Principles of Chemistry Chapter 7 .pptx

Burning a dry wood…
Solar energy to
Chemical energy
𝐶 6 𝐻 12𝑂6(𝑠)+6𝑂2(𝑔)→6𝐶𝑂2(𝑔)+6 𝐻 2𝑂(𝑙)+h𝑒𝑎𝑡

Energy Conversation and Energy Transfer
• Law of Conservation of Energy:
(First law of Thermodynamics)
Energy can neither be created nor destroyed
But… it can be converted
from one form to the other
What happens when it hit the ground?

Internal energy (E): the sum of all kinetic and potential energies of
an object (system)
• State function:
Depends on the state of the system, now how the system arrived at
that state
ΔE = Efinal – Einitial
Analogy:
Either path you take, change in elevation will be 7000 ft
Δelevation=
=10000 ft - 3000 ft
=7000
You start at 3000 ft

ΔE initial
ΔE final
Change in energy (ΔE) occurs due to the difference in internal energy
between reactants and products
ΔE = Eproducts – Ereactants
𝐶 6 𝐻 12𝑂6(𝑠)+6𝑂2(𝑔)→6𝐶𝑂2(𝑔)+6 𝐻 2𝑂(𝑙)+h𝑒𝑎𝑡

First Law of Thermodynamics: The total energy of the universe
is constant (conti)
ΔE univ = ΔE sys + ΔE surr = 0
ΔE sys = - ΔE surr
Universe = System + Surroundings
𝐶 6 𝐻 12𝑂6(𝑠)+6𝑂2(𝑔)→6𝐶𝑂 2(𝑔)+6 𝐻 2𝑂(𝑙)+h𝑒𝑎𝑡

First Law of Thermodynamics: The total energy of the universe is
constant
System Surroundings
Energy
Player 1 Player 2
$$$
Analogy:
𝐶6 𝐻 12𝑂6(𝑠)+6𝑂2(𝑔)→6𝐶𝑂2(𝑔)+6 𝐻 2𝑂(𝑙)+h𝑒𝑎𝑡

System
The sub-part of the universe - the focus of a thermodynamic study
Isolated / open / closed
Has a lid on it

Exothermic process
Exo – release
Heat flows out of system to surroundings
q is negative (q < 0)
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

Endothermic process
Endo = absorb
Heat flows into system from surroundings
q is positive (q > 0)

Changes in energy takes place in two forms : Work (w) and
Heat (q)
System Surroundings
Energy
w & q
Fuel and air
Consider a piston in a gasoline
engine….
- Work is done on the system by
turning on the engine
- Part of the energy escapes to the
surrounding as heat

Energy is exchanged between the system and surroundings
through heat (q) and work (w).
ΔE = q + w

Analogy:
Checking
account
Deposit : ‘+’ withdrawal: ‘-’
Sign Convention

A cylinder and piston assembly (defined as the system) is warmed by an external flame. The
contents of the cylinder expand, doing work on the surroundings by pushing the piston
outward against the external pressure. If the system absorbs 559 J of heat and does 488 J of
work during the expansion. What is the value of E?
Δ

Calculate the change in the internal energy for a process in which a system
releases 1422J of heat to the surrounding and does 855 J of work on the
surrounding.

Units of Energy
Joule (J) (often use KJ)
The SI unit of energy
calorie (cal)
4.184 J = 1 cal

Enthalpy and Calorimetry
(conti)

Specific heat capacity: The energy required to raise the temperature
of 1 gram of a substance by 1°C at constant pressure
𝑞=𝑚𝑐𝑃 ∆ 𝑇
Substance cp (J/g °C)
Copper 0.376
Stainless Steel 0.502
Plastic 1.3
• Heat capacity (CP) : (regardless of mass)
• Molar heat capacity (cp, n) : Unit (J/mol °C )

Ammonia reacts with oxygen according to the equation:
4NH3 (g) + 5O2 (g) → 4 NO(g) + 6H2O (g) H
Δ = -906 kJ
Calculate the heat (in kJ) associated with complete reaction of 155 g of NH3.

C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l) + Energy H
Δ = -1418 kJ/mol
How much energy is emitted if we burn 25.6 g of ethanol in excess oxygen?

Suppose you find a penny in the snow.
a) How much heat is absorbed by the penny as it warms from the temperature of the snow,
which is -8.00°C , to the temperature of your body, 37.0°C? Assume the penny is pure
copper and has a mass of 3.10g. (cp : 0.376 J/g °C)
b) What is the molar heat capacity of the penny?

A 55.0 g aluminum block initially at 27.5°C absorbs 725 J of heat. What is the final
temperature? (cp : 0.903 J/g °C)

Calorimetry: measuring heat flow during physical or chemical
change
• Calorimeter: Device used to measure
the absorption or release of heat
• Coffee-Cup Calorimetry: Two nested
styrofoam cups
• Can be used to determine specific heat
capacity and heat of a reaction (ΔHrxn)
for aqueous solution
qrxn = −qsolution
= − mcwater ΔT

Bomb Calorimetry: measures the heat of reaction for combustible
samples
• Consists of heavy steel container,
water insulated container and minor
components
• Ccalorimeter: The heat capacity of the
bomb calorimeter and all other
insulated components
• Constant volume Calorimetry
ΔErxn rxn qrxn = −qcalorimeter
= − Ccalorimeter ΔT

Different Ways to Determine Enthalpy Change of Chemical Reaction

Hess’s Law: The change in enthalpy for a stepwise process is the sum of the
enthalpy changes of the steps

Enthalpy is a state function
Hess’s Law: ΔHrxn = ΔH1 + ΔH2 + ……. + ΔHn
CH4 (g) + 2O2 (g)  CO2(g) + 2H2O (g) ΔH1 = -802 kJ
CH4 (g) + 2O2 (g)  CO2(g) + 2H2O (l) ΔH = ?
2H2O (l)  2H2O (g) ΔH2 = 88kJ (Flip the equation)
ΔHoverall = ΔH1 + ΔH2 = -890kJ

© 2009, Prentice-Hall, Inc.
Hess’s Law
• H is well known for many reactions, and it is
inconvenient to measure H for every reaction
in which we are interested.
• However, we can estimate H using published
H values and the properties of enthalpy.

Hess’s Law
Hess’s law states that “[i]f
a reaction is carried out
in a series of steps, H
for the overall reaction
will be equal to the sum
of the enthalpy changes
for the individual steps.”

Hess’s Law
Because H is a state
function, the total
enthalpy change depends
only on the initial state of
the reactants and the final
state of the products.

Using the following data, calculate the Hrxn for the following reaction.
C2H4 (g) + H2 (g)  C2H6 (g)
H2 (g) + ½ O2 (g)  H2O (ℓ) - 285.8 kJ
C2H4 (g) + 3 O2 (g)  2H2O (ℓ) + 2CO2 (g) -1411 kJ
C2H6 (g) + 7/2 O2 (g)  3H2O (ℓ) + 2CO2 (g) -1560 kJ

Principles of Chemistry Chapter 7 .pptx

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