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Thermodynamics
Terms/ Definitions• Thermodynamics
– Deals with the interconversion of heat an other forms of energy
• Thermochemistry– Deals with heat change in chemical reactions
• State Function– Function that depends only on the conditions (state) not on how the state was obtained
• Energy(E)– Internal energy = kinetic + potential energy
– Kinetic energy comes from molecular motion, electron motion etc.
– Potential energy comes from attractive and repulsive forces in nuclei, and interactions between molecules
• Enthalpy(H)– H = E + PV– Extensive property– State function– Can only measure the difference in enthalpy between two states
• Heat (Q,q)– Transfer of thermal energy between two bodies at different temperatures
• Work (W,w)– Form of energy, can be mechanical or non mechanical
– Mechanical work is normally pressure - volume work
– W = -PexV– Work done by the system on the surrounding = negative
– Work done by the surroundings on the system = positive
• System– Specific thing we are looking at
• Surroundings– Everything outside the system
• Universe– System + surroundings
• Open system– Can exchange both matter and energy with the surroundings
• Closed system– Can exchange energy but not matter with the surroundings
• Isolated system – No exchange of matter or energy with the surroundings
– Q = ?– Q = 0
• Thermochemical Equation– Chemical equation that includes the enthalpy change
– e.g. H2O(s) --> H2O(l) H = 6.01 kJ
• Heat flow– Endothermic, Q = +, heat absorbed by the system
– Exothermic, Q = -, heat given off by the system
• Units of energy– Joules(J), kilojoules(kJ)– Calories(cal), kilocalories(kcal)– 1 cal = 4.184 J– (Liter)(atmosphere) 1 L atm = 101.3 J
• Enthalpy– State function– Heat energy of state– Look at change between states H
• Extensive Property - depends on the amount
1 mole of the reaction as written
12
N2(g) 12
O2(g) NO(g) H90.4 kJ/mol
N2(g) O2(g) 2NO(g)
H?
2(90.4 kJ/mol)=180.8 kJ/mol
• Changes sign as reverse reaction:
• Follows Hess’ Law
12
N2(g) 12
O2(g) NO(g) H90.4 kJ/mol
NO(g) 12
N2(g) 12
O2(g) H?
90.4 kJ/mol
Hess’ LawIf a reaction can be considered to go by a series of steps, H of the reaction is the sum of H of the steps.
Reaction kJ/mol
1/2 N2(g) + 1/2 O2(g) NO(g) +90.4
NO2(g) NO(g) + 1/2 O2(g) +56.5
What is the value of H for the reaction:1/2 N2(g) + O2(g) NO2(g)
NO(g) + 1/2 O2(g) NO2(g) -56.5
1/2 N2(g) + 1/2 O2(g) NO(g) +90.4
1/2 N2(g) + O2(g) NO2(g)+33.9
Standard StatesMost stable form at 1 atmosphere and usually at 25oC
Indicate standard state by Ho
AllotropesMore than one stable form at 1 atmosphere
O2 and O3C(s) graphite and C(s) diamond
Standard heat of formation (Standard enthalpy of formation)
Heat change that occurs when 1 mole of a single product is formed from elements in their standard states.
C(s) graph + O2(g) CO2(g) -393.5 kJ
Hfo
HfoCO2 = -393.5 kJ/mol
e.g. C(s)graphiteH2(g) N2(g) Br2(l)
Fe(s) S(s) Hg(l) I2(s)
Write the formation reaction for the following:
NH4NO3(s)
benzene
aniline
N2(g) + H2(g) + O2(g)NH4NO3(s)2 3/2
C6H6(l)
C6H6(l)C(s)graphite + H2(g)6 3
C6H5NH2(l)
C6H5NH2(l) C(s)graphite + H2(g) + N2(g)6 7/2 1/2
Value of standard state of elements:
By definition, Ho formation of an element in its standard state = 0.
Which of the following will NOT have H formation = 0
H2(g) Ne(g) Cl2(g) I2(l)
Hg(s) Br2(l) Ca(s) Fe(l)
Calculate Heat of a reaction:
Hrxno
2. Heat of reaction can be calculated using Hess’ Law
1. Heat of reaction can be calculated from Heat of Formation data
Using Heat of formation dataIf data for all the reactants and products is given as , then is equal to the sum of of the products minus the sum of of the reactants.
