Transcript AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

AP CHEMISTRY
CHAPTER 6 NOTES
THERMOCHEMISTRY
Energy- the capacity to do work
or to produce heat
1st Law of Thermodynamics:
Law of Conservation of Energy
Energy can be converted from
one form to another but it can be
neither created nor destroyed.
The total amount of energy in
the universe is constant.
Potential energy- energy due to
position or composition
Kinetic energy- energy due to
the motion of an object
-depends on mass and velocity of
an object
KE = ½
2
mv
m = mass in kg
v = velocity in m/s
units are J, since J = kg m2
s2
Heat- involves a transfer of
energy between two objects due
to a temperature difference.
Work- force acting over a
distance
-involves a transfer of energy
Temperature- a property that
reflects random motions of the
particles of a particular substance
Exothermic- reaction which
releases heat
• energy flows out of the system
• potential energy is changed to thermal energy
• products have lower potential energy than reactants
Heat term is on the right
side of the equation.
2C8H18 + 25O2  16CO2 + 18H2O + 5076 kJ
OR
2C8H18 + 25O2  16CO2 + 18H2O ΔH= 5076 kJ/molrxn
For an exothermic
reaction, ΔH is negative.
Endothermic- reaction which
absorbs heat
• energy flows into the system
• thermal energy is changed into potential energy
• products have higher PE than reactants
2C + 2H2 + 52.3 kJ  C2H4
Heat term is on the left
OR
side of the equation.
2C + 2H2  C2H4
ΔH = 52.3 kJ/molrxn
For an endothermic
reaction, the ΔH is POSITIVE!
The system is our reaction. The
surroundings are everything else.
Internal energy (E) of a system
is the sum of the kinetic and
potential energies of all the
particles in a system.
E = q + w
• E is the change in the system’s
internal energy
• q represents heat
• w represents work
• usually in J or kJ
Thermodynamic quantities
always consist of a number and a
sign (+ or ). The sign represents
the systems point of view.
(Engineers use the surroundings
point of view)
Exothermic q
(systems energy is decreasing)
Endothermic +q
(systems energy is increasing)
Example:
Calculate E if q = 50 kJ and
w = +35kJ.
E = q + w
= 50 + 35
= 15 kJ
For a gas that expands or is
compressed, work can be
calculated by:
w = PV
units:
Latm = (atm)(L)
1 Latm = 101.325 J (not tested)
Example:
Calculate the work if the volume
of a gas is increased from 15 mL to 2.0
L at a constant pressure of 1.5 atm.
w = PV
w = 1.5 atm (1.985L)
w = 3.0L . atm
At constant pressure, the terms heat
of reaction and change in enthalpy
are used interchangeably.
Example:
For the reaction
2Na + 2H2O  2NaOH + H2 ,
H = 368 kJ/molrxn
Calculate the heat change that occurs when
3.5 g of Na reacts with excess water.
3.5g Na 1 mol Na 368 kJ =
23.0g Na 2 mol Na
H = 28 kJ (or 28 kJ is released)
Calorimetry - the
science of measuring
heat flow
• based on observing
the temperature
change when a
body absorbs or
discharges heat.
• instrument is the
calorimeter
Calorimetry can be used to find
the ΔH for a chemical reaction,
the heat involved in a physical
change, or the specific heat of a
substance.
Specific Heat Capacity
C= J
or C =
J
(g)oC
(mol)oC
-specific heat capacity of H2O is
4.18 J/ g oC
Energy released as heat =
(mass of solution ) ×
(specific heat capacity) ×
(increase in temp)
q = mCT
J = ( g)
o
(J/g C
) (T)
Example:
A coffee cup calorimeter contains 150 g
H2O at 24.6 oC. A 110 g block of molybdenum is
heated to 100oC and then placed in the water in
