2.
Energy is...
• The ability to do work.
• Conserved.
• Work is done when energy is transferred
– At macroscopic level work is done as force
acting over a distance to move an object
– At microscopic level work is done on the
particles as HEAT
• a state function.
– independent of the path, or how you get from
point A to B.
3.
Energy
• Kinetic Energy MOTION
» Thermal  Heat
» Mechanical
» Electrical
» Sound
• Potential Energy  POSITION
» Chemical
» Gravitational
» Electrostatic
4.
K.E. kinetic energy motion
• Magnitude of K.E. or EKof an object is
dependent on its mass(m) and velocity (v)
• EK = 1/2mv2
• Mass and velocity determine how
much work an object can do
5.
Potential Energy
• Potential energy: stored energy as a result
of the energy an object possesses by its
position relative to other objects and some
force acting between
• Charged particles have P.E. due to attractive
and repulsive forces (electrostatic forces)
between them
6.
The Universe
• is divided into two halves.
• the system and the surroundings.
• The system is the part you are concerned with.
• The surroundings are the rest.
q into system = q from surroundings
7.
State Function
• A system has a fixed value of energy for a
given set of conditions
• Value independent of pathway
• Value depends only on present condition
not how it got there
• Example – PE of bowling ball at top of
stadium
8.
Energy is transferred as ...
• Work is a force acting over a distance.
• Heat is energy transferred between objects
because of temperature difference.
9.
First Law of Thermodynamics
Energy of the Universe is Constant
E = q + w
q = heat. Transferred between two bodies due to
difference in temperature (q = mCΔT)
w = work. Force acting over a distance (F x d)
11.
Energy Exchange
• A system can exchange energy with its
surroundings through work or heat
• Internal energy = heat energy + work done
ΔE = q + w
• q is pos. gains heat from surroundings q>0
• w pos. work done on system by surroundings w>0
• q neg. value heat lost to surroundings q<0
• w neg. work done on surroundings w<0
12.
Change internal energy = difference between
internal energy of system at the completion of
a process and that of the beginning
∆E = Efinal – Einitial ∆E= Ef – Ei
• However, it is impossible to know the
internal energy at the beginning and the end
• So, we may only determine the change in
internal energy by measuring energy
transferred as heat and work
16.
Work
• Gas pushes back the
surrounding
atmosphere
• Does work on its
surroundings
• Exerts a force over a
distance to push up the
piston
h
17.
Expansion work
• Work = force x distance
• Pressure = Force /Area
• Δ Volume = Area x height
F
18.
work = force × distance (h)
since pressure = force / area,
and volume = area x distance (h)
work = pressure × volume
Force (area x height)
area
wsystem = −P∆V
Work
20.
The piston, moving a
distance ∆h against a
pressure P, does work
on the surroundings.
Since the volume of a
cylinder is the area of
the base times its
height, the change in
volume of the gas is
given by ∆h x A = V.
21.
Bomb CalorimetryBomb Calorimetry
(ConstantVolume Calorimetry)(ConstantVolume Calorimetry)
CalorimetryCalorimetry
No Δ in volume
No work done
Energy
transfer
only as heat
E (internal energy)
can be measured in
bomb calorimeter,
since no energy is
transferred as
work. Measuring
the heat transfer
alone is enough.
22.
Constant Volume Calorimetry
• Constant Volume
• NO MOVEMENT – NO DISTANCE
• NO WORK
• So the ONLY energy transfer is as heat.
• Thus measuring the heat transfer is a
measure of change in internal energy
23.
ConstantPressure CalorimetryConstantPressure Calorimetry
CalorimetryCalorimetry
Enthalpy is measured
at constant pressure.
The work done by the
system will be ignored.
24.
Constant Pressure Calorimetry
• The change in volume is ignored
• Thus the work done is ignored
• Only heat transfer will be measured
• The change in enthalpy is defined as the
heat transferred at constant pressure thus
ignores work done
25.
ConstantPressure CalorimetryConstantPressure Calorimetry
Atmospheric pressure is constant!
∆H = qP
qsystem = qsurroundings
 The surroundings are composed of the water in the
calorimeter and the calorimeter.
 For most calculations, the qcalorimeter can be ignored.
qsystem =  qwater
csystemmsystem ∆Tsystem =  cwatermwater ∆Twater
CalorimetryCalorimetry
26.
NH4NO3(s) NH4
+
(aq) + NO3

(aq)
∆Twater = 16.9o
C – 22.0o
C = 5.1o
C
mwater = 60.0g
cwater = 4.184J/go
C
msample = 4.25g
Now calculate ΔH in kJ/mol
qsample = qwater
qsample = cwatermwater ∆Twater
qsample = (4.184J/go
C)(60.0g)(5.1o
C)
qsample = 1280.3J
CalorimetryCalorimetry
27.
Problem: Heats of Chemical
Reaction
100 ml solutions of 1.00 M NaCl and 1.00
M AgNO3 at 22.4 o
C are mixed in coffee cup
calorimeter and the resulting temperature
rises to 30.2 o
C. What is the heat per mole
of product? Assume the solution density
and specific heat are the same as pure
water.
28.
Problem: Heats of Chemical
Reaction
• 100 ml solutions of 1.00 M NaCl and 1.00 M AgNO3 at 22.4 o
C are
mixed in coffee cup calorimeter and the resulting temperature rises to
30.2 o
C. What is the heat per mole of product? Assume the solution
density and specific heat are the same as pure water.
• Write balanced chemical reaction:
• Net ionic: Ag+(aq) + Cl(aq) AgCl(s)→
• Determine heat of reaction:
• qrxn= qcal = m×c× T∆
• m = 200 ml × 1.0g/ml = 200g
• c = cH2O = 4.18 J/goC
= 200g × 4.18 J/go
C × (30.222.4)
• = 6,520 J
• Determine heat per mole of product:
• stoichiometric reactants, 0.1 mol in 100 ml
• qrxn/mol = 6.52 kJ/0.1 mol
• = 65.2 kJ/mol
29.
Practice calorimetry calculation
When 50.0 mL of 0.25 M Ba(NO3)2 solution at 25.1°
C are
mixed with 50.0 mL of 0.25 M Na2SO4 at 25.1°
C in a
calorimeter, the white solid barium sulfate forms and the
temperature increases to 26.6°
C. Assuming the calorimeter
absorbs only minimal heat and that the specific heat and
density of the solution are equivalent to water, calculate
the enthalpy per mole of barium sulfate.
• Write the balanced and net ionic equation.
• Show the work to solve for the q rxn
• Show the work to find the molar enthalpy
30.
Definition of Enthalpy
• Thermodynamic Definition of Enthalpy
(H):
H = E + PV
E = energy of the system
P = pressure of the system
V = volume of the system
31.
Changes in Enthalpy
• Consider the following expression for a chemical
process:
∆H = Hproducts  Hreactants
If ∆H >0, then qp >0. The reaction is endothermic
If ∆H <0, then qp <0. The reaction is exothermic
32.
Thermodynamic State Functions
• Thermodynamic State Functions:
Thermodynamic properties that are dependent on
the state of the system only. (Example: ∆E and
∆H)
• Other variables will be dependent on pathway
(Example: q and w). These are NOT state
functions. The pathway from one state to the other
must be defined.
33.
Hess’ Law Defined
• Enthalpy is a state function. As such, ∆H for
going from some initial state to some final state is
pathway independent.
• Hess’ Law: ∆H for a process involving the
transformation of reactants into products is not
dependent on pathway. Therefore, we can pick
any pathway to calculate ∆H for a reaction.
34.
Hess’ LawHess’ Law : if a reaction is carried out in a
number of steps, ∆H for the overall reaction is
the sum of ∆H for each individual step.
Note that:
∆H1 = ∆H2 + ∆H3
35.
Calculations via Hess’s Law
1. If a reaction is reversed, ∆H is also reversed.
N2(g) + O2(g) → 2NO(g) ∆H = 180 kJ
2NO(g) → N2(g) + O2(g) ∆H = −180 kJ
2. If the coefficients of a reaction are multiplied by
an integer, ∆H is multiplied by that same integer.
3[2NO(g) → N2(g) + O2(g)] ∆H = 3[−180 kJ]
6NO(g) → 3N2(g) + 3O2(g) ∆H = −540 kJ
36.
• A consequence of Hess’ Law is that
reactions can be added just like algebraic
equations.
• For example:
Hess’ LawHess’ Law
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = 802 kJ
2H2O(g) → 2H2O(l) ∆H = 88 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = 890 kJ
37.
Hess’s Law
Hess‘s Law is particularly useful for calculating ∆fHo
which would not
be easy to measure experimentally. ∆Ho
f for CO cannot be measured
as CO2 is also formed when graphite is burned
C(s) + 1/2O2 CO ∆Ho
f = x
Standard heats of formation are the enthalpy change associated with forming one
mole of the compound from its elements in their standard states.
Rearrange the equations to calculate the for ∆Ho
f CO
CO + 1/2O2 CO2 ∆rxnHo
= 283 kJmol1
C(s) + O2 CO2 ∆fHo
= 393.5 kJmol1
38.
Rearrange the equations to calculate the for ∆Hof CO
CO + 1/2O2 CO2 ∆Ho
= 283 kJmol1
C(s) + O2 CO2 ∆Ho
= 393.5 kJmol1
39.
Using Hess’ Law
• When calculating ∆H for
a chemical reaction as a
single step, we can use
combinations of
reactions as “pathways”
to determine ∆H for our
“single step” reaction.
2NO2 (g)
N2 (g) + 2O2 (g)
q
2NO2 (g)N2 (g) + 2O2 (g)
41.
1st
Example (cont.)
• If we take the previous two reactions and add
them, we get the original reaction of interest:
N2 (g) + O2 (g) 2NO(g) ∆H = 180 kJ
2NO (g) + O2 (g) 2NO2(g) ∆H = 112 kJ
N2 (g) + 2O2 (g) 2NO2(g) ∆H = 68 kJ
42.
1st
Example (cont.)
