#37 Notes
Unit 5: Thermodynamics/Thermochemistry/Energy
Ch. Thermochemistry (Thermodynamics)
-is the study of energy.
I. Energy is the capacity to do work or produce heat.
Law of Conservation of Energy
Energy cannot be created or destroyed, just transferred.
1st Law of Thermodynamics
The energy in the universe is constant.
State Functions
-depend only on the current state of the system. P, V, T, energy
Heat
Heat flows from hot to cold: high KE, high velocity particles collide with cold/slower
particles, making them speed up, increasing their KE.
Exothermic
Heat energy flows out of a system into its surroundings.
(hot pack) **burning paper
Endothermic
Heat energy flows into a system from its surroundings.
(cold pack) **ice cube melting
ΔE = q + w
w = -PΔV
ΔE = change in internal energy of a system (joules)
q = heat, w = work
P = pressure
ΔV = Vf - Vi (change in volume)
** 1 L·atm = 101.3 J
Ex. 1) A system gives off 196 kJ of heat to its surroundings and the surroundings do
4.20 X105 J of work on the system. Find ΔE of the system.
4.20 X105 J │1 kJ
= 420 kJ
3
│1 X10 J
ΔE = q + w = -196 kJ + 420 kJ = 224 kJ
(-) q, since heat is lost,
(+) w, since work gained by system (work done on it)
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Ex. 2) A balloon is being inflated. The volume of the balloon changes from 400 L to 750 L
by the addition of 2.40 X106 J of heat. Assume the balloon expands against a constant
pressure of 1.00 atm. Find ΔE of the balloon.
ΔE = q + w
w = -PΔV
ΔV = Vf - Vi
w = -(1.00 atm) (750 L – 400 L)
w = -350 L·atm
w = -350 L·atm │101.3 J = -3.54 X104 J
│ 1 L·atm
ΔE = q + w = 2.40 X106 J – 3.54 X104 J
ΔE = 2.36 X106 J
II. Enthalpy (H)
-is the heat energy.
q = H at constant P
Exothermic
Heat lost.
Endothermic
Heat gained.
We can measure the difference in heat for the formation of compounds from their elements.
See Textbook for ΔHof values (o for standard state, f for formation) cr = crystalline solid
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ΔHreaction = Σ n ΔHproducts(final) – Σ n ΔHreactants(initial)
↑
Gr. letter sigma (summation)
Ex. 1) Find ΔHreaction for Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
reactants
products
ΔHreaction = Σ n ΔHproducts(final) – Σ n ΔHreactants(initial)
ΔHreaction = [(1 mol ZnCl2) (-488.19 kJ/mol) + (1 mol H2) ( 0 kJ/mol)]
- [(1 mol Zn) ( 0 kJ/mol) + ( 2 mol HCl) ( -167.159 kJ/mol)]
ΔHreaction = -488.19 kJ + 0 – 0 + 334.318 kJ = -153.87 kJ exothermic
**make sure subscripts match! ( (cr ) = (s) )
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#38 Notes
III. Calorimetry
-is the science of measuring heat.
ΔE = q + w
w = -PΔV
q = smΔT
where s = specific heat capacity,
m = mass, and ΔT = Tf - Ti
Specific heat capacity (s)
-is the energy required to raise the temperature one degree Celsius for one gram of the
substance.
s for H2O = 4.184 J/(g·oC) = 1 calorie, s for Fe = 0.45 J/(g·oC)
Molar heat capacity
-is the same except for 1 mol of the substance.
J /(mol·oC)
**so the “q” equation would have mols for “m”, instead of mass
Ex. 1a) How much heat is required to raise 8.77 g CCl4 from 37.1 oC to 56.4 oC?
s for CCl4 = 0.856 J/(g·oC)
q = smΔT = (0.856 J/(g·oC)) ( 8.77 g) ( 56.4 oC – 37.1 oC)
= (0.856 J/(g·oC)) ( 8.77 g) ( 19.3 oC) = 1.45 X102 J
Ex. 1b) Calculate the molar heat capacity.
0.856 J │ 154.13115 g CCl4 = 132 J/(mol·oC)
g·oC │ 1 mol CCl4
Ex.1c) Calculate the heat of raising 24.1 mol CCl4 from 25 oC to 325 oC.
q = smΔT = (132 J/(mol·oC)) ( 24.1 mol) ( 325 oC – 25 oC)
= (132 J/(mol·oC)) ( 24.1 mol) ( 300 oC) = 9.5 X105 J
Ex. 2) A 25.0 g sample of metal was heated to 90.0 oC and added to 50.0 cm3 of water at
20.0 oC. What is the specific heat of the metal, if the final temperature was 22.0 oC?