Hfo
Hfo
Hfo
Hrxno
Hrxno Hf
o(products) Hfo(reactants)
1. Compound Ho
f kcal/mol C2H6 (g) - 20.2
CO2 (g) - 90.1 H2O (l) - 68.3
Calculate Horxn for the reaction:
C2H6(g) + O2(g) CO2(g) + H2O(l) 2 37/2
Hrxno Hf
o(products) Hfo(reactants)
Hrxno {2Hf
o(CO2(g))3Hfo(H2O(l))} {Hf
o(C2H6(g))+(72Hf
o(O2(g) )}
= {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (?)}= {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (0)}
= -364.9 kcal/ mol
Using Hess’ LawIf all information is not given as Hf, then need to go a different way
Use Hess’ LawGiven: Reaction Ho
(kJ/mol)
C(s)graph O2(g) CO2(g) 393.5
H2(g) 12
O2(g) H2O(l) 285.8
H2(g) 12
O2(g) H2O(g) 241.8
2C2H2(g) 5O2(g) 4CO2(g)+2H2O(l) 2598.8
Find H for the formation of C2H2 1. Write a target reaction: --> C2H2(g)2C(s) graph + H2(g)
2. Rearrange reactions so that when added together you get the target
4CO2(g)+2H2O(l) 2C2H2(g) 5O2(g) 2598.8
4C(s)graph 4O2(g) 4CO2(g) 4( 393.5)
2H2(g) O2(g) 2H2O(l) 2( 285.8)
--> 2C2H2(g)4C(s) graph + 2H2(g) +453.2
--> C2H2(g)2C(s) graph + H2(g) +453.2/2 = 226.6 kJ
Which of the following would have = 0?
H fo
A) C(s) diamondB) Hg(g)C) Xe(g)D) Br2(g)E) Cl - (aq)
Energy(E) - Internal Energy (KE + PE)Kinetic energy: comes from the
molecular motion and the electronic motion
Potential Energy: comes from attractive and repulsive forces in nuclei and from interactions between
molecules.First Law:Energy can be converted from one form to another, it can not be created or destroyed. i.e. The energy of the universe is constant.
Euniverse = Esystem + Esurroundingss = 0
Esystem = - Esurroundingss
All energies do not have to be the same form
Since we are primarily interested in what happens to a chemical system, we use the form:
E = q + w Q = heat
W = work
Sign convention
- +Heat(q) from system to from surroundings
surroundings to system (exothermic) (endothermic)
Work(w) by the system on by surroundingsthe surroundings on the system
Work includes all kinds of work, both mechanical and non-mechanical. We limit to mechanical work at this time.
Mechanical work:W = -PexV = - PV at constant P A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done?
A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work(in J) is done?
A) 10.0 JB) - 10.0 JC) 1.01 x 103 JD) 20.0 JE) - 1.01 x 103 J
A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done?
= - (5.0 atm)(4.0 L - 2.0 L) = - 10.0 L atm
1 L atm = 101.3 J
= (- 10.0 L atm) (101.3 J/ Latm) = - 1013 J = - 1.01 x 103 J
W = - PV
Relationship between E and q
E = q + w
w = - PexV E = q - PexV At constant volume DV = ?= 0
E = qv
Relationship between E and H
By definition:H = E + PVSo: H = E + (PV)At constant P: (PV) = P V
H = E + P VWhen you are looking at a reaction dealing with gases:
PV = nRT So: P V = nRT
And: H = E + nRT
Relationship between H and q
H = E + P V E = q + wH = q + w + P VW = - PexVAt constant P: Pex = P
W = - PVH = q - PV + P V
H = qP
A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is E for the process?
A) + 1000 JB) + 400 JC) - 400 JD) - 1000 J
A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is E for the process?
