lecture 304/18/07. solid/liquid heat of fusion solid liquid endothermic ice water (333 j/g or 6...
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Lecture 30 4/18/07
Solid/Liquid
Heat of fusion Solid Liquid Endothermic ice Water (333 J/g or 6 KJ/mol)
Heat of crystallization Liquid Solid Exothermic Water ice (- 333 J/g or - 6 KJ/mol)
Liquid/Gas
Heat of vaporization Liquid Gas Endothermic water water vapor (40.7 KJ/mol)
Heat of condensation Gas Liquid Exothermic vapor Water (- 40.7 KJ/mol)
Solid/Gas
Heat of sublimation Solid Gas Endothermic
Heat of deposition Gas Solid Exothermic
What is the minimum amount of ice at 0 °C that must be added to 340 mL of water to cool it from 20.5°C to 0°C?
A rainstorm deposits 2.5 x 1010 kg of rain. Calculate the quantity of thermal energy in joules transferred when this much rain forms.
(∆Hcond = - 40.7 KJ/mol)
Exothermic or endothermic?
1st Law of Thermodynamics revisited
∆E = q + w
Change in Energy content
heat
work
work
work (w) = - F x d w = - (P x A) x d w = - P∆V
if ∆V = 0, then no work
State function
property of a system whose value depends on the final and initial states, but not the path driving to Mt Washington
route taken vs. altitude change
∆E is a state function q and w are not
Change in Enthalpy (∆H or qp)
equals the heat gained or lost at constant pressure
∆E = qp + w ∆E = ∆H + (-P∆V) ∆H = ∆E + P∆V
∆E vs. ∆H
Reactions that don’t involve gases 2KOH (aq) + H2SO4 (aq) K2SO4 (aq) + 2H2O (l) ∆V ≈ 0, so ∆E ≈ ∆H
Reactions in which the moles of gas does not change N2 (g) + O2 (g) 2NO (g) ∆V = 0, so ∆E = ∆H
Reactions in which the moles of gas does change 2H2 (g) + O2 (g) 2H2O (g) ∆V > 0, but often P∆V << ∆H, thus ∆E ≈ ∆H
Enthalpy is an extensive property Magnitude is proportional to amount of reactants
consumed H2 (g) + ½ O2 (g) H2O (g) ∆H = -241.8 KJ
2H2 (g) + O2 (g) 2H2O (g) ∆H = -483.6 KJ
Enthalpy change for a reaction is equal in magnitude (but opposite in sign) for a reverse reaction H2 (g) + ½ O2 (g) H2O (g) ∆H = -241.8 KJ
H2O (g) H2 (g) + ½ O2 (g) ∆H = 241.8 KJ
Enthalpy change for a reaction depends on the state of reactants and products H2O (l) H2O (g) ∆H = 88 KJ
Constant pressure calorimetry(coffee cup calorimetry)
heat lost = heat gained
Measure change in temperature of water
Constant pressure calorimetry(coffee cup calorimetry)
heat lost = heat gained
Measure change in temperature of water
10 g of Cu at 188 °C is added to 150 mL of water in a coffee cup calorimeter and the temperature of water changes from 25 °C to 26 °C. Determine the specific heat capacity of copper.
Bomb calorimetry
Mainly for combustion experiments ∆V = 0 qrxn + qbomb + qwater = 0
combustion chamber
Bomb calorimetry
Mainly for combustion experiments ∆V = 0 qrxn + qbomb + qwater = 0
Often combine qbomb + qwater into 1 calorimeter term with qcal = Ccal∆T
combustion chamber
Bomb calorimeter math
qrxn + qbomb + qwater = 0
qrxn + Cbomb∆T + Cwatermwater∆T = 0
In the lab: qrxn + qcalorimeter = 0
qcalorimeter = qbomb + qwater
qrxn + Ccalorimeter∆T = 0
empirically determined
same value
On the exam
Bond enthalpies
Enthalpies of formation
Hess’ Law
Example
A hot plate is used to heat two 50-mL beakers at the same constant rate. One beaker contains 20.0 grams of graphite (C=0.79 J/g-K) and one contains 10 grams of ethanol (2.46 J/g-K).
Which has a higher temperature after 3 minutes of heating?
Standard heat of reaction (∆H°rxn)
Same standard conditions as before: 1 atm for gas 1 M for aqueous solutions 298 K For pure substance – usually the most stable form of
the substance at those conditions
Standard heat of formation (∆H°f)
Enthalpy change for the formation of a substance from its elements at standard state
Na(s) + ½ Cl2 (g) NaCl (s) ∆H°f = -411.1 kJ
Three points An element in its standard state has a ∆H°f = 0
∆H°f = 0 for Na(s), but ∆H°f = 107.8 KJ/mol for Na(g)
Most compounds have a negative ∆H°f formation reaction is not necessarily the one done in lab
Using ∆H°f to get ∆H°rxn
2 ways to look at the problem
Calculate ∆H°rxn for:
C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)
Given:
3 C(s) + 4 H2 (g) C3H8 (g) ∆H°f = -103.85 KJ/mol
C(s) + O2 (g) CO2 (g) ∆H°f = -393.5 KJ/mol
O2 (g) + 2 H2 (g) 2H2O (l) ∆H°f = -285.8 KJ/mol
Using Hess’s Law and ∆H°f to get ∆H°rxn 1st way: Hess’s Law
C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)∆H°rxn = ∆H1 + ∆H2 + ∆H3
Reverse 1st equation:
C3H8 (g) 3 C(s) + 4 H2 (g) ∆H1 = - ∆H°f = 103.85 KJMultiply 2nd equation by 3:3C(s) + 3O2 (g) 3CO2 (g) ∆H2 = 3x∆H°f = -1180.5 KJMultiply 3rd equation by 2:2O2 (g) + 4 H2 (g) 4H2O (l) ