Download Thermochemistry

Thermochemistry
Chapter 6!
By James Lauer and David Miron
6.1 The Nature of Energy
Key Terms
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First law of thermodynamics - The energy of the universe is constant
Energy - the capacity to do work or to produce heat
Law of Conservation of Energy - energy can be converted from one form to another but can be neither created nor
destroyed
Potential vs Kinetic energy - energy due to position or composition and energy due to motion (KE=1/2mv2)
Heat - the transfer of energy between two objects due to temperature difference
Work - force acting over a distance
Pathway - can affect energy transfer but the total energy remains constant
State Function/Property - property dependent on present state of object
System - the part of the universe on which we wish to focus on
Surroundings - everything else in the universe
Exo- vs Endothermic - energy flows out of the system (combustion) and heat flows into the system
Thermodynamics - study of energy and its interconversions
Internal Energy (E) - the sum of kinetic and potential energy of all particle in the system (∆E = q + w)
6.1 Energy Flow
In the final photo ball A’s potential energy has _______ and
ball B’s potential energy has ______
Ball B is not as high (in the final photo) as Ball A was in the
initial photo. Why?
Where has some of the energy gone? What did it form?
6.1 Equations
KE=1/2mv2
∆E = q + w q=heat w=work ∆E=change in energy
∆E>0 endo/absorbed
∆E<0 exo, energy released
+q=endo, flow in
-q=exo, flow out
+w=endo, work done on
-w=exo, work done by
Work = force X distance ---> w= -P∆V V=volume P=external pressure
6.1 Example
Calculate ∆E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and
where 1.4 kJ of work is done on the system. (∆E = q + w)
∆E = 15.6 kJ + 1.4 kJ = 17.0 kJ
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external
pressure of 15 atm. (w = -P∆V)
w = -15 atm X 18 L = -270 L x atm
6.2 Enthalpy and Calorimetry
Key Terms
Enthalpy(H) - H=E+PV state function
Calorimeter - device used to determine heat associated with a chemical reaction
Heat Capacity(C) - C=heat absorbed/increase in temperature
Specific/molar heat capacity - heat capacity per gram(J/°C x g or J/K x g), heat capacity per mole(J/°C x mol or J/K x
mol)
Constant Pressure Calorimetry - basic calorimeter, to determine changes in enthalpy for reactions in solutions
Constant Volume Calorimetry - rigid container (bomb) and reactants are ignited inside
6.2 Equations
H = E + PV
H= enthalpy E= internal energy of system PV= pressure x volume
• Chemical reaction ∆H = Hproducts - Hreactants
• Constant pressure ∆H = qp qp= heat at constant pressure
Energy released by reaction = s x m x ∆T (specific heat capacity, mass of solution, increase in temperature)
• Constant volume ∆E = q + w = qv
Energy released by reaction = temperature increase x heat capacity of calorimeter
6.2
Constant Pressure
Constant Volume
6.2 Example
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution
at 25.0°C in a calorimeter, the white solid BaSO4 forms, and the temperature of the mixture
increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that
the specific heat capacity of the solution 4.18 J/°C x g, and that the density of the final solution is
1.0 g/mL, calculate the enthalpy change per mole of BaSO4. (∆H=specific heat x mass of solution
x increase in temperature)
Net ionic equation: Ba2+(aq) + SO42-(aq) ---> BaSO4(s)
Mass of solution: 2L=2000ml=2000g
Temperature increase: 28.1-25.0= 3.1°C
4.18 x 2000 x 3.1= 2.6 x 104 J
∆H = -26 kJ/mol
all units cancel except J 1000J = 1 kJ
The reaction is exothermic.
6.3 Hess’ Law
Key terms:
Hess’ Law- states that in going from a particular set of reactants to a particular set of products, the change in enthalpy
is the same whether the reaction takes place in one step or a series of steps.
Hess’ Law allows scientists to determine the enthalpy of formation in a reaction they are unable to complete using a
series of reactions containing the same reactants and products that are within the main reaction.
Enthalpy of synthesis = - Enthalpy of Decomposition
6.3 Hess’ Law example
What is the value for ΔH for the
following reaction?
CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g)
Given:
C(s) + O2(g) → CO2(g); ΔHf = -393.5
kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8
kJ/mol
C(s) + 2 S(s) → CS2(l); ΔHf = 87.9
kJ/mol
Using this info we can infer from hess’
law
CO2(g)--> C(s) + O2(g)→→→; ΔHf =
393.5 kJ/mol
SO2(g)--> S(s) + O2(g)→; ΔHf =
296.8 kJ/mol
CS2(l)--> C(s) + 2 S(s); ΔHf = -87.9
kJ/mol
6.3 Hess’ Law example
What is the value for ΔH for the
following reaction?
CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g)
Given:
C(s) + O2(g) → CO2(g); ΔHf = -393.5
kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8
kJ/mol
C(s) + 2 S(s) → CS2(l); ΔHf = 87.9
kJ/mol
Using this info we can infer from hess’
law
CO2(g)--> C(s) + O2(g)→→→; ΔHf =
393.5 kJ/mol
SO2(g)--> S(s) + O2(g)→; ΔHf =
296.8 kJ/mol
CS2(l)--> C(s) + 2 S(s); ΔHf = -87.9
kJ/mol
Focus on compounds
CS2(l)--> C(s) + 2 S(s); ΔHf = -87.9 kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol
C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol
CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g); ΔHf = -1075.0
kJ/mol
6.4 Standard Enthalpies of
Formation
Key Terms
Standard enthalpy of formation (Δ Hfo ) - the change in enthalpy that accompanies the formation of one mole of a
compound from its elements with all substances in their standard states.
Degree symbol (o) - indicates that the process is carried out under standard conditions
Standard State - a precisely defined reference state.
Compound:
Gas: 1 atm
Liquid or solid: Pure liquid or solid
Solution: 1 M
Element: 1 atm and 25oC
6.4
Things to wrap your head around:
You cannot measure absolute absolute values for enthalpy...you can only determine changes in
enthalpy (hence Δ )
Because Enthalpy is a state function, you can use Hess’ Law and manipulate it so you can use it.
so
Synthesis = -Decomposition etc.
6.4 Equations
Δ Horeaction = ΣnpΔHof(products) - ΣnpΔHof(reactants)
Public Relations
P-R
6.4 examples
Calculate ΔH for the following reaction:
ΔH = 4 (-1669.8) - 3(-1120.9) = -3316.5 kJ/mol
8 Al(s) + 3 Fe3O4(s) --> 4 Al2O3(s) + 9 Fe(s)
Given
ΔHf Al2O3(s) = -1669.8 kJ/mol
ΔHf Fe3O4(s) = -1120.9 kJ/mol
First ignore the elements, because they are formed, and
you don’t need them to find the change in enthalpy
3 Fe3O4(s) --> 4 Al2O3(s)
Public Relations… P-R so
ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s)