Transcript Thermochemistry

Thermochemistry

Study of energy and energy transfer
◦ All physical changes and chemical changes are
accompanied by a change in energy
Thermochemistry

The total energy of the universe is
constant
ΔEuniverse = 0
◦ Ie. Energy cannot be created nor destroyed

Energy CAN, however, be transferred from
one form to another!
Law of Conservation of Energy

To interpret energy changes, we need to
first define some terms:
◦ The system is the part of the universe being
studied/observed (ex. reactants and products)
◦ The surroundings is everything else in the
universe.
The relationship between system and
surroundings:
 ΔEuniverse = ΔEsystem + ΔEsurroundings = 0

System and Surroundings

Any change in the energy of a system
must be accompanied by an equal and
opposite change in the energy of the
surroundings
ΔEsystem = -ΔEsurroundings
Surroundings = rest
of the universe!
System
The system is more likely
to only significantly
influence the immediate
surroundings, so this is
what we consider.
Law of Conservation of Energy

Open system = open to surroundings- both
matter and energy can be exchanged

Closed system= no matter can be
exchanged, however energy still can be

Isolated system = completely isolated from
surroundings, both in terms of matter and
energy
Types of Systems

We know that:
◦ Potential energy = stored energy
◦ Kinetic energy = energy of motion

SI units for both are joules (J)
◦ = 1kg·m2/s2

Temperature is a measure of the average
kinetic energy of the particles that make up a
system
◦ Celcius = relative scale (designed relative to H2O)
◦ Kelvin = absolute scale (0 = absolute 0, molecules
have no kinetic energy!)
◦ 273K = 0oC
Types of Energy
Kelvin vs Celcius

Heat (q) is the transfer of kinetic energy
between objects with different
temperatures (units = J)
◦ That is why when a substance absorbs heat,
it’s particles speed up- they have really
absorbed kinetic energy!

According to the first law of
thermodynamics:
qsystem = -qsurroundings
What is Heat?

What factors are important to consider
when measuring changes in energy?
◦ Temperature
◦ Mass
◦ Specific heat capacity (c)
 Reflects the nature of a given species to retain
heat (aka. store energy)
 The energy required to change 1 gram of a
substance by 1 degree Celcius
 Units = J/g·oC

For example, the specific heat capacity of
water is 4.184 J/g·oC
◦ Helps to moderate Earth’s temperature
Factors that Affect Change in
Energy

Remember, heat is an energy
◦ Need to include temperature, mass and specific
heat capacity:
Change in
temperature
(K or oC)
Heat (J)
Mass (g)
Type of
substance
Specific heat
capacity (J/goC)
Depends
on
State of
substance
Okay, So How do we Calculate
Heat Transfer??

10.0 g of ice is added to 60.0 g of water that
has an initial temperature of 26.5oC. The final
temperature of the mixture is 9.7oC. How
much heat is lost by the water?
◦ m = 60.0 g (we are only concerned about the
water!)
◦ Ti = 26.5oC
◦ Tf = 9.7oC
◦ c = 4.184 J/g oC
q= mcΔt
= (60.0 g)(4.184 J/g oC) (26.5oC – 9.7 oC)
= - 4.22 x 103 J (or -4.22 kJ)
Therefore, the water lost 4.22 kJ of heat

Okay, let’s do an example

Which has a greater heat capacity?

Water has the same specific heat capacity
in both cases- but different masses have
different heat capacities
Heat Capacity

Relates the heat of a sample to it’s
change in temperature
C= c x m
◦ C = heat capacity (kJ/oC)
◦ c = specific heat capacity (kJ/kg oC)
◦ m = mass (kg)

And, therefore
q = CΔT
Heat Capacity

A bathtub contains 100 kg of water. What is
the heat capacity of the water in the tub?
We need to use units of kJ and kg for the specific
heat capacity
C = 4.184 kJ/kgoC x 100 kg
= 418 kJ/oC

How much heat is transferred to the
surroundings as the water cools from 60oC to
20oC?
q = 418 kJ/oC x (20oC – 60oC)
= -1.67 x 104 kJ
qsystem = - qsurroundings
Therefore the surroundings gained 1.67 x 103 kJ of
heat
For Example

What about the teacup?

The teacup contains 0.100 kg of water.
What is the heat capacity of the water in
the teacup? What is the energy
transferred to the surroundings if the
water cools from 60oC to 20oC?
For Example

What is heat?

