Transcript Thermochemistry

Thermochemistry
Heat
• energy transferred between two objects as a
result of the temperature difference between
them.
Temperature
• A measure of kinetic energy
1st Law of Thermodynamics
• The energy of the universe is constant.
• i.e. the energy of the universe is conserved
E = Efinal  Einitial
•  E if energy leaves system
• + E if energy enters system
• Note the E of a system doesn’t depend on
how system got there -- i.e. it is a state
function
State Function
• A function or property whose value depends
only on the present state (condition) of the
system, not on the path used to arrive at that
condition.
•E = q + w
Heat gain or loss
Work done = -PV
Matches our earlier
convention that Ein is + and
Eout is –
Enthalpy
• H = qP = E + PV
• H = Hfinal  Hinitial
•
= Hproducts  Hreactants
Quantity of heat supplied
q
Specific Heat Capacity or Specific Heat (C) 
m  T
Tells how much heat is
required to change the
temp of a substance.
Some specific heats are
Al
0.902 J/g oK
Cu
0.385 J/g oK
H2O 4.184 J/g oK
Temperature change
(always Tf-Ti)
• A 55.0 g piece of metal was heated in
boiling water to a temperature of 99.8oC
and dropped into an insulated beaker with
225 mL of water (d = 1.00 g/ml) at 21.0 oC.
The final temperature of the metal and
water is 23.1oC. Calculate the specific heat
of the metal assuming that no heat was lost
to the surroundings.
• Octane, C8H18, a primary constituent of
gasoline, burns in air.
• C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(l)
• Suppose that a 1.00 g sample of octane is
burned in a calorimeter that contains 1.20
kg of water. The temperature of the water
and the bomb rises from 25.00oC to
33.20oC. If the specific heat of the bomb,
Cbomb, is known to be 837 J/oC, calculate the
molar heat of reaction of C8H18.
A quantity of ice at 0oC is added to 90.0 g of
water at 80oC. After the ice melted, the
temperature of the water was 25oC. How
much ice was added?
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specific heat of ice
specific heat of water
specific heat of steam
heat of fusion
heat of vaporization
2.06 J/goC
4.184 J/goC
2.0 J/goC
333 J/g
2226 J/g
0.91 kJ/moloC
7.54 kJ/moloC
0.92 kJ/moloC
6.01 kJ/mol
40.67 kJ/mol
• 50.0 g of ice at -20.0 oC are added to 342.0
g of water at 86.0 oC. What will be the final
temperature of the sample?
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specific heat of ice
specific heat of water
specific heat of steam
heat of fusion
heat of vaporization
2.06 J/goC
4.184 J/goC
2.0 J/goC
333 J/g
2226 J/g
0.91 kJ/moloC
7.54 kJ/moloC
0.92 kJ/moloC
6.01 kJ/mol
40.67 kJ/mol
• A 33.14 g sample of copper and aluminum
was heated to 119.25oC and dropped into a
calorimeter containing 250.0 g of water at
21.00oC. The temperature rose to 23.05oC.
Assuming no heat was lost to the
surroundings, what is the percent copper in
the sample?
Enthalpy
• Enthalpy transferred out of reactants 
exothermic  H = 
• Enthalpy transferred into products 
endothermic  H = +
Enthalpy
• Hforward = Hreverse
(For reversible reactions)
• H2O(g)  H2(g) + 1/2 O2(g)
H = +241.8 kJ
• H2(g) + 1/2 O2(g)  H2O(g)
H = 241.8 kJ
Enthalpy
• The H is proportional to the amount of
substance undergoing change.
• H2O(g)  H2(g) + 1/2 O2(g)
H = +241.8 kJ
• 2 H2O(g)  2 H2(g) + 1 O2(g)
H = +483.6 kJ
Enthalpy
• The physical state of reactants and products
is important.
• H2O(g)  H2(g) + 1/2 O2(g)
H = +241.8 kJ
• H2O(l)  H2(g) + 1/2 O2(g)
H = +285.8 kJ
Enthalpy
• Enthalpy is a state function -- it doesn’t
matter how you go from one place to
another -- enthalpy and enthalpy changes
are the same!!
• The H value is the same no matter how
you get from AB
Hess’s Law
• The overall enthalpy change for a reaction
is equal to the sum of the enthalpy changes
for the individual steps in the reaction.
• Valid because enthalpy is a state function.
