Chapter 17 - Thermochemistry -is the study of energy relationships in chemical and physical...

Chapter 17 - Thermochemistry

-is the study of energy relationships in chemical and physical reactions

almost nothing happens in chemistry unless there is an energy advantage; all reactions are an attempt to arrive at a lower system energy.

a) 1st law of Thermodynamics: Law of conservation of energy.

Energy is never lost or gained , it only changes form.

This law states that energy can change form but is never really lost in any closed system.

b) 2nd law of Thermodynamics: Heat Energy always travels spontaneously

from a warmer body (body with higher temperature) to a colder body. -droping a hot rock in water -frostbite

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OBJECTIVES:

• Explain the relationship between energy and heat.

• Distinguish between heat capacity and specific heat.

temperature is simply an indirect measure of the average kinetic energy of the particles in a sample of matter. The scale of measurement is arbitrary:

i) Celsius scale sets the zero value at the freezing point of water at 101.325 kPa and 100 at the boiling point of water at the same pressure.

ii) Kelvin scale sets the zero value at absolute zero.

The degrees are the same size as celsius degrees.

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Energy and Heat Thermochemistry - concerned with

heat changes that occur during chemical reactions

Energy - capacity for doing work or supplying heat

• weightless, odorless, tasteless

• if within the chemical substances- called chemical potential energy

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Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.

• only changes can be detected!

• flows from warmer cooler object

Exothermic and Endothermic Processes

Essentially all chemical reactions, and changes in physical state, involve either:

• release of heat, (exothermic) or

• absorption of heat (endothermic)

In studying heat changes, think of defining these two parts:

• the system - the part of the universe on which you focus your attention

• the surroundings - includes everything else in the universe

Together, the system and it’s surroundings constitute the universe

Thermochemistry is concerned with the flow of heat from the system to it’s surroundings, and vice-versa.

The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed.

• All the energy is accounted for as work, stored energy, or heat.

heat flowing into a system from it’s surroundings:

• defined as positive

• q has a positive value

• called endothermic

–system gains heat as the surroundings cool down

heat flowing out of a system into it’s surroundings:

• defined as negative

• q has a negative value

• called exothermic

–system loses heat as the surroundings heat up

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Every reaction has an energy change associated with it

Exothermic reactions release energy, usually in the form of heat.

Endothermic reactions absorb energy

Energy is stored in bonds between atoms

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Heat Capacity and Specific Heat

A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.

• Used except when referring to food

• a Calorie, written with a capital C, always refers to the energy in food

• 1 Calorie = 1 kilocalorie = 1000 cal.

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Heat Capacity and Specific Heat

The calorie is also related to the joule, the SI unit of heat and energy

• named after James Prescott Joule

• 4.184 J = 1 cal Heat Capacity - the amount of heat

needed to increase the temperature of an object exactly 1 oC

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Heat Capacity and Specific Heat Specific Heat Capacity - the amount

of heat it takes to raise the temperature of 1 gram of the substance by 1 oC often called simply “Specific Heat”

mass; measured in g or kg. The greater the mass, the more heat required to change the temperature.

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For water, C = 4.18 J/(g oC), Thus, for water:

• it takes a long time to heat up, and

• it takes a long time to cool off! Water is used as a coolant!

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Heat Capacity and Specific Heat

To calculate, use the formula: q = m C T heat abbreviated as “q” m = mass T = change in temperature (Tf - Ti) C = Specific Heat Units are either J/(g oC) or kJ/(kg oC)

Table of Specific Heats

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• Ex: Determine the heat required to raise the temperature of 100g of water from 298.0 K to 373.0 K .

Q = m c ΔT

Q = 100g (4.184 J/g K)( 373.0 K – 298.0 K)

Q = 418.4 J/K (75 K)

Q = 31350 J

Q = 31.4 kJ

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Calorimetry Calorimetry - the accurate and

precise measurement of heat change for chemical and physical processes.

The device used to measure the absorption or release of heat in chemical or physical processes is called a Calorimeter

Calorimetry Foam cups are excellent heat

insulators, and are commonly used as simple calorimeters

A Cheap Calorimeter

For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system

Calorimetry Changes in enthalpy = H q = H These terms will be used

interchangeably in this textbook Thus, q = H = m x C x T H is negative for an exothermic

reaction H is positive for an endothermic

reaction

Calorimetry Calorimetry experiments can be

performed at a constant volume using a device called a “bomb calorimeter” - a closed system

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C + O2 CO2E

nerg

y

Reactants Products

C + O2

C O2

395kJ

+ 395 kJ

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In terms of bonds

COO C

O

O

Breaking this bond will require energy.

CO

OOO C

Making these bonds gives you energy.In this case making the bonds gives you more energy than breaking them.

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Exothermic The products are lower in energy

than the reactants Releases energy

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CaCO3 CaO + CO2E

nerg

y

Reactants Products

CaCO3

CaO + CO2

176 kJ

CaCO3 + 176 kJ CaO + CO2

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Endothermic The products are higher in energy

than the reactants Absorbs energy

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Chemistry Happens in

MOLES An equation that includes energy is

called a thermochemical equation CH4 + 2O2 CO2 + 2H2O + 802.2 kJ

1 mole of CH4 releases 802.2 kJ of energy.

When you make 802.2 kJ you also make 2 moles of water

What is the molar enthalpy of CO2 (g) in the reaction for the burning of butane below?