the calorimeter. The contents of the calorimeter
come to a temperature of 28.0oC. What is the
heat capacity per g of molybdenum?
q = mCT
q = 150g(4.18J/goC)3.4oC = 2132J
Drawing
2132J = 110g (C)(72oC)
pictures may
oC
C
=
0.27J/g
help to answer
the question.
Example:
4.00g of ammonium nitrate are added to 100.0 mL of water in a
polystyrene cup. The water in the cup is initially at a temperature of
22.5°C and decreases to a temperature of 19.3°C. Determine the heat
of solution of ammonium nitrate in kJ/mol. Assume that the heat
absorbed or released by the calorimeter is negligible.
q = mCT
q = 104g(4.18J/goC)3.2oC = 1391J absorbed = 1.39 kJ
4.00g NH4NO3 x 1 mol NH4NO3 = 0.0500 mol
80.06g NH4NO3
Drawing
pictures may
help to answer
the question.
1.39 kJ/0.0500 = 28 kJ/mol
extensive property - depends on
the amount of substance
intensive property - doesn’t
depend on the amount of
substance
heat of reaction is extensive
temperature is intensive
Hess’s Law -the change in
enthalpy (H) is the same
whether the reaction occurs in
one step or in several steps.
H is not dependent on the
reaction pathway.
The sum of the H for each step
equals the H for the total reaction.
1. If a reaction is reversed, the sign
of H is reversed.
2. If the coefficients in a reaction are
multiplied by an integer, the value of
H is multiplied by the same integer.
N2 + O2  2NO
2NO + O2  2NO2
N2 + 2O2  2NO2
Example:
Given the following reactions and their
respective enthalpy changes, calculate H for the
reaction: 2C + H2 C2H2.
C2H2 + 5/2 O2  2CO2 + H2O
C + O2  CO2
H2 + ½ O2  H2O
2C + 2O2  2 CO2
H2 + ½ O2  H2O
2CO2 + H2O  C2H2 + 5/2 O2
2C + H2 C2H2
 H = 1299.6 kJ/molrxn
H = 393.5 kJ/molrxn
H = 285.9 kJ /molrxn
H = 2(393.5) kJ/molrxn
H = 285.9 kJ/molrxn
 H = +1299.6 kJ/molrxn
 H = 226.7kJ/molrxn
Example:
The heat of combustion of C to CO2 is 393.5
kJ/mol of CO2, whereas that for combustion of CO
to CO2 is 283.0 kJ/mol of CO2. Calculate the heat
of combustion of C to CO.
C + O2  CO2
CO + 1/2O2  CO2
H = 393.5 kJ
H = 283.0 kJ
C + O2  CO2
H = 393.5 kJ
CO2  CO + 1/2 O2 H = +283.0 kJ
C + 1/2O2  CO
H = 110.5 kJ
Standard enthalpy of formation (Hof)
-change in enthalpy that accompanies
the formation of one mole of a
compound from its elements with all
substances in their standard states at
25oC.
Standard States• for gases, pressure is 1 atm
• for a substance in solution, the
concentration is 1 M
• for a pure substance in a condensed state
(liquid or solid), the standard state is the
pure liquid or solid.
• for an element, the standard state is the form
in which the element exists under
conditions of 1 atm and 25oC.
Values of H are found in
Appendix 4
Horeaction = Hfoproducts - Hforeactants
Example:
Consider the reaction:
2ClF3(g) + 2NH3(g)  N2(g) + 6HF(g) + Cl2(g)
 Ho = 1196 kJ/molrxn. Calculate the Hof for
ClF3(g).
Compound
Hof
NH3
-46 kJ/mol
HF
-271 kJ/mol
[0 + 6(271) + 0][2 Hof ClF3 + 2(46)] = 1196 kJ
1626 [2Hof ClF3 92] = 1196 kJ
2Hof ClF3 = 2730 kJ
Hof for ClF3 = 1365 kJ
One version of the First Law of Thermodynamics is
expressed as
∆E = q + w
Which gives the sign convention for this relationship that is
usually used in chemistry?
Heat, q
Added to
the
system
A)
B)
C)
D)

+
+
+
Heat, q
added to the
surroundings
+
+
+

Work, w
done on
the
system

+
+
+
Work, w
done on the
surroundings

+


When one mole of liquid compound X is
vaporized, it is observed that the flexible
container containing X expands.
What is the sign of q and the sign of w?
q
w
A)
+
+
B)
+

C)
−
+
D)
−
−