Formation
All elements in their standard states.
What is heat of reaction of AB CD?
23Hess' Law of Summation
- ?Hfo values are given in Table 6.2 p.237 and will
be provided on exams.
24Example 6.11 p.238
- Some chefs keep baking soda, NaHCO3, handy to put
out grease fires. when thrown on the fire, baking
soda partly smothers the fire and the heat
decomposes it to give CO2, which further smothers
the flame. The eqn is - 2NaHCO3 (s) ?? Na2CO3 (s) H2O (l) CO2 (g)
- Use the data in Table 6.2 (p.236) to calc the ?Ho
for this reaction in kilojoules. - http//www.youtube.com/watch?vz7JzJlgNuU8
25Calculating ?H For Reactions Using ?Hf
- 2Fe(s) 6H2O(l) ? 2Fe(OH)3(s)
3H2(g) - ?Hf -285.8 -696.5
- kJ mol-1
- CO2(g) 2H2O(l) ? 2O2(g) CH4(g)
- ?Hf -393.5 -285.8 -74.8
- kJ mol-1
- Do Pract Exer 18, 19, 20 p.239
26SUMMARY
- What are the different ways you can determine the
?H of a reaction? - Measure q from calorimetry experiment. In open
containers, q ?H - Calculate by manipulating given thermochemical
equations. - Calculate from ?Hfo
- Calculate from bond energies (handout)