Determine the H for the sublimation of ice to
water vapor at 0oC.
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H2O(s)  H2O(l)
H = 6.02 kJ/reaction
H2O(l)  H2O(g)
H = 40.7 kJ/reaction
----------------------------------------------------H2O(s)  H2O(g)
H = 46.7 kJ/reaction
• Calculate the enthalpy change for the
formation of methane, CH4, from solid
carbon (as graphite) and hydrogen gas.
•
C(s) + 2 H2(g)  CH4(g)
• The enthalpies for the combustion of
graphite, hydrogen gas and methane are
given.
• C(s) + O2(g)  CO2(g)
393.5 kJ
• H2(g) + ½ O2(g)  H2O(l)
285.8 kJ
• CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
890.3 kJ
• Calculate the enthalpy change for the
reaction
•
S(s) + O2(g)  SO2(g)
• given
• 2 SO2(g) + O2(g)  2 SO3(g) H = 196 kJ
• 2 S(s) + 3 O2(g)  2 SO3(g) H = 790 kJ
Standard Heat of Formation
• The enthalpy change, Hfo, for the
formation of 1 mol of a substance in the
standard state from the most stable forms of
its constituent elements in their standard
states.
Hf
o
superscript o means standard state
= 25oC and 1 atm pressure
subscript f means formation from most stable elements
o
Hrxn
o
 Hf products
o
 Hf reactants
• Benzene, C6H6, is an important
hydrocarbon. Calculate its enthalpy of
combustion; that is, find the value of Ho
for the following reaction.
• C6H6(l)+15/2 O2(g)  6 CO2(g)+3 H2O(l)
• Given
•
Hfo [C6H6(l)] = +49.0 kJ/mol
•
Hfo [CO2(g)] = 393.5 kJ/mol
•
Hfo [H2O(l)] = 285.8 kJ/mol
• Nitroglycerin is a powerful explosive,
giving four different gases when detonated.
•
2 C3H5(NO3)3(l)  3 N2(g) + ½
O2(g) + 6 CO2(g) + 5 H2O(g)
• Given the enthalpy of formation of
nitroglycerin, Hfo, is 364 kJ/mol,
calculate the energy liberated when 10.0 g
of nitroglycerin is detonated.
Enthalpies from Bond Energies
 Calculate the enthalpy of formation of water
vapor from bond energies.
•
• 2 H2(g) + O2(g)  2 H2O(g)
•
(The experimental value is 241.8kJ/mol)
• Oxygen difluoride, OF2, is a colorless, very
poisonous gas that reacts rapidly and
exothermically with water vapor to produce
O2 and HF. Calculate the Hof for OF2.
• OF2(g) + H2O(g)  2 HF(g) + O2(g)
Horxn = -318 kJ
• The heats of formation for H2O(g) and
HF(g) are -241.8 kJ/mol and -271.1 kJ/mol
respectively.
Stoichiometry using Enthalpy
• Consider the following reaction:
• 2 Na(s) + Cl2(g) 2 NaCl(s) H = 821.8 kJ
• Is the reaction exothermic or endothermic?
• Calculate the amount of heat transferred
when 8.0 g of Na(s) reacts according to this
reaction.
• We generally expect that reactions evolving heat
should proceed spontaneously and those that
absorb heat should require energy to occur.
– Mix barium hydroxide and ammonium chloride
Ba(OH)28H2O(s) + 2 NH4Cl(s)
 BaCl2(aq) + 2 NH3(g) + 10 H2O(l)
Determination of H using
Hess’s Law
 Hrxn is well known for many reactions, but it is
inconvenient to measure Hrxn for every reaction.
 However, we can estimate Hrxn for a reaction of interest
by using Hrxn values that are published for other more
common reactions.
 a series of
 The Standard Enthalpy of Reaction (Hrxn) of
reaction steps are added to lead to reaction of interest
(indirect method).
 Standard conditions (25°C and 1.00 atm pressure).
(STP for gases T= 0°C)
Hess’s Law
“If a reaction is carried
out in a series of steps,
H for the overall
reaction will be equal to
the sum of the enthalpy
changes for the
individual steps.”
- 1840, Germain Henri Hess
(1802–50), Swiss
Calculation of H by Hess’s Law
3 C(graphite) + 4 H2 (g)  C3H8 (g)
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
H= -104
H= +104
H=-1181
H=-1143
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
•
Appropriate set of Equations with their H values are obtained (or given), which containing
chemicals in common with equation whose H is desired.
•
These Equations are all added to give you the desired equation.
•
These Equations may be reversed to give you the desired results (changing the sign of H).
•
You may have to multiply the equations by a factor that makes them balanced in relation to
each other.
•
Elimination of common terms that appear on both sides of the equation .
Calculation of H by Hess’s Law
C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104
3 C(graphite) + 3 O2 (g)  3 CO2 (g) H=-1181
4 H2 (g) + 2 O2 (g)  4 H2O (l)
H=-1143
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Hrxn =
+ 104 kJ
-1181 kJ
- 1143 kJ
- 2220 kJ
Calculation of H by Hess’s Law
Calculate heat of reaction
W + C (graphite)  WC (s)
Given data:
2 W(s) + 3 O2 (g)  2 WO3 (s)
½(2 W(s) + 3 O2 (g)  2 WO3 (s) )
W(s) + 3/2 O2 (g)  WO3 (s) )
ΔH = ?
ΔH = -1680.6 kJ
½(ΔH = -1680.6 kJ)
ΔH = -840.3 kJ
C (graphite) + O2 (g)  CO2 (g)
C (graphite) + O2 (g)  CO2 (g)
ΔH = -393.5 kJ
ΔH = -393.5 kJ
2 WC (s) + 5 O2 (g)  2 WO3 (s) + CO2 (g)
ΔH = -2391.6 kJ
½(2 WO3 (s) + CO2 (g)  2 WC (s) + 5 O2 (g))
WO3 (s) + CO2 (g)  WC (s) + 5/2 O2 (g)
W + C (graphite)  WC (s)
½ (ΔH = +2391.6 kJ)
ΔH = + 1195.8 kJ)
ΔH = - 38.0
Hess’s Law
Problem: Chloroform, CHCl3, is formed by the following reaction:
Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g)
Determine the enthalpy change for this reaction (ΔH°rxn), using the
following:
2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g) ΔH°f = – 103.1 kJ/mol
CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g)
ΔH°rxn = – 890.4 kJ/mol
2 HCl (g) → H2 (g) + Cl2 (g)
ΔH°rxn = + 184.6 kJ/mol
C (graphite) + O2 (g) → CO2(g)
ΔH°rxn = – 393.5 kJ/mol
H2 (g) + ½ O2 (g) → H2O (l)
ΔH°rxn = – 285.8 kJ/mol
answers: a) –103.1 kJ b) + 145.4 kJ
c) – 145.4 kJ
305.2 kJ
e) – 305.2 kJ f) +103.1 kJ
d) +
Methods of determining H
1. Calorimetry (experimental)
2. Hess’s Law: using Standard
Enthalpy of Reaction (Hrxn) of a
series of reaction steps (indirect
method).
3. Standard Enthalpy of Formation

(Hf ) used with Hess’s
Law (direct
method)
4. Bond Energies used with Hess’s
Law
Experimental
data combined
with
theoretical
concepts
(3) Determination of H using
Standard Enthalpies of Formation (Hf )
Enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in
which a compound is made from its constituent elements in their elemental forms.
C + O2  CO2
∆Hf= -393.5 kJ/ 
 measured under standard conditions
Standard Enthalpy of formation Hf are
(25°C and 1.00 atm pressure).
Calculation of H
CH4(g) + O2(g)  CO2(g) + H2O(g)
ΔHf = -74.8 kJ/ŋ
ΔHf = -393.5 kJ/ŋ
ΔHf = -241.8 kJ/ŋ
C + 2H2(g)  CH4(g)
C(g) + O2(g)  CO2(g)
2H2(g) + O2(g)  2H2O(g)
We can use Hess’s law in this way:


H = nHf(products)
- mHf(reactants)
where n and m are the stoichiometric coefficients.
n CO2(g) + n H2O(g)
-
n CH4(g) + n O2(g)
H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)]
= - 560.5 kJ
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = nHf(products) - mHf(reactants)
H
= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) - (-103.85 kJ)
= -2219.9 kJ
Table of Standard Enthalpy of formation, Hf
(4) Determination of H
using Bond Energies
• Most simply, the strength of a bond is measured
by determining how much energy is required to
break the bond.
• This is the bond enthalpy.
• The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is measured to be 242 kJ/mol.
Average Bond Enthalpies (H)
• Average bond enthalpies are positive, because bond
breaking is an endothermic process.