Aluminum has a specific heat capacity of
0.902 J/goC. Copper has a specific heat
capacity of 0.389 J/goC. If an equal
amount of heat is transferred to similar
masses of each metal, which will increase
more in temperature?
Quick Review

okay, so we can see a change in kinetic
energy through a change in temperature

What happens during a phase change?
◦ Ie. solid  liquid  gas

What happens during a chemical reaction?

These changes are due to changes in the
potential energy of the system
Enthalpy Change

Enthalpy changes refer to changes in the
potential energy of a system
◦ Symbol ΔH
◦ Measured at constant pressure
◦ Units kJ/mol

Recall: Potential Energy Diagrams!
◦ Enthalpy change of a reaction represents the
difference between the potential energy of the
reactants and the products
Enthalpy Change

What’s going on here?
◦ Chemical bonds are
sources of stored energy
(PE)
◦ Breaking a chemical bond
always requires energy
◦ Recall: Reactivity of reactants
◦ Forming a bond always releases energy
+ Energy
Enthalpy Change

How much energy is required to break a bond
depends on the strength of the bond
◦ Consider a synthesis reaction:
N2 + O2  2NO
◦ If more energy is required to break the triple bonds
of N2 and double bonds of O2 than is given off in
forming N-O bonds, the reaction has a net
absorption of energy
 ENDOTHERMIC
◦ If more energy was given off during the formation
of N-O bonds, the reaction would have a net
release of energy
 EXOTHERMIC
Enthalpy Change

The internal energy of a substance (at constant
pressure) = enthalpy
◦ Chemists never care about the absolute enthalpy
◦ We study the change in enthalpy that accompanies a
process (relative)

Enthalpy of reaction ΔHrxn
◦ Depends on temperature and pressure
◦ Aka. heat of reaction

Standard enthalpy of reaction ΔH0rxn
◦ Occurs at SATP (25oC and 100 kPa)

Standard molar enthalpy of formation ΔH0f
◦ Quantity of energy absorbed or released when one mole
of a compound is formed directly from it’s elements in
their standard states.
Enthalpy Change

Three ways:
◦ 1. Easiest way is to write a thermochemical
equation (ie. include heat in the balanced
equation
 (we did this with Le Chatelier’s Principle!)
H2(g) + ½O2(g)  H2O(l) + 285.8 kJ/mol
◦ 2. Write as a separate expression
H2(g) + ½O2(g)  H2O(l)
ΔH = -285.8 kJ/mol
Enthalpy, H
◦ 3. Enthalpy diagram
H2(g) + ½ O2(g)
ΔH = -285.8 kJ/mol
H2O(l)
Representing Enthalpy Change

Show the standard molar enthalpies of
formation for the following substances
using a balanced thermochemical
equation and by writing it as a separate
expression.
◦
◦
◦
◦

H2O
CaCl2
CH4
C6H6
Draw an enthalpy diagram for the
standard molar enthalpy of formation of
sodium chloride
TRY IT

Enthalpy changes also depend on the
amount of reactants you have
moles

So:
Heat and Enthalpy

How much heat is released when 50.00 g
of methane (natural gas) is formed from
it’s elements?
◦ Remember that to find moles of an element we
have to divide the mass of the substance by
the molar mass (Y diagram)
n = 50.00g CH4 / (12.01 g/mol + 4(1.01 g/mol)
= 3.13 mols CH4
q = nΔH0f
= 3.13 mols x -74.6 kJ/mol
= -232 kJ
Example

Hydrogen gas and oxygen gas react to
form 0.534 g of liquid water. How much
heat is released to the surroundings?

How my heat is released if 0.543g of
gaseous water is produced?
Try This
Try some enthalpy
calculations now

In general, enthalpy changes for physical
changes are smaller than chemical
◦ Intermolecular forces are weaker than bonds

PE difference between liquid and gas is
much greater than a liquid and solid
Enthalpy Changes and Change of
State
ΔHvap : enthalpy change for to change one
mole from liquid to gas
 ΔHcond : enthalpy change for to change
one mole from gas to liquid
 ΔHmelt : enthalpy change for to change one
mole from solid to liquid (aka. fusion,
ΔHfus)
 ΔHfre : enthalpy change for to change one
mole from liquid to solid
ΔHvap = -ΔHcond
ΔHmelt = -ΔHfre

Molar Enthalpy

ΔHsoln :Heat transfer when 1 mole of
solute dissolves in solvent
◦ Can be endothermic or exothermic
 Ex. hot and cold packs

Calculate heat changes for state changes
the same way as chemical changes
◦ Ex: An ice cube with a mass of 8.2 g is placed
in some lemonade. The ice cube melts
completely. How much heat does the ice cube
absorb from the lemonade as it melts?
Enthalpy of Solution

Show temperature changes as heat is
removed or added to a substance
Heating and Cooling Curves

During the phase change, the
temperature is constant!
Heating and Cooling Curves
Okay, lets say you want to make some
spaghetti. You put water on the stove to
boil.
 Meanwhile, you get distracted by an
episode of Vampire Diaries and water
boils off completely.

