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PPT – Energy Part III: Calculation of ?H from a) Thermochemical Equations b) Heat of Formation Chapter 7 Sec 6 PowerPoint presentation | free to download - id: 69d915-MjNkZ

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Energy Part III Calculation of ?H from a)

Thermochemical Equations b) Heat of

Formation Chapter 7 Sec 6 Sec 8 of Jespersen

6TH ed

- Dr. C. Yau
- Spring 2013

1

Standard Heat of Reaction

?H enthalpy of reaction or heat of

reaction heat transferred in a rxn (usually

in kJ, not kJ/mol) ?Ho standard heat of

reaction ?H at standard conditions

o means standard conditions (1 atm, 1M if

aq soln, usually 25oC) Remember these conditions!

2

Thermochemical 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. The amt of heat transferred is

directly proportional to the moles in

thermochemical eqn shown.

3

Thermochemical Equations

- Example Magnesium burns in air to produce a

bright light and is often used in fireworks

displays. - 2 Mg (s) O2 (g) 2MgO (s) ?Ho -1203 kJ
- How many grams of Mg is needed to produce 400.

kJ of heat? - p.297 7.62
- How much heat (in kJ) is liberated by the

combustion of 6.54 g of Mg? - Set these problems up in dimensional analysis.

Thermochemical 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 - 802.3kJ

5

Example 7.7 p.276

- 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 9 10 p.277

Thermochemical 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

- Determination of ?H by manipulation of Eqns.
- Use the two equations below to determine the

standard enthalpy change for the reaction - H2O2 (l) ?? H2O (l) ½ O2 (g)
- (1) H2 (g) O2 (g) ?? H2O2 (l) ?H ?188 kJ
- (2) H2 (g) ½ O2 (g) ?? H2O (l) ?H ?286 kJ
- Strategy
- We need H2O2 on the left side, so Eqn 1 must be

reversed. - Eqn 2 probably can stay as is. WE MUST CHECK.

How? - On my exams you are expected to show your work

as we are doing here in class. Take notes!

Determination of ?H by manipulation of Eqns.

- Ethylene glycol, HOCH2CH2OH, is used as

antifreeze. It is produced from ethylene oxide,

C2H4O, by the reaction - (1) C2H4O (g) H2O (l) ?? HOCH2CH2OH (l)
- What is the heat of reaction of this reaction...
- Given
- (2) 2C2H4O (g) 5O2 (g) ?? 4CO2 (g) 4H2O (l)
- ?H ?2612.2 kJ
- (3) HOCH2CH2OH(l) 5/2 O2(g) ??2CO2(g) 3H2O(l)
- ?H ?1189.8 Kj
- What is the strategy?
- Do Pract Exer 13, 14, 15 p.283

- Determination of ?H by manipulation of Eqns.
- 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 manipulating the given eqns.
- ANS -141 kJ

Determination of ?H by manipulation of Eqns.

- Example 7.9 p.282
- 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)

Enthalpy 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.

Enthalpy Diagrams

- N2 (g) O2 (g) ?? 2 NO (g) ?H 181 kJ
- Draw the enthalpy diagram for this reaction.
- We are skipping Example 7.8, Pract Exer 11 12.

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.

Hess' 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.

Enthalpy 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

?H1 ?H2 ?H3 ?H4?H5 ?H6

Hess Law is extremely useful because if ?H1

cannot be measured, it can be calculated from ?H2

and ?H3.

- For example
- A B C
- D E

?H1?

?H3

?H2

?H1 ?H2 ?H3

Hess' Law

- Watch out for the directions of the arrows.
- What is ?H1 in terms of ?H2 and ?H3?
- F G J
- K L

?H1

?H2

?H3

?H1 ?H2 - ?H3

Standard 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.

- Example 7.10 p.283
- 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 16 17 p.235

Standard 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.

- ?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 7.11 and Pract Exer 18 19 on p.286

?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.

Applying Hess' Law to Heats of Formation

- A B C

D

All elements in their standard states.

What is heat of reaction of AB CD?

Hess' Law of Summation

- ?Hfo values are given in Table 7.2 p. 285 and

will be provided on exams.

Example 7.12 p.287

- 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 7.2 (p.285) to calc the ?Ho

for this reaction in kilojoules. - Do not put out a kitchen fire with water!
- http//www.youtube.com/watch?vsZGzbd0IvUEfeature

related

Calculating ?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 20, 21 22 p. 288

SUMMARY

- 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)
- Also, remember how to calculate amt of heat from

stoichiometry thermochemical eqn.