NOTE: These are
average bond
enthalpies, not
absolute bond
enthalpies; the
C—H bonds in
methane, CH4,
will be a bit
different than the
C—H bond in
chloroform,
CHCl3.
Enthalpies of Reaction (H )
• Yet another way to
estimate H for a reaction
is to compare the bond
enthalpies of bonds
broken to the bond
enthalpies of the new
bonds formed.
• In other words,
Hrxn = (bond enthalpies of bonds broken) 
(bond enthalpies of bonds formed)
Hess’s Law:
Hrxn = (bonds broken)  (bonds formed)
CH4(g) +
Cl2(g)  CH3Cl(g) +
HCl(g)
Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)]
= (655 kJ)  (759 kJ)
= 104 kJ
Bond Enthalpy and Bond Length
• We can also measure an average bond length for
different bond types.
• As the number of bonds between two atoms
increases, the bond length decreases.
2003 B Q3
2005 B
2002
Entropy
• The amount of randomness, or molecular
disorder, in a system.
• S = more positive to indicate greater
disorder.
Predict which has greater entropy
• O2(g) at 5 atm of O2 at 0.5 atm
• Br2(l) or Br2(g)
• 1 mol N2 (g) in 22.4 L or 1 mol N2(g) in
2.24 L
• CO2(g) or CO2(aq)
Predict entropy changes for
• freezing of one mole of water
• evaporation of 1 mol of Br2
• precipitation of BaSO4 upon mixing of
aqueous solutions of Ba(NO3)2 and H2SO4
• 2 C(s) + O2(g)  2 CO(g)
• 2 K(s) + Br2(l)  2 KBr(s)
• 2 MnO2(s)  2 MnO(s) + O2(g)
• O(g) + O2(g)  O3(g)
• + S is entropy favored
•  S is entropy disfavored
•  H is enthalpy favored
• + H is enthalpy disfavored
Gibbs Free Energy, G
• Determines whether a reaction is
spontaneous and at what temperature it
becomes spontaneous.
• Spontaneous -- A process that proceeds on
its own with out any continuous external
influence.
Energy Units
• 1 calorie = 4.184 J
• 1 food calorie = 1 Cal = 1 kcal = 1000 cal
• Given the reaction below for the combustion
of glucose to form carbon dioxide and water,
calculate the Calories/g for carbohydrates.
• C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
Hrxn = -2801.6 kJ
• M & M candies consist of 70%
carbohydrates, 21% fat, and 4.6% protein as
well as other ingredients that do not have
caloric value. What quantity of energy is
generated if 47.9 g of M&Ms (1 small
package) were burned in a bomb
calorimeter? How long will a I need to
walk to use up the value of the M&Ms if 1
hour of walking uses up 400 Cal?
• 4 Cal/g carbs
• 4 Cal/g protein
• 9 Cal/g fat
G = H  TS
• If H = + and S =  never spontaneous
G = +
• If H =  and S = + always spontaneous
G = 
• If H = + and S = + or if H =  and S = 
temperature determines spontaneity
• At T where G =  reaction is spontaneous
• At T where G = + reaction is nonspontaneous
• Ca(s) + Cl2(g)  CaCl2(s)
H = 59.8 kJ
S = 273J/K
• G = H  TS
spontaneous at low T nonspontaneous at high T, entropy takes precedence
• Reaction becomes spontaneous at temperature
where G becomes zero -- or when G = zero
reaction is spontaneous in neither direction -equilibrium!
• 0 = 59.8 kJ  T(0.273kJ/K)
• T = 219K or 53oC
• Reaction is spontaneous below 53oC
• NH3(g) + 2 O2(g)  HNO3(aq) + H2O(l)
H = 413 kJ S = 386J/K
• 0 = 413kJ  T(0.386J/K)
• T = 1069K = 796oC
• Reaction is spontaneous below 796oC
• C6H12O6(s)  2 C2H5OH(l) + 2 CO2(g)
H = 70 kJ S = +780J/K
• 0 = 70kJ  T(+0.78J/K)  T = 90K
• spontaneous at all temperatures -- would
need an impossible temperature to become
non-spontaneous!!
• C6H12(l) + 6 O2(g)  3 CO2(g) + 6 H2O(g)
H = 
S = +
• 6 CO2(g)+6 H2O(g) C6H12O6(s)+9 O2(g)
H = +
S = 