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Units of Energy Like we saw with pressure, many different units are used throughout the world for energy. 4.184 Joules (J) = 1 calorie (cal) 1 kcal = 1000 calories = 1 Nutritional Calorie (Cal) Joule (J) calorie (cal) erg (erg) electron volts (eV) British thermal unit (Btu) Amount of energy it takes to raise one gram of water 1ºC SI unit for energy 2 2 s m kg 1 J 1 = Energy in Thermochemistry Heat (q): energy transferred from an object with a higher temperature to an object with a lower temperature Work (w): energy transferred against a force 100º C 50º C heat Heat cannot be transferred in the opposite direction! Example: Combustion reactions allow your car to move...performing work Vroom... Thermochemistry In thermochemistry, the universe is divided into two parts: System Surroundings Surroundings Universe System + Surroundings System Surroundings Surroundings The system : The physical process or chemical reaction in which we are interested. We can define the system anyway we like. The surroundings : Everything else in the universe. Energy flow between system and surroundings System Surroundings Surroundings Energy (both heat and work) can be transferred between the system and surrounds. System Surroundings Surroundings _ + The sign “+” or “- ” tells us the direction of energy flow. After I add 2.0kJ of heat energy to each sample, will they still be the same temperature? 5.0 g Al 25°C 5.0 g Cu 25°C copper

Chapter 8 Spring 2010 Steward

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Energy in Thermochemistry

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Page 1: Chapter 8 Spring 2010 Steward

Units of Energy

Like we saw with pressure, many different units are used throughout the world for energy.

4.184 Joules (J) = 1 calorie (cal)

1 kcal = 1000 calories = 1 Nutritional Calorie (Cal)

Joule (J) calorie (cal) erg (erg)

electron volts (eV) British thermal unit (Btu)

Amount of

energy it takes

to raise one gram of water

1ºC

SI unit for energy

2

2

s

m kg 1 J 1

⋅=

Energy in Thermochemistry

Heat (q): energy transferred from an object with a higher temperature to an object with a

lower temperature

Work (w): energy transferred against a force

100º C 50º Cheat

Heat cannot be transferred in the opposite

direction!

Example: Combustion reactions allow your car to move...performing work

Vroom...

Thermochemistry

In thermochemistry, the universe is divided into two

parts:

System

Surroundings

Surroundings

UniverseSystem +

Surroundings

System

Surroundings

Surroundings

The system: The physical process or chemical reaction in

which we are interested.

We can define the system anyway we like.

The surroundings: Everything else in the

universe.

Energy flow between system and

surroundings

System

Surroundings

Surroundings

Energy (both heat and work) can be

transferred between the system and surrounds.

System

Surroundings

Surroundings

_

+

The sign “+” or “-” tells us the direction of energy flow.

After I add 2.0kJ of heat energy to each sample, will they still be the same temperature?

5.0 g Al

25°C

5.0 g Cu

25°C

copper

Page 2: Chapter 8 Spring 2010 Steward

Calorimetry

The energy released from the reaction (or whatever is in the calorimeter) is gained by the water. So…

-qrxn = qwater

Notice that they have opposite signs, which tells us the direction of transfer.

However, assuming no heat loss, the magnitudes are the same. The amount of heat lost by the reaction is

the same amount gained by the water.

Example

• A 28.2 gram sample of nickel is heated to 99.8oC and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5oC. After the metal cools, the final temperature of the metal and water is 25.0oC.

• Which substance absorbed heat?

• Which substance released heat?

• Calculate the heat absorbed by the substance you indicated above.

Exothermic vs. Endothermic processes

System

Surroundings

Surroundings

System:

chemical reaction or process

Surroundings

Exothermic

Endothermic

Feels warm to the touch

q = -

Feels cool to the touch

q = +

Enthalpies of Reaction

• Determine if the following processes are endothermic or exothermic…

– Combustion of methane

– Neutralization of HCl

– Boiling

– Melting

– CaCO3 (s) � CaO (s) + CO2 (g)

1111

Heat and Enthalpy

So how are q and ∆H related?

At constant pressure:

qp = ∆H

So, using a calorimeter, the ∆H of a chemical reaction can be calculated from the change in

heat in the calorimeter.

Page 3: Chapter 8 Spring 2010 Steward

Enthalpies of Physical/Chemical Changes

∆H = Hproducts – Hreactants

Energy is released if: ∑Energyreactants > ∑Energyproducts

Energy is absorbed if: ∑Energyreactants < ∑Energyproducts

Enthalpy (∆H) describes the change in heat from a chemical reaction or process.

