Transcript Chpt 5: First Law of Thermodynamics

Chapter 5: Thermochemistry
Law of Conservation of Energy: Energy is neither created nor
destroyed during a chemical or
physical change.
It can be transformed into a different type.
Kinetic energy: energy associated with motion.
Potential energy: stored energy that will be
released if the object falls.
Electrostatic potential energy: energy
stored as a result of interactions between
charged particles.
 = 8.99 × 109 J-m/C2
EK = ½ mv2
EP = mgh
E el 
 Q1Q 2
Qe = -1.60 x 10-19 C
r
2
Energy is usually defined as the ability to do work (w) or to
transfer heat (q).
The surroundings
If we heat the system (+q), the expansion
of the gas will lift the piston and do work
on the surroundings (-w).
(w = F•d)
As a result of the addition of heat and
the loss of energy from the system, the
internal energy (sum of all types of
energy within the system) has changed.
DE = Ef - Ei
The system
Note: It is very difficult to know the
absolute internal energy of a system.
Fortunately, DE can be found from q and w 
DE = q + w
Two gases, A(g) and B(g), are confined in a cylinder-and-piston
arrangement like that in Figure 5.3. Substances A and B react to form a
solid product: A(g) + B(g)→ C(s). As the reactions occurs, the system loses
1150 J of heat to the surroundings. The piston moves downward as the
gases react to form a solid. As the volume of the gas decreases under the
constant pressure of the atmosphere, the surroundings do 480 J of work
on the system. What is the change in the internal energy of the system?
DE = q + w
q = -1150 J
w = +480 J
DE = -1150 J + 480 J = -670 J
Recall:
In an endothermic process, the system gains heat.
(Heat enters the system.)
In an exothermic process, the system loses heat.
(Heat exits the system.)
State functions depend only on the initial and final state, and
are independent of the path taken.
DEP = Ef - Ei = -10 J
This system has lost 10 J of
potential energy to the
surroundings.
Examples of state functions
Initial
Final
EP = 10 J
EP = 0 J
DE, DH, DS, DG, DT
Q: Are q and w state functions?
No, both are dependent upon the path taken. (Think about
trying to move a box across the floor.)
Enthalpy: the amount of heat energy transferred when the
system is under constant pressure.
Thus, DH = qp
DE = qp + w
DE = DH + w
The only time w is nonzero is when gases are involved.
P
DE = DH - PDV
w = F•d = -PDV
DH = DE + PDV
Recall:
d  DV
+DH
endothermic
-DH
exothermic
Enthalpies of reaction (DHrxn):
If 2 mol of H2 and 1 mol of O2 are combusted, DHrxn = -483.6 kJ
2 H2 (g) + O2 (g)  2 H2O (g)
DH = -483.6 kJ
Q: What would DH be if 4 mol of H2 and 2 mol of O2 were combusted?
A: -967.2 kJ
Note: While DH is a state function, it is NOT an intensive property. The
enthalpy change depends upon the quantity of matter involved in
the reaction.
Q: What would DH be if 6 g of H2 was combusted in an excess of O2?
6 g H2/(2 g/mol) = 3 mol H2
 483.6 kJ
2 mol H
2

