Transcript Slide 1

The Nature of Energy
• Energy is the ability to do work or produce
heat.
• weightless, odorless, tasteless
• Two forms of energy exist, potential and
kinetic.
• Potential energy is due to composition or
position.
• Chemical potential energy is energy stored in
a substance because of its composition.
• Kinetic energy is energy of motion.
The Nature of Energy (cont.)
• The law of conservation of energy
states that in any chemical reaction or
physical process, energy can be
converted from one form to another,
but it is neither created nor destroyed.
• Heat is energy that is in the process
of flowing (transferring) from a
warmer object to a cooler object.
• q is used to symbolize heat.
Endothermic and Exothermic Processes
•
Essentially all chemical reactions
and changes in physical state
involve either:
• release of heat, or
• absorption of heat
Endothermic and Exothermic Processes
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
Endothermic and Exothermic Processes
•Heat flowing into a system from it’s
surroundings:
•defined as positive
•q has a positive value
•called endothermic
•system gains heat (gets warmer) as the
surroundings cool down
Endothermic and Exothermic Processes
•Heat flowing out of a system into it’s
surroundings:
•defined as negative
•q has a negative value
•called exothermic
•system loses heat (gets cooler) as the
surroundings heat up
Measuring Heat
• A calorie is defined as the amount of
energy required to raise the temperature of
one gram of water one degree Celsius.
• Food is measured in Calories, or 1000
calories (kilocalorie).
• A joule is the SI unit of heat and energy,
equivalent to 0.2390 calories.
• 1 calorie = 4.184 J or 1 J = 0.2390 calories
Measuring Heat (cont.)
• Example:
• A candy bar has 245 Calories. Convert this to calories
and then to Joules of energy.
Specific Heat
• The specific heat of
any substance is the
amount of heat required
to raise one gram of that
substance one degree
Celsius.
• Some objects require
more heat than others to
raise their temperature.
Specific Heat (cont.)
• Calculating heat absorbed and released
– q = c × m × ΔT
– q = heat absorbed or released (in Joules)
– c = specific heat of substance
– m = mass of substance in grams
– ΔT = change in temperature in Celsius
Specific Heat (cont.)
• Examples:
• How much heat does a 20.0 g ice cube absorb as its
temperature increases from (-27.0oC) to 0.0oC? Give
your answer in both joules and calories.
• q = c × m × ΔT
• Specific Heat of Ice = 2.03 J/goC
• 1 calorie = 4.184 J
Specific Heat (cont.)
• Example Cont.




q=?
c = 2.03 J/goC
m = 20.0 grams
ΔT = FinalTemp(0.0oC) – InitialTemp (-27.0oC) = Change (27.0oC)
 q = c × m × ΔT
q = (2.03 J/goC)(20.0g)(27.0oC)=
• Example 2:
• A 5.00 gram sample of a metal is initially at 55.0 ºC.
When the metal is allowed to cool for a certain time, 98.8
Joules of energy are lost and the temperature decreases
to 11.0º C. What is the specific heat of the metal? What
metal is it?
• q = c × m × ΔT
To make the problem
easier, solve for the
unknown BEFORE you
plug in the numbers.
Measuring Heat
• For Water during a phase change:
– The Heat of Fusion (melting) is 334 j/g
– The Heat of Solidification (freezing) is 334 j/g
• They are the same value (energy in or out)
– The Heat of Vaporization is 2260 j/g
– The Heat of Condensation is 2260 j/g
• They are the same value (energy in or out)
• Example – Phase change
• Calculate the amount of energy needed to
convert 55.0 grams of ice to all liquid water at its
normal melting point.
• Using the same amount of water calculate the
energy needed to completely vaporize the water
at its normal boiling point.
• Why is there such a large difference in energy
needed?
The liquid is boiling at
100o
120
C; no temperature change
(use q = mass x ΔHvap.)
The gas temperature is rising
from 100 to 120 oC
(use
The Heat Curve for Water,
q =going
massfrom
x ΔT -20
x C)to 120 oC,
The liquid temperature is rising
from 0 to 100 oC
(use q = mass x ΔT x C)
The solid is melting at 0o C; no temperature change
(use q = mass x ΔHfus.)
The solid temperature is rising from -20 to 0 oC
(use q = mass x ΔT x C)
Calorimetry
• Calorimetry - the measurement of the heat
into or out of a system for chemical and
physical processes.
• A calorimeter is an insulated device used
for measuring the amount of heat absorbed
or released in a chemical reaction or
physical process.
• Based on the fact that the heat released =
the heat absorbed
Chemical Energy and the Universe (cont.)
• Chemists are interested in changes in
energy during reactions.
• Enthalpy is the heat content of a system at
constant pressure.
• Enthalpy (heat) of reaction is the change in
enthalpy during a reaction symbolized as
ΔHrxn.
ΔHrxn = Hfinal – Hinitial
ΔHrxn = Hproducts – Hreactants
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
Chemical Energy and the Universe (cont.)
Chemical Energy and the Universe (cont.)
Exothermic
• The products are lower in energy than
the reactants
• Thus, energy is released.
• ΔH = -395 kJ
– The negative sign does not mean negative
energy, but instead that energy is lost.
Endothermic
• The products are higher in energy than
the reactants
• Thus, energy is absorbed.
• ΔH = +176 kJ
– The positive sign means energy is
absorbed
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 and 1 mole of
CO2
Thermochemical Equations
The
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 (SC) for the
reaction is (1 atm.) and 25 oC (different
from STP)
1 CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) + 802.2 kJ
 If
10.3 grams of CH4 are burned
completely, how much heat will be
produced? Convert to moles Convert moles to desired unit
Start with known value
10. 3 g CH4
1 mol CH4
16.05 g CH4
802.2 kJ
1 mol CH4
Ratio from balanced equation
= 514 kJ
ΔH = -514 kJ, which means the heat is
released for the reaction of 10.3 grams CH4