Where do these energies come from?

Enthalpies of Phase Changes

• Heat of fusion (∆Hfus): Amount of heat required to melt (solid � liquid)

• Heat of vaporization (∆Hvap): Amount of heat required to evaporate (liquid � gas)

• Heat of sublimation (∆Hsub): Amount of heat required to sublime (solid � gas)

Why are there no values for ∆Hfreezing, ∆Hcondendsation, or ∆Hdeposition?

Thermodynamic Standard State

– A gas at 1 atm

– An aqueous solution (1 M conc.) at a pressure of 1 atm

– Pure liquids and solids

– The most stable form of elements at 1 atm and 25oC (298 K)

∆H˚The “˚” (knot)

denotes enthalpy under standard

states

Standard States

Amount of reactant, and ∆H

In chemistry, we usually specify energy in terms of a specific amount of reactant or product. For example:

2H2(g) + O2(g) → 2H2O(g) ∆H =-241.8 kJ

∆H =-483.6 kJ

Which means if we react 2 mole of H2 with 1 mole of O2, 241.8kJ of heat energy will be

released.

But what if we react 4 moles of H2 with 2 moles of O2?

This can be read as:

-241.8 kJ per 2 moles H2

-241.8 kJ per 1 mole O2

-241.8 kJ per 2 moles of H2O

Direction of reaction and ∆H

2H2(g) + O2(g) → 2H2O(g) ∆H =-241.8 kJ

What about the reverse reaction of:

2H2O(g) → 2H2(g) + O2(g) ∆H =+241.8 kJ

All we do is change the sign. Instead of energy being released (as in the first reaction), energy is absorbed when breaking water molecules

apart.

Thermochemical Equations and Stoichiometry

• 2Al (s) + Fe2O3 (s) � 2Fe (s) + Al2O3 (s) ∆Ho = -852 kJ

• How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al?– 10.0 g Fe2O3 � x mol Fe2O3

– x mol Fe2O3 � 852 kJ / 1 mol Fe2O3

Page 4: Chapter 8 Spring 2010 Steward

Hess’s Law

• If a compound cannot be directly synthesized from its elements, we must use multiple reactions to calculate the enthalpy of reaction

• Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps

• Look at direction of reaction and amount of reactants/products

Hess’s Law

2020

These two steps (solid →liquid → gas)

should have the same change in energy as…

…this one step (solid → gas)

∆H is a state function…it doesn’t matter what route we take to get there. Same change in

energy, ∆H.

Hess’s Law

• Values of enthalpy change– For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign• Reverse direction, heat flow changes; endothermic becomes exothermic (and vice versa)

– Proportional to the amount of reactant consumed • Twice as many moles = twice as much heat

– Depends on the physical state of the reactants and products (i.e., elemental state)

2121

Hess’s Law

• Does the energy change in a reaction depend on the number of steps in the reaction?

• H2(g) + I2(s) → 2HI(g) ∆H = 53.00 kJI2(s) → I2(g) ∆H = 62.44 kJ

• H2(g) + I2(g) → 2HI(g) ∆H = ?

• How can we add the first two reactions to get the third?

• Hess’s Law: ∆Hsum of steps = ∆H1 + ∆H2

2222

Application of Hess’s Law

• We can use known values of ∆Ho to calculate unknown values for other reactions

• P4 (s) + 3 O2 (g) � P4O6 (s) ∆H = -1640.1 kJ

• P4 (s) + 5 O2 (g) � P4O10 (s) ∆H = -2940.1 kJ

• What is ∆Ho for the following reaction?P4O6 (s) + 2 O2 (g) � P4O10 (s) ∆H = ?

2323

Hess’ Law Problem

• Given:

• CH4 (g) + 2O2 (g) � CO2 (g) + 2H2O (g) ∆H = -802 kJ

• H2O (l) � H2O (g) ∆H = 88 kJ

• Find the enthalpy for the following reaction:

• CH4 (g) + 2O2 (g) � CO2 (g) + 2H2O (l) ∆H = ?

Page 5: Chapter 8 Spring 2010 Steward

Standard Enthalpies of Formation

• Standard enthalpy of formation (∆Hof): heat needed

to make 1 mole of a substance from its stable elements in their standard states

� ∆Hof = 0 for a stable (naturally occurring) element

• Which of these have ∆Hof = 0?

– CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s)

• Do the following equations represent standard enthalpies of formation? Why or why not?