• Our reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ
• This reaction can also be carried out in two
steps:
N2 (g) + O2 (g) 2NO(g) ∆H = 180 kJ
2NO (g) + O2 (g) 2NO2(g) ∆H = 112 kJ
43.
1st
Example (cont.)
• Note the important things about this example, the
sum of ∆H for the two reaction steps is equal to
the ∆H for the reaction of interest.
• We can combine reactions of known ∆H to
determine the ∆H for the “combined” reaction.
44.
Hess’ Law: Details
• One can always reverse the direction of a
reaction when making a combined reaction.
When you do this, the sign of ∆H changes.
N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ
2NO2(g) N2(g) + 2O2(g) ∆H = 68 kJ
45.
Details (cont.)
• The magnitude of ∆H is directly proportional to
the quantities involved (it is an “extensive”
quantity).
• As such, if the coefficients of a reaction are
multiplied by a constant, the value of ∆H is also
multiplied by the same integer.
N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ
2N2(g) + 4O2(g) 4NO2(g) ∆H = 136 kJ
46.
Using Hess’ Law
• When trying to combine reactions to form a
reaction of interest, one usually works backwards
from the reaction of interest.
• 2nd
Example:
What is ∆H for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)
C(gr) indicates that it is carbon in the form of graphic
rather than diamond.
47.
2nd
Example (cont.)
3C (gr) + 4H2 (g) C3H8 (g) ∆H = ?
• You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) ∆H = 394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) ∆H = 2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) ∆H = 286 kJ
48.
Example (cont.)
• Step 1. Only reaction 1 has C (gr).
Therefore, we will multiply by 3 to get the
correct amount of C (gr) with respect to our
final equation.
3[C (gr) + O2 (g) CO2 (g)] 3[ ∆H = 394 kJ]
3C (gr) + 3O2 (g) 3CO2 (g) ∆H = 1182 kJ
Given:
After manipulating:
49.
Example (cont.)
• Step 2. To get C3H8 on the product side of
the reaction, we need to reverse reaction 2.
3CO2 (g) +4H2O (l) C3H8 (g)+5O2 (g) [∆H =  2220 kJ]
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) ∆H = 2220 kJ
Given:
After manipulating:reverse
50.
Example (cont.)
• Step 3: Add two “new” reactions together
to see what is left:
3C (gr) + 3O2 (g) 3CO2 (g) ∆H = 1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) ∆H = +2220 kJ
2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ
51.
Example (cont.)
• Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ
H2 (g) + 1/2O2 (g) H2O (l) ∆H = 286 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4
52.
Example (cont.)
• Step 4: Compare previous reaction to final
reaction, and determine how to reach final
reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 ∆H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) ∆H = 1144 kJ
3C (gr) + 4H2 (g) C3H8 (g)
55.
Calculate the change in enthalpy for the reaction.
56.
Practice Calculate ∆H for :
• 2C(s) + H2(g) C2H2(g) given the following:
• Correct moles and divide so function of coefficients moles for wanted product is one
• C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(l) ∆H = 1299.6kJ
• C(s) + O2(g) CO2(g) ∆H = 393.5kJ
• H2(g) + 1/2O2(g) H2O(l) ∆H = 285.9kJ
• Rearrange the equations so similar to algebra:
58.
Another Example
• Calculate ∆H for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) ∆H = 176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) ∆H = 92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) ∆H = 629 kJ
59.
Changes in Enthalpy
• Consider the following expression for a chemical
process:
∆H = Hproducts  Hreactants
If ∆H >0, then qp >0. The reaction is endothermic
If ∆H <0, then qp <0. The reaction is exothermic
60.
Standard Heat of Formation (ΔHf
°
)
• Standard heat of formation is the enthalpy change
associated with forming one mole of compound
from its elements in their standard states
• Standard heats of formations are given in a table
of thermodynamic values
• Finding the sum of the heat formations for
products minus reactants is an application of
Hess’s Law ΔH°
= Σ ΔHf
°
products Σ ΔHf
°
reactants
61.
Standard States
Compound
For a gas, pressure is exactly 1 atmosphere.
For a solution, concentration is exactly 1 molar.
Pure substance (liquid or solid), it is the pure liquid or
solid.
Element
The form [N2(g), K(s)] in which it exists at 1 atm and
25°C.
62.
Changes in Enthalpy
• Consider the following expression for a chemical
process:
∆H = Hproducts  Hreactants
If ∆H >0, then qp >0. The reaction is endothermic
If ∆H <0, then qp <0. The reaction is exothermic
63.
Using the equation
ΔH°
= Σ ΔHf
°
products  Σ ΔHf
°
reactants
We will calculate the theoretical ΔH
for the reaction
of hydrochloric acid and sodium hydroxide.
64.
Experimental ΔH
Heat of reaction at constant pressure!
Use a “coffeecup” calorimeter
to measure it
Practice:
When 50mL of 1M HCl is
mixed with 50mL of 1M NaOH
in a coffeecup calorimeter, the
temperature increases from
21o
C to 27.5o
C.
What is the enthalpy change, if
the density and specific heat of
the solution are assumed to be
the same as water?