2 C4H10 +13 O2 8 CO2 +10 H2O

∆H=-5315 kJ

Answer: Molar enthalpy is the enthalpy change in equation divided by the balance of CO2

Molar enthalpy, H substance = 5315 kJ ÷ 8 mol

= 664 kJ / mol.

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Thermochemical Equations A heat of reaction is the heat

change for the equation, exactly as written• The physical state of reactants

and products must also be given.

• Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 oC

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CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ

If 10. 3 grams of CH4 are burned completely, how much heat will be produced?

10. 3 g CH4

16.05 g CH4

1 mol CH4

1 mol CH4

802.2 kJ

= 514 kJ

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CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ

How many liters of O2 at STP would be required to produce 23 kJ of heat?

How many grams of water would be produced with 506 kJ of heat?

How much heat will be released if 65 grams of butane is burned in a lighter according the equation:

2 C4H10 +13 O2 8 CO2 +10 H2O

∆H=-5315 kJ

= 2976.4 kJ

= 3.0 MJ

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104104 2

5315

14.58

165

HmolC

kJ

g

HmolCHgC

Calculate the heat released when 120 grams of Iron (III) oxide is formed by the following equation 

2 Fe2O3 (s) → 4 Fe(s)+3 O2 (g)

∆H=1625 kJ

mol

kJ

g

OmolFeOgFe

2

1625

70.159

1120 32

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= 610.5 kJ= 610 kJ

Q = n ∆H (substance)

Where n = # of moles

What mass of carbon dioxide must form to create 1200 kJ of heat when the following reaction occurs?

C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l) 

∆H=- 2808kJ

Answer: 110 grams

3) What mass of oxygen is needed to completely react and release 550 kJ of heat in the following reaction?

4Fe (s)+3O2 (g) → 2 Fe2O3 (s)

∆H=- 1625 kJ

Answer: 32 grams

Summary, so far...

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Enthalpy The heat content a substance has at a

given temperature and pressure Can’t be measured directly because

there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy

changes

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Enthalpy Symbol is H Change in enthalpy is H (delta H) If heat is released, the heat content of

the products is lower

H is negative (exothermic) If heat is absorbed, the heat content

of the products is higher

H is positive (endothermic)

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Ene

rgy

Reactants Products

Change is down

H is <0

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Ene

rgy

Reactants Products

Change is upH is > 0

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Heat of Reaction The heat that is released or absorbed in a

chemical reaction Equivalent to H C + O2(g) CO2(g) + 393.5 kJ

C + O2(g) CO2(g) H = -393.5 kJ

In thermochemical equation, it is important to indicate the physical state

H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ

H2(g) + 1/2O2 (g) H2O(l) H = -285.8 kJ

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Heat of Combustion The heat from the reaction that

completely burns 1 mole of a substance

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OBJECTIVES:

• Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.

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OBJECTIVES:

• Calculate heat changes that occur during melting, freezing, boiling, and condensing.

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Heats of Fusion and Solidification

Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid

Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies

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Heats of Fusion and Solidification

Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies

• Thus, Hfus = -Hsolid

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Heats of Vaporization and Condensation

When liquids absorb heat at their boiling points, they become vapors.

Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.

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Heats of Vaporization and Condensation

Condensation is the opposite of vaporization.

Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses

Hvap = - Hcond

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Heats of Vaporization and Condensation

The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous

• You can receive a scalding burn from steam when the heat of condensation is released!

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Heats of Vaporization and Condensation

H20(g) H20(l) Hcond = - 40.7kJ/mol

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Heat of Solution Heat changes can also occur when

a solute dissolves in a solvent. Molar Heat of Solution (Hsoln) -

heat change caused by dissolution of one mole of substance

Sodium hydroxide provides a good example of an exothermic molar heat of solution:

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Heat of Solution

NaOH(s) Na1+(aq) + OH1-

(aq)

Hsoln = - 445.1 kJ/mol The heat is released as the ions

separate and interact with water, releasing 445.1 kJ of heat as Hsoln thus becoming so hot it steams

H2O(l)

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OBJECTIVES:

• Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.

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OBJECTIVES:

• Calculate heat changes using standard heats of formation.

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Hess’s Law If you add two or more

thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

Called Hess’s law of heat summation

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Why Does It Work? If you turn an equation around, you change

the sign: If H2(g) + 1/2 O2(g) H2O(g) H=-285.5 kJ

then, H2O(g) H2(g) + 1/2 O2(g) H =+285.5 kJ

also, If you multiply the equation by a number,

you multiply the heat by that number: 2 H2O(g) H2(g) + O2(g) H =+571.0 kJ

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You make the products, so you need their heats of formation

You “unmake” the products so you have to subtract their heats.

How do you get good at this?

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Standard Heats of Formation The H for a reaction that produces 1

mol of a compound from its elements at standard conditions

Standard conditions: 25°C and 1 atm. Symbol is H f

0

The standard heat of formation of an element = 0

This includes the diatomics

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What good are they? Table 11.6, page 316 has standard

heats of formation The heat of a reaction can be calculated

by:

• subtracting the heats of formation of the reactants from the products

Ho = (H f0 H f

0Products) - ( Reactants)

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Examples CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

H f0

CH4 (g) = - 74.86 kJ/molH f

0O2(g) = 0 kJ/mol

H f0

CO2(g) = - 393.5 kJ/mol

H f0

H2O(g) = - 241.8 kJ/mol H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]

H= - 802.4 kJ

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Examples