– 2Ag (l) + Cl2 (g) � 2AgCl (s)

– Ca (s) + F2 (g) � CaF2 (s)

2525

Standard Enthalpies of Formation

• Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book)

� ∆Horxn = Σn∆Ho

f (products) – Σn∆Hof (reactants)

− Σ = sum

– n = number of moles (coefficients)

Heats of Formation

� ∆Horxn = Σ n∆Ho

f,products - Σ n∆Hof,reactants

• Calculate values of ∆Ho for the following rxns:

• 1) CaCO3 (s) � CaO (s) + CO2 (g)

• 2) 2C6H6 (l) + 15O2 (g) � 12CO2 (g) + 6H2O (l)

2727

∆H˚ of (in kJ/mol):CaCO3 (-1207.1) CaO (-635.5) CO2 (-393.5)C6H6 (49.0) CO2 (-393.5) H2O (-285.8)

Group Work

• Use Standard Heat of Formation values to calculate the enthalpy of reaction for:

• C6H12O6(s) � 2 C2H5OH(l) + 2 CO2(g)

� ∆Hof (C6H12O6(s)) = -1260.0 kJ/mol

� ∆Hof (C2H5OH(l)) = -277.7 kJ/mol

� ∆Hof (CO2(g)) = -393.5 kJ/mol

Bond Dissociation Energies• Bond dissociation energy (BDE). Energy needed to

break one 1 mol of a specific bond (eg. H-H, C-H, C=C, C-C, etc.)

• Bond dissociation energy varies somewhat from one molecule to another, or even within one molecule, so in most cases, we have use an average bond energy (D).

• For example:• H-OH 502 kJ/mol

• H-O 427 kJ/mol• H-OOH 431 kJ/mol• Average = 453.3 kJ/mol for O-H

• Note: there are certain bonds where the BDE is not and average. These are bonds that only exist within diatomic molecules (eg. H-H, H-Br, Cl-Cl, etc.)

Bond Energy

3030

Page 6: Chapter 8 Spring 2010 Steward

Bond Dissociation Energies

� ∆Horxn = ΣBDE (reactants) + - ΣBDE (products)

endothermic exothermic

energy input energy released

• Or: ∆Horxn = ΣBDE (reactants) – ΣBDE (products)

� ΣBDE(react) > ΣBDE(prod) � endothermic

� ΣBDE(react) < ΣBDE(prod) � exothermic

• We use only when heats of formation are not available, since bond energies are average values for gaseous molecules

3131

Enthalpy of Reaction

• Use bond energies to calculate the enthalpy change for the following reaction:

• N2(g) + 3H2(g) → 2NH3(g)

� ∆Hrxn = [BEN ≡ N + 3BEH-H] – [6BEN-H]∆Hrxn = [945 + 3(436)] – [6(390)] = -87 kJ

• measured value = -92.2 kJ

• Why are the calculated and measured values different?

3232

NΞN = 945 kJ/mol

H-H = 436 kJ/mol

N-H = 390 kJ/mol

Enthalpy of Reaction

• Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride:NCl3 (g) � N2 (g) + Cl2 (g) (hint: balance first!!)

� ∆Hrxn?

3333

N-Cl = 200 kJ/mol N≡N = 945 kJ/mol Cl-Cl = 243 kJ/mol

Practice Problems

• Which amount of energy is higher?

– Energy that is released when 55g of Al is cooled from 5˚C to -25 ˚C

– Energy that is absorbed when 64g of Cu is heated from 298K to 315K?

• 8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature. What is the specific heat of the metal? Which metal is it?

Practice Problems

• Practice: Worked Example 8.10, page 320….

• Identify how to set up the following problems:

• Calculate the ∆Ho of reaction for:

– C3H8 (g) + 5O2 (g) � 3CO2 (g) + 4H2O (l)

� ∆Hof C3H8(g): -103.95 kJ/mol; ∆Ho

f CO2(g): -393.5 kJ/mol; ∆Ho

f H2O(l): -285.8 kJ/mol

• 8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature. What is the specific heat of the metal? Which metal is it?

Practice Problems

• C2H4(g ) + 6F2(g) � 2CF4(g) + 4HF(g) ∆Ho = ?

– H2 (g) + F2 (g) � 2HF (g) ∆Ho = -537 kJ

– C (s) + 2F2 (g) � CF4 (g) ∆Ho = -680 kJ

– 2C (s) + 2H2 (g) � C2H4 (g)∆Ho = 52.3 kJ

• Use average bond energies to determine the enthalpy of the following reaction (from Table 7.1).

– CH4 (g) + Cl2 (g) � CH3Cl (g) + HCl (g)

– (BEC-Cl = 328 kJ/mol)