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Metal
Water
Loses heat Gains heat
-q = + q
- smΔT = + smΔT
D = m/v
(1.0 g/cm3) =
m___
(50.0 cm3)
50.0 g = m
↓
-s ( 25.0 g) ( 22.0 C – 90.0 C) = (4.184 J/(g· C)) ( 50.0 g) ( 22.0 oC – 20.0 oC)
-s (25.0 g) (-68 oC) = (4.184 J/(g·oC)) ( 50.0 g) ( 2.0 oC)
s (1700) = (418.4)
s = 0.246 J/(g·oC)
o
o
o
Ex. 3) Calculate the final temperature of the mixture when 300.0 ml of 1.00 M HCl is mixed
with 350.0 ml of 0.750 M Ca(OH)2. (Ti = 25 oC, final mass of mixture = 650.0 g,
s = 4.18 J /(g·oC)
2 HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2 H2O(l) ΔH = -112 kJ
↑
for the whole reaction, i.e. for 1 mol CaCl2, for 2 H2O etc.
Find limiting reagent:
HCl
Ca(OH)2
M = mol/L
M= mol/L
1.00 M HCl = mol
0.750 M Ca(OH)2 = mol
0.300 L
0.350 L
0.300 = mol HCl
0.263 = mol Ca(OH)2
0.300 mol HCl │1 mol CaCl2 = 0.150 mol CaCl2 ←limiting reagent
│2 mol HCl
0.263 mol Ca(OH)2 │1 mol CaCl2 = 0.263 mol CaCl2
│1 mol Ca(OH)2
0.150 mol CaCl2 │ -112 kJ
= -16.8 kJ heat produced
│1 mol CaCl2
ΔE = q + w, no work, so ΔE = q = smΔT
**since heat is produced, Tf should be larger than Ti
So change heat to positive
q = smΔT
16.8 kJ │1 X10 J = (4.18 J /(g·oC)) (650.0 g) (Tf – 25 oC)
│ 1 kj
3
6.18 = Tf – 25oC
31 oC = Tf
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#39 Notes
IV. Hess’s Law
The enthalpy change (ΔHro) for a reaction is the sum of the enthalpy changes for a series
of reactions, that add up to the overall reaction.
Steps:
For each reaction:
1) Check to see, if the compounds are on the correct sides of the reaction.
**If not, reverse the entire reaction, and change the sign of ΔH.
2) Check to see, if all of the unwanted compounds will cancel completely.
**If not, multiply an entire reaction by a number so that they do cancel completely and
multiply ΔH by that same number.
2 HF(g)→ H2(g) + F2(g)
½ C(s) + F2(g) → ½ CF4(g)
2 C(s) + 2 H2(g) → C2H4(g)
Ex.1) Given:
ΔH = +537 kJ
ΔH = -340. kJ
ΔH = 52 kJ
Find ΔHro for: C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g)
H2(g) + F2(g) → 2 HF(g)
ΔH = -537 kJ (F2 & HF were on wrong side, now -537))
½ C(s) + F2(g) → ½ CF4(g)
ΔH = -340. kJ (F2 & CF4 on correct sides already)
C2H4(g) → 2 C(s) + 2 H2(g) ΔH = -52 kJ (C2H4 was on wrong side, now -52)
2 ( H2(g) + F2(g) → 2 HF(g) )
ΔH = -537 kJ X 2 (need to cancel H2) = -1074 kJ
4 (½ C(s) + F2(g) → ½ CF4(g))
ΔH = -340. kJ X 4 (need to cancel C) = -1360 kJ
C2H4(g) → 2 C(s) + 2 H2(g)
ΔH = -52 kJ
C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g) ΔH = -2486 kJ
Ex. 2) Given the following data:
SO3(g) → S(s) + 3/2 O2(g)
2 SO2(g) + O2(g) → 2 SO3(g)
ΔH =+395 kJ
ΔH = -198 kJ
Calculate ΔH for the reaction: S(s) + O2(g) → SO2(g)
S(s) + 3/2 O2(g) → SO3(g)
2 SO3(g) → 2 SO2(g) + O2(g)
ΔH = -395 kJ (S & O2 on wrong side, now -395)
ΔH = +198 kJ (SO2 was on the wrong side, now
+198)*
*Now for the second reaction O2 is on the wrong side, but there is other O2 in the first reaction!
2 (S(s) + 3/2 O2(g) → SO3(g) )
ΔH = -395 kJ X 2 (need to cancel SO3) = -790 kJ
2 SO3(g) → 2 SO2(g) + O2(g) ΔH = +198 kJ
2S(s) + 2O2(g) * → 2SO2(g)
ΔH = -592 kJ
*(3 O2 on the left reduces to 2 O2 by cancelling the 1 O2 on the right)
S(s) + O2(g) → SO2(g)
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ΔH = -296 kJ (Divided by 2 to get answer.)
#40 Notes
Ch. Energy
I. Entropy (S)
-is the disorder in a system.
In nature systems get more messed up (gain entropy).
