Dr. C. Yau Fall 2013

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1 1 Energy Part III: Calculation of ΔH from a) Thermochemical Equations b) Heat of Formation Chapter 7 Sec 6 – Sec 8 of Jespersen 6 TH ed Dr. C. Yau Fall 2013

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Energy Part III: Calculation of Δ H from a) Thermochemical Equations b) Heat of Formation Chapter 7 Sec 6 – Sec 8 of Jespersen 6 TH ed. Dr. C. Yau Fall 2013. 1. Standard Heat of Reaction. Δ H = “enthalpy of reaction” or “heat of reaction” = heat transferred in a rxn - PowerPoint PPT Presentation

Transcript of Dr. C. Yau Fall 2013

Page 1: Dr. C. Yau Fall 2013

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Energy Part III:Calculation of ΔH froma) Thermochemical Equationsb) Heat of Formation

Chapter 7 Sec 6 – Sec 8of Jespersen 6TH ed

Dr. C. Yau

Fall 2013

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Standard Heat of ReactionΔH = “enthalpy of reaction”

or “heat of reaction”

= heat transferred in a rxn

(usually in kJ, not kJ/mol)

ΔHo= standard heat of reaction

= ΔH at standard conditions o means standard conditions (1 atm, 1M if aq soln, usually 25oC)Remember these conditions!

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Thermochemical Equations

N2 (g) + 3H2 (g) 2NH3 (g) ΔHo = -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 ΔHo is in kJ and not kJ/mol.

The amt of heat transferred is directly proportional to the # moles in thermochemical eqn shown.

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Thermochemical EquationsExample: 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.

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Thermochemical Equations

N2 (g) + 3H2 (g) 2NH3 (g) ΔHo = -92.39 kJ

The value of ΔHo depends on the coefficients in the equation.

If coefficients are doubled, ΔHo would be doubled:

2N2 (g) + 6H2 (g) 4NH3 (g) ΔHo = -92.39x2 kJ

Note also that ΔHo is dependent on the physical states as stated in the equation.

CH4 (g) + 2O2 (g) CO2(g) + 2H2O(l) ΔHo = - 890.5kJ

CH4 (g) + 2O2 (g) CO2(g) + 2H2O(g) ΔHo = - 802.3kJ

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Example 7.7 p.276The following thermochemical equation is for the

exothermic reaction of hydrogen and oxygen that produces water.

2H2 (g) + O2 (g) 2H2O (l) ΔHo = -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

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Thermochemical Equations

If we reverse a reaction, the magnitude of ΔHo 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

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Determination of ΔH 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!

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Determination of ΔH 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 KjWhat is the strategy?

Do Pract Exer 13, 14, 15 p.283

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Determination of ΔH by manipulation of Eqns.2Cu (s) + O2 (g) 2CuO (s) H° = 310 kJ2Cu (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

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Determination of ΔH by manipulation of Eqns.

Example 7.9 p.282Fe2O3 (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)

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Enthalpy DiagramsC (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.

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Enthalpy DiagramsN2 (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.

You should know how H diagrams differ between an endo- and exothermic reactions.

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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.

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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

H1 = H2 + H3 + H4+H5 + H6

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For example:

A + B C

D + E

H1=?

H2H3

H1 = H2 + H3

Hess’ Law is extremely useful because if H1 cannot be measured, it can be calculated from H2 and H3.

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Hess' LawWatch 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

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Standard Heat of Combustion

ΔHoc = 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, ΔHo

c is always negative.

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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 ΔHo

c is -890 kJ/mol

Do Pract Exer 16 & 17 p.235

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Standard Enthalpy of Formation

LEARN THIS DEFINITION!

ΔHfo

= 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.

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ΔHfo

for solid potassium sulfate is -1433.7 kJ. Write the thermochemical equation corresponding to this value.

ΔHfo

for solid ammonium chloride is – 315.4 kJ. Write the thermochemical equation corresponding to this value.

ΔHfo

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

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ΔHfo of Elements

Write the thermochemical equation corresponding to the ΔHf

o of chlorine gas.

What do you think the value of ΔHfo

would be for chlorine gas?

What about the value of ΔHfo

of solid silver? of liquid mercury?

Remember this! You will not be provided ΔHf

o elements in their standard states.

They are always ZERO kJ.

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Applying Hess' Law to Heats of Formation

A + B C + D

All elements in their standard states.

What is heat of reaction of A+B C+D?

Hrxn = ?

Hf (A) Hf (B) Hf (C) Hf (D)

Hrxn = -Hf (A) -Hf (B) + Hf (C) +Hf (D)

Hrxn = - {Hf (A) +Hf (B) } + {Hf (C) +Hf (D)}

Hrxn = +{Hf (C) +Hf (D)} - {Hf (A) +Hf (B) }

Hrxn = SumH (products) - Sum H(reactants)

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Hess' Law of Summation

ΔHfo values are given in Table 7.2 p. 285

and will be provided on exams.

rxn f fH n H (Products) n H (Reactants)

Remember Hf is per mole.You must multiply it with n (the number of moles from the coefficients of the balanced equation).

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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 ΔHo for this reaction in kilojoules.

Do not put out a kitchen fire with water!http://www.youtube.com/watch?v=sZGzbd0IvUE&feature=related

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Calculating ΔH For Reactions Using ΔHf°

2Fe(s) + 6H2O(l) → 2Fe(OH)3(s) + 3H2(g)

ΔH°f = -285.8 -696.5

in kJ mol-1

CO2(g) + 2H2O(l) → 2O2(g) + CH4(g)

ΔH°f = -393.5 -285.8 -74.8

in kJ mol-1

Do Pract Exer 20, 21 & 22 p. 288

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SUMMARY

What are the different ways you can determine the ΔH of a reaction?

• Measure q from calorimetry experiment. In open containers, q = ΔH

• Calculate by manipulating given thermochemical equations.

• Calculate from ΔHfo

• Calculate from bond energies (handout)

Also, remember how to calculate amt of heat from stoichiometry & thermochemical eqn.