E = q + w q = + 300 J w = + 700 J
E = 300J + 700 J = 1000 J
CalorimetryDeals with the transfer of heat energy
Heat Capacity - Capacity of a system to store heat- Heat needed to raise the temperature of a system
by 1oC ( 1K) = J/Kq = (capacity of system)(T)
Specific Heat (c) -Heat needed to raise the temperature of 1 gramof material by 1oC (1K) = J/g K
Molar Heat Capacity (C) -
Heat needed to raise the temperature of 1 moleof material by 1oC (1K) = J/ mol K
q = msT
q = nCT
Constant volume CalorimetryBomb CalorimeterQ = E
Isolated system
qsystem = ? = 0
qsystem = qsample + qcalorimeter
qsample = - qcalorimeter
qsample not directly measurable
qcalorimeter can be measured
qcalorimeter = heat capacity of bomb(T) = - qsample
Since reaction carried out at constant volume qsample = E
Constant Pressure Calorimetry
Coffee Cup CalorimeterQ = H Isolated systemqsystem = ? = 0
Generally look at two differentkinds of systems1. Reaction in aqueous solution2. Metal solid in water
qsystem = qreaction + qsolution qreaction = - qsolution
qreaction not directly measurable qsolution can be measured
1.
qsolution = mcT = -qreaction
2.qsystem = qmetal + qwater
- qmetal = qwater
-(m)(c)T = (m)(c)T metal water
Since at constant pressureqreaction = Hreaction
0.500 L of 1.0 M Ba(NO3)2 at 25oC was mixed with 0.500 L of 1.0 MNa2SO4 at 25oC. A white precipitate forms (BaSO4) and the Temperature rises to 28.1oC. What is H/ mol of BaSO4 formed?Specific heat mix = 4.184 J/g K d of mix = 1.0 g/ mL
Assume a coffee cup calorimeter, P = constant
q transfer from system to surroundings = ? = 0Why?Coffee cup calorimeter assumes an isolated system
qsys = qrxn + qmix = 0
qrxn = - qmix Can’t measure heat of the ppt’n directly but canmeasure heat of mixing
qmix = msT = (1000mL)(1.0g/mL)(4.184J/g K)(28.1-25.0)
= 12970J = 12.970kJ
qppt = - qmix = - 12.970kJBut this is only for the amount ofBaSO4 made in this reaction and wewant per mole
0.500 L of 1.0 M Ba(NO3)2 at 25oC was mixed with 0.500 L of 1.0 MNa2SO4 at 25oC. A white precipitate forms (BaSO4) and the Temperature rises to 28.1oC. What is H/ mol of BaSO4 formed?Specific heat mix = 4.184 J/g K d of mix = 1.0 g/ mL
Need to determine the amount of BaSO4 that was formed in the reaction
Rxn: Ba(NO3)2 + Na2SO4 --> BaSO4 + 2NaNO3
Mole Ba(NO3)2 = VM = (0.500L)(1.0M) = 0.50 mol Mole Na2SO4 = VM = (0.500L)(1.0M) = 0.50 mol
Ba(NO3)2 + Na2SO4 --> BaSO4 + 2NaNO3 0.50 mol 0.50 mol
--> 0.50 molSo, the reaction produced 0.50 mol BaSO4 and that released - 12.97kJ
Therefore the production of 1 mol of BaSO4 will have H = ?
H = 2(- 12.97) = - 25.94 kJ = - 25.9 kJ
28.2 g Ni metal at 99.8 oC was placed in 150. g water at 23.50 oC. The final temperature at equilibrium was 25.00 oC. Find the specificheat of Ni in cal/ g K.
A) 0.107 cal/ g KB) 0.1066 cal/ g KC) - 0.1066 cal/ g KD) - 0.107 cal/g KE) 0.1267 cal/g K
28.2 g Ni metal at 99.8 oC was placed in 150. g water at 23.50 oC. The final temperature at equilibrium was 25.00 oC. Find the specificheat of Ni in cal/ g K.
qsystem = qmetal + qsolution qmetal = mcT for the Ni
qsolution = mcT for the water
qmetal = (28.2 g)coC
qsolution = (150 g)(1 cal/g oC)(25.0-23.5oC)
- qmetal = qsolution
- (28.2 g)coC = (150 g)(1 cal/g oC)(25.0-23.5oC)
cNi = 0.1066 cal/goC = 0.107 cal/ g K