Excess = the amount of O2 doesn’t matter
x
3 mol H
x = -725.4 kJ
2
Hess’s Law states that if a reaction is carried out in a series of
steps, ΔH for the overall reaction will equal the sum of the
enthalpy changes for the individual steps.
EX: The enthalpies of reaction for the combustion of C to CO2
and for CO to CO2 are shown below:
flip
C + O2  CO2
DH = -393.5 kJ
CO + ½ O2  CO2
DH = -283.0 kJ
What is the DHrxn for C + ½O2  CO ?
0.5
C + O2  CO2
+ CO2  CO + ½ O2
C + ½ O2  CO
DH
-393.5 kJ
+ +283.0 kJ
-110.5 kJ
An Enthalpy Diagram
C + O2
DH = —110.5 kJ
CO + ½ O2
DH = -393.5 kJ
DH = —283.0 kJ
CO2
2 AlBr3 + 3 Cl2
2 AlCl3 + 3 Br2
2 AlBr3 + 3 Cl2
But how do we
determine the heat
content in the first
place?
2 AlCl3 + 3 Br2
Energy
DHrxn = Heat content of products – heat content reactants
DHrxn < 0
Reaction is exothermic
Heat of formation, DHf
• The DHf of all elements in their standard state equals zero.
• The DHf of all compounds is the molar heat of reaction for
synthesis of the compound from its elements
DHrxn (AlBr3): 2 Al + 3 Br2
2 AlBr3
Are forming 2 moles of AlBr3 so the DHrxn is two times
larger than the DHf for one mole of AlBr3
Therefore,
DHrxn
DHf(AlBr3) =
2
• Since the DHrxn can be used to find DHf, this means that DHf
can be used to find DHrxn WITHOUT having to do all of the
calorimetric measurements ourselves!!
The Law of Conservation of Energy strikes again!!
Hess’s Law:
DHrxn = S DHf(products) – S DHf(reactants)
6 CO2 (g) + 6 H2O (l)
C6H12O6 (s) + 6 O2 (g)
DHrxn = [DHf(C6H12O6) + 6 DHf(O2)] – [6 DHf(CO2) + 6 DHf(H2O)]
From DHf tables:
DHf(C6H12O6) = -1250 kJ/mol
DHf(CO2) = -393.5 kJ/mol
DHf(H2O) = -285.8 kJ/mol
DHrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]
DHrxn = +2825.8 kJ/mol
Calorimetry: The measurement of heat flow
Calorimeter: A device to measure heat flow.
Heat capacity: the amount of heat needed to make the
temperature an object increase by 1C.
Q: Is heat capacity an intensive or extensive property?
A: Extensive. Think about how long it takes a large
amount of water to heat up as compared to a much
smaller amount with the same starting temperature.
Specific heat: the amount of heat needed to increase the
temperature of 1 gram of a substance by 1C.
Q: Is specific heat an intensive or extensive property?
A: Intensive because it is now on a per gram basis.
Q: What do you think the molar heat capacity is?
A: The amount of heat needed to increase the temperature
of one mole of a substance by 1C.
q = mCpDT
q = heat, m = mass, Cp = specific heat, DT = Tf - Ti
Ex. How much heat is needed to warm 250 g of water (about 1
cup) from 22°C to near its boiling point, 98°C? The specific heat
of water is 4.18 J/g−K. (b) What is the molar heat capacity of
water?
q = mCpDT = (250 g)(4.18 J g-1K-1)(98C – 22C)
q = 7.9 x 104 J
b)
Cp 
4.18 J
g K
x
18.015 g H 2 O
1 mol

75.2 J
mol  K
How much heat is required by a 100 g paraffin candle to
increase the temperature by 5 °C?
Cp(paraffin) = 2.1 J/g°C
q = Cp(mass)(DT)
q = (2.1 J/g°C)(100 g)(5 °C)
q = 1050 J
If the same amount of heat was used to heat 100 g of
water [Cp(liquid water) = 4.184 J/g°C], what would be the
DT of the water?
q = Cp(mass)(DT)
1050 J = (4.184 J/g°C)(100g)(DT)
DT = 2.5 °C
For the same amount of heat and
mass, DT decreases as the specific
heat of the substance increases
Constant Pressure Calorimetry
100 kg Pb
If the temperature of the lead
is 327°C before it hits the
water, what is the final
temperature of the lead and
the water?
q = mCpDT
200 ft
DT = Tf - Ti
-qPb = qH2O
-(1x105 g)(0.13 J/g°C)(Tf – 327°C) =
10 kg H2O
Ti = 20 °C
Cp(Pb) = 0.13 J/g°C
(1x104 g)(4.18 J/g°C)(Tf – 20°C)
Tf = 93 °C
Cp (H2O) = 4.18 J/g°C
Constant Volume Calorimetry: Bomb Calorimetry
• The heat from the reaction
(usually a combustion) causes the
temperature of the Bomb and the
surrounding water to increase.
• The heat capacity of the bomb
calorimeter, Ccal, must be
determined using a reaction for
which the DHcomb is known.
qrxn = -CcalDT
Ex: A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a
calorimeter whose heat capacity is 4.812 kJ/°C. The
temperature increases from 23.10°C to 24.95°C. Calculate the
heat of combustion of lactic acid (a) per gram and (b) per mole.
Ccal = 4.812 kJ/C
DT = Tf – Ti = 24.95C – 23.10C = 1.85C
qrxn = -CcalDT = (-4.812 kJ/C)(1.85C) = -8.90 kJ
(a) Per gram:
-8.90 kJ/0.5865 g = -15.2 kJ/g
(b) Per mole:
(-15.2 kJ/g)(90.079 g/mol) = -1370 kJ/mol
FW(HC3H5O3) = 90.079 g/mol