◦ What would the heating curve for this look
like?
◦ How much heat was required to completely
vaporize all of the water?

Heat Curve:

To find the heat required, we have 2
Use this for a change in kinetic energy
equations:
(ie. temperature is changing)
Use this for a change in potential
energy (ie. phase is changing)

How much heat is absorbed by 5.00 kg of
water at 10oC if it is heated until all of the
water has evaporated?
◦ Step 1: Heating the water to it’s boiling point
q = mcΔT
= (5000 g)(4.184 J/goC)(100oC-10oC)
=1.88 x 106 J or 1880 kJ
◦ Step 2: Change liquid water into water vapour
q = nΔHvap
= (5000 g/18 gmol-1)(40.7 kJ/mol)
= 1.13 x 104 kJ
◦ Finally, 1880 kJ + 11300 kJ = 13180 kJ

So, the total heat absorbed by the water was
1.32 x 104 kJ
Example
Now practice calculating
heat over phase changes

We measure the heat transferred in a
reaction by monitoring the temperature
change
◦ We need to minimize the heat transferred to
surroundings that we are not measuring
Surroundings = rest
of the universe!
System
But the water is the
only part of the
surroundings we
want to study!
Calorimetry
We need an isolated system
We use devices called calorimeters to
study changes in kinetic energy
 Calorimeters are typically composed of:

◦ Water
◦ A thermometer
◦ And isolated system
A simple version of a
calorimeter is a coffee-cup calorimeter:

Calorimetry


Recall the law of conservation of energy:
Assumes:
qsystem = -qsurroundings
◦ The system is isolated
◦ The amount of heat exchanged with the
calorimeter itself is negligible
◦ The properties of the water remains the same

The system is at thermal equilibrium
when all components have the same
temperature
Calorimetry

A 70.0g piece of metal was heated to
95oC in a hot water bath. This was quickly
transferred to a coffee cup calorimeter.
The calorimeter contained 100 g of water
at 19.8oC. The final temperature of the
contents was 22.6oC. What is the specific
heat capacity of the metal?
◦ Recall: I already gave you a question similar to
this when we started this unit!
Example

Consider the reaction:
CuSO4(aq) + 2NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq)

50 mL of 0.3 M CuSO4 is mixed with 50 mL of
0.6 M NaOH. The initial temperature of both
solutions is 21.4oC. After mixing the solutions
in a calorimeter the final temperature is
24.6oC. What is the enthalpy change for this
reaction? (assume the final solution has a
density of 1g/mL)

HINT: qrxn = -qsolution
Heat of Reaction

What we know:
◦ Volume CuSO4 solution = 50 mL, [CuSO4] =
0.3 M
◦ Volume NaOH = 50 mL, [NaOH] = 0.6 M
◦ Ti = 21.4oC
◦ Tf = 24.6oC
Bomb calorimeters are
designed to measure the
enthalpy changes of
combustion reactions at a
constant volume
 Many parts which can
absorb or release small
amounts of energy- but
not small enough to
assume the quantities are
negligible

Bomb Calorimeter

To get precise measurements we need to
know the heat capacity of the bomb
calorimeter
Ctotal = Cwater + Cthermometer + Cstirrer + Ccontainer

Bomb calorimeters are calibrated using
constant masses, so the heat capacities
do not include mass units
qcal = CΔT
Bomb Calorimeter


A laboratory wants to test the energy content
of peanut butter. A 16 g sample is placed in a
steel bomb calorimeter (calibrated to C =
8.28 kJ/oC). During the experiment the
temperature increased by 50.5oC. What is the
heat released by the sample?
qcal = (8.28 kJ/oC)(50.5oC)
=418 kJ
qcal = -qsample
Therefore the sample released 418 kJ of
energy.
Example- Bomb Calorimeter