Physical versus Chemical Change of State
Chemical
composition
does not change
Chemical changes occur
e.g., chemical
Thermodynamics : focus on initial and final states: IGNORE what happens
on the way
An idealized schematic diagram of the change of state that occurs when
gasoline is burned in an automobile cylinder.
2-1
State and Path Functions
Properties describing system
Ebbing p 232
Two types
Depend only on
state of system:
difference
between start and
end (STATE
FUNCTION)
Depend on how
you get from
start to finish
Distance between
two cities is a
state function, but
distance
traveled is not a
state function.
2-2
Energy absorption and release
+)U
-)U
-)U
+)U
Schematic diagrams showing energy flows that accompany reactions.
(a) When a reaction releases energy, the surroundings gain energy. For
these reactions, changes in energy()
)U) is negative for the system and
positive for the surroundings. (b) When the reaction absorbs energy, the
surroundings lose energy. For these reactions, ) U is positive for the
system and negative for the surroundings.
Change in Internal Energy is written as
)U
2-3
The internal energy, U, is the sum of the kinetic and potential energies of the
particles making up the system.
Internal energy is a state function. That is, it is a property of a system that
depends only on its present state, which is completely determined by
variables such as temperature and pressure.
Internal energy can change by heat loss (or gain) and/or by work
done by (or on the system; heat is given symbol q : work w we
will look at them in more detail soon.
Increased Internal Energy of system
surroundings
)Usys
Decreased Internal Energy of system
2-4
Energy and Heat
How do we measure heat energy and work? We need to get this
value )U by adding q and w
• first :heat flow (q) : Units Joules
Experimentally this is done by
measuring a temperature change.
The temperature change is then
converted from Celsius to joules
(energy)
This is done using a calorimeter
2-5
Block diagram of a calorimeter.
Note: )T is in 0C. The value of T
can therefore be in C or K
(Kelvin)
2-6
In order to determine how much energy flowed by measuring
temperature changes the calorimeter must be calibrated
23.55O
q
Chemical reaction
In this simple example the chemical reaction is seen to
produce 10.00kJ of energy
Ccal is the heat capacity of the calorimeter : here it is
10.00/23.55 = 0.424 kJ/OC
once you have Ccal you can get q
i.e., q = Ccal x )T : = 0.424 x 23.55 = 10.00kJ
2-7
Important calorimetry equations:
q = m x C x ∆T
∆T = Tfinal - T initial
Heat energy equals the product of mass, heat capacity
and temperature change
NOTE: ∆T gives the sign of q. If temperature
goes up then qcal is +ve.
This means that qchemicals must be -ve. The chemicals lost energy
to heat up the calorimeter
We will see that C can be expressed as a molar
heat capacity (J/mol) or as a specific heat
capacity (J/g). You must be careful with units
2-8
Example: a styrofoam cup contains 100g of water at 25OC. 900J of work is done by
stirring. As a result:
1. The water gets cooler
2. Water warms up by 2 degrees
3. No effect.
q = m x C x ∆T
so ∆T = q/mC
= 900/100 x 4.184
= 2O C
Coffee cup calorimeter
UNITS: C is in Jg-1K-1:it
is the heat capacity of
water
2-9
So : Amount of Energy to raise 1 g of water through 1CE or
1 K Units J g-1 K-1
e.g. H2O(R) Specific Heat = 4.184 Jg-1 deg-1
Example. 50.0 g Cu (Molar Ht. Cap 24.5 J K -1 mol-1) is heated to 100.EC
and then placed in 250. g H2O at 0.0EC (Sp Ht H2O = 4.184 J g-1K-1).
What is final temp of H2O (in EC)?
NOTE: here the heat capacity of copper is given as a molar quantity
so we have to convert the amount of copper into moles
Copper loss of energy = - water gain of energy.
Let final Temp be Tf EC (copper and water end up at the same temperature)
)T = Tfinal - Tinitial)
moles Cu = 50.0 g/63.55 g mol-1 = 0.787 mol
loss by Cu = -(gain by H2O)
Cp × mol × )T
=
24.5 JK-1mol-1 × 0.787 mol × (Tf -100)deg
(recall
Sp.Ht × mass × )T
= (4.184 Jg-1deg-1 × 250 g × (Tf-0.0)deg)
19.3Tf -1928 = -1046Tf
ˆ -1065.3Tf = -1928
Also see examples 6.5 and 6.6 Ebbing
Tf = 1.81EC.
2-10
Example
1.00g (CH3)2N2H2 :MW 60.1) is burned in a Calorimeter
(heat capacity 1840. J deg-1)
5.00 kg H2O in bath goes from 24.62 6 26.07EC
Calculate the energy released when 1 mole is burned.
heat produced 6 warms calorimeter + mass H2O
6 (1840 J deg-1 × 1.45 deg)
+ (4.184 J g-1 deg-1 × 5.00 × 103g × 1.45 deg)
6 2668 + 30334
= 33002 J = 33.0 kJ evolved by 1.00/60.1 mol
= 0.01664 mol
0.0166 mol evolves 33.0 kJ
1 mol evolves 33.0/0.0166 = 1.98 x 103 kJ
2-11
Recall : energy change results in flow of energy between system and surroundings
• can be heat (q) or work (w)
•now we add work to the equation
Note internal energy of
system decreases as
work is done on
surroundings
w = F x d (force times distance)
UNITS: Nm = Joules
Expanding gas in an automobile cylinder displaces the piston. In this
displacement, work is done against the force exerted on the area of the
piston by the external pressure.
2-12
Expansion work: important for const pressure calorimeters
P = F/A (pressure = force /unit area) : F =PA
W = F x d = Pexternal x A x d
and: )V = A.d
so wsurr = Pext )V :
Work is done on the surroundings
so wsys = -Pext . )V sys
2-13
Homework
1. List the four ways in which the internal energy of a system can be changed.
2. A system absorbs 6 kJ of heat, what is the change in the internal energy.
3. In compressing a system, 4.5 kJ of work is done on it, what is the change in the internal energy.
4. A machine does 1000 kJ of work, what is the change in the internal energy? Explain whether the
sign should be positive or negative.
Solutions are given on the first page of next lecture---
first try to solve these questions before looking
2-14