Ex. 1) Is entropy increasing or decreasing?
a) melting Al
Al(s) → Al(l)
↑
↑
More ordered
More messy
(lower entropy) (higher entropy)
I2(s) → I2(g)
↑
↑
More ordered
More messy
(lower entropy) (higher entropy)
ΔS = (+) increasing
b) sublimating I2
ΔS = (+) increasing
c) 2 CO(g) + O2(g) → 2 CO2(g)
all gases, so look at amounts
3mols
2 mols
Three objects can get more disorganized than 2 objects.
High entropy
Low entropy
ΔS = (-) decreasing
d) Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
↑
↑ aqueous to aqueous is no change
↑
↑
Solid
Gas
More messy, higher entropy
ΔS = (+) increasing
Ex. 2) Which has more entropy?
a) C6H12O6(s) or H2O(s)
C6H12O6(s) is more complex
b) He at 0 oC or He at 200 oC
200 oC is a higher temperature, more motion
c) He at 1 atm or He at 2 atm
1 atm has lower pressure, less organized
** The higher the temperature and the lower the pressure, the higher the entropy!
STP = 273 K & 1 atm
II. 2nd Law of Thermodynamics
In any spontaneous process there is an increase in the Universe’s entropy.
**spontaneous immediately reacts, but it could be a fast or slow process.
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ΔSuniverse = ΔSsystem + ΔSsurroundings
(+) ΔSuniverse is spontaneous, since it gains entropy,
ΔSsurroundings is usually determined by heat flow to/from the system.
For example: Melting Ice
For the system: the ice gains heat (ΔH = (+)),
the ice goes from solid to liquid (it gains entropy, ΔS = (+))
For the surroundings: it loses heat to the ice (ΔH = (-)),
so it loses entropy as well (ΔS = (-))
ΔSuniverse
= ΔSsystem + ΔSsurroundings
(+) to be
↑
↑
spontaneous.
(+)
(-) Which is stronger? Will the ice melt?
This is determined by temperature.
When the surroundings is > 0 oC, the ice melts. (ΔSsystem is stronger. ΔSuniverse be (+).)
When the surroundings is < 0 oC, the ice doesn’t melt.
(ΔSsurroundings is stronger. ΔSuniverse be (-).)
ΔSsurroundings = -ΔHsystem
T
but ΔSsystem = +ΔHsystem
T
T is in Kelvin
Ex. 1) The melting point of iron is 1535 oC and the enthalpy of fusion is 13.8 kJ /mol.
What is the entropy of fusion (ΔSsystem)? What is the ΔSsurroundings?
ΔSsystem = 13.8 kJ/mol = 7.63 X10-3 kJ │1 X103 J = 7.63 J/mol∙K
1808 K
mol∙K │1 kj
So, ΔSsurroundings = -7.63 J/mol∙K
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#41 Notes
III. 3rd Law of Thermodynamics
The entropy of a perfect crystal at 0 Kelvin is ZERO.
(Absolute Zero temperature = no vibrations, no random movement, no Kinetic Energy)
IV. Gibb’s Free Energy (G)
-is the chemical potential energy.
ΔG (-) = spontaneous
ΔG (+) = nonspontaneous
ΔG = ΔH – TΔS
ΔG in kj/mol, ΔH in kj/mol, T in Kelvin,
ΔS in j/(mol·K) **so change to kj/(mol·K)
Ex.1) Will a reaction be spontaneous at 25 oC, if ΔH = 542 kJ/mol and ΔS = -14 J/(mol·K)?
-14 J │1 kJ
= -0.014 kJ/(mol·K)
3
mol•K│1 X10 J
ΔG = ΔH – TΔS = (542 kJ/mol) – [(298 K) ( -0.014 kJ/(mol·K))]
ΔG = 542 kJ/mol + 4.172 kJ/mol = 546 kJ/mol (+) so nonspontaneous
ΔGreaction = Σ n ΔGproducts(final) – Σ n ΔGreactants(initial)
ΔSreaction = Σ n ΔSproducts(final) – Σ n ΔSreactants(initial)
Ex. 2a) Calculate ΔGreaction for: 2 HF(g) → H2(g) + F2(g)
ΔGreaction = Σ n ΔGproducts(final) – Σ n ΔGreactants(initial)
ΔGreaction = [(1 mol H2) ( 0 kJ/mol) + (1 mol F2) ( 0 kJ/mol) ]
– [( 2 mol HF) ( -273.2 kJ/mol)]
ΔGreaction = 0 + 0 + 546.4 kJ = +546.4 kJ nonspontaneous
Ex. 2b) Calculate ΔSreaction
ΔSreaction = Σ n ΔSproducts(final) – Σ n ΔSreactants(initial)
ΔSreaction = [(1 mol H2) ( 130.684 J/(mol·K)) + (1 mol F2) ( 202.78 J/(mol·K)) ]
– [( 2 mol HF) ( 173.779 J/(mol·K))]
ΔSreaction = 130.684 J/K + 202.78 J/K – 347.558 J/K = -14.094 J/K decreasing
*End of Notes*
(Assignments #42-43 are Review Assignments. There are no notes for these assignments.)
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