1. Energy and Chemical Change

2. Some terms Energy is the ability to do work and supply heat
Work is motion against an opposing force
Kinetic energy is the energy an object has because of its motion
For an object of mass m with velocity v
Potential energy (PE) is the energy of position or internal arrangement
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S.A. McFarland©2006

3. The law of conservation of energy states that energy cannot be created or destroyed.
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4. SI unit of energy is the joule (J)
A 2 kg object moving at 1 meter per second has 1 J of kinetic energy
You may also encounter the calorie (cal)
The dietary Calorie (note capital), Cal, is actually 1 kilocalorie
11/29/2018
S.A. McFarland©2006

5. Calories
Children need at least 30 g of fat each day, enough to supply about 1100 kJ of energy when metabolized. How many dietary Calories does this correspond to?
Analysis: this is just a unit conversion problem
1100 kJ x (1 Cal/4.184 kJ) = 260 Cal
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S.A. McFarland©2006

6. 11/29/2018
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7. Thermodynamics energy , pressure, volume, temperature
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy
, pressure, volume, temperature
ΔE = Efinal - Einitial
ΔP = Pfinal - Pinitial
ΔV = Vfinal - Vinitial
ΔT = Tfinal - Tinitial
When the state of a system changes, the magnitude of change in any state function depends only on the initial and final states of the system and not on how the change is accomplished.
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
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S.A. McFarland©2006

8. First law of thermodynamics
energy can be converted from one form to another, but cannot be created or destroyed
ΔEsystem + ΔEsurroundings = 0 or
ΔEsystem = -ΔEsurroundings
C3H8 + 5O CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
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S.A. McFarland©2006

9. Another form of the first law for ΔEsystem
ΔE = q + w
ΔE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PΔV when a gas expands against a constant external pressure
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S.A. McFarland©2006

10. Heat and work are not state functions!
…but remember that q + w is a state function
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11. Work is not a state function!
Work done by the system
w = Fd (mechanical)
w = -P ΔV
ΔV > 0
-PΔV < 0
wsys < 0
P x V = x d3 = Fd = w F d2
Work is not a state function!
Δw = wfinal - winitial
initial
final
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S.A. McFarland©2006

12. W = -0 atm x 3.8 L = 0 L•atm = 0 joules
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?
w = -P ΔV
(a)
ΔV = 5.4 L – 1.6 L = 3.8 L
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
See… work is
NOT a state function!
(b)
ΔV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = L•atm
w = L•atm x
101.3 J
1L•atm
= J
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S.A. McFarland©2006

13. Enthalpy, a new term
Most reactions are run under the constant pressure of the atmosphere and are free to change volume
They may transfer energy as heat and expansion work
This is inconvenient for calculating internal energy changes; in order to do so, we’d have to measure volume changes
Solution: approximate ΔE as enthalpy, ΔH, where ΔH = ΔE + PΔV
ΔH = qp = heat of reaction
ΔH is a good approximation of ΔE when ΔV is tiny, reactions involving only solids and liquids
ΔH is not such a good approximation of ΔE when reactions produce or consume gases, when ΔV is substantial
ΔH is a state function
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14. Enthalpy and the first law of thermodynamics
ΔE = q + w
q = ΔH and w = -PΔV
At constant pressure:
ΔE = ΔH - PΔV
ΔH = ΔE + PΔV
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S.A. McFarland©2006

15. ΔH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
ΔH = H (products) – H (reactants)
ΔH = heat given off or absorbed during a reaction at constant pressure
Heat content. It is equal to the sum of the internal energy of the system plus the product of the pressure-volume work done on the system.
Hproducts < Hreactants
Hproducts > Hreactants
ΔH < 0
ΔH > 0
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S.A. McFarland©2006

16. Thermochemical equations
Is ΔH negative or positive?
System absorbs heat
Endothermic
ΔH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
ΔH = 6.01 kJ
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S.A. McFarland©2006

17. Thermochemical equations
Is ΔH negative or positive?
System gives off heat
Exothermic
ΔH < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
ΔH = kJ
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S.A. McFarland©2006

18. Thermochemical equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
ΔH = 6.01 kJ
If you reverse a reaction, the sign of ΔH changes
H2O (l) H2O (s)
ΔH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
ΔH = 2 x 6.01 = 12.0 kJ
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S.A. McFarland©2006

19. Thermochemical equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
ΔH = 6.01 kJ
H2O (l) H2O (g)
ΔH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) ΔH = kJ
1 mol P4
123.9 g P4 x
3013 kJ
1 mol P4 x
266 g P4
= kJ
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20. 2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) ΔH = -367.5 kJ/mol
Comparison of ΔH and ΔE
2Na (s) + 2H2O (l) NaOH (aq) + H2 (g) ΔH = kJ/mol
ΔE = ΔH - PΔV
At 25 0C, 1 mole H2 = 24.5 L at 1 atm
PΔV = 1 atm x 24.5 L = 2.5 kJ
ΔE = kJ/mol – 2.5 kJ/mol = kJ/mol
Compare to 22.4 L
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21. Heat (q) absorbed or released:
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = msΔt
q = CΔt
Δt = tfinal - tinitial
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S.A. McFarland©2006

22. How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = J/g • 0C
Δt = tfinal – tinitial = 50C – 940C = -890C
q = msΔt
= 869 g x J/g • 0C x –890C
= -34,000 J
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S.A. McFarland©2006

23. Constant-volume calorimetry
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = msDt
qbomb = CbombDt
Reaction at constant V
DH = qrxn
DH ~ qrxn
No heat enters or leaves!
11/29/2018
S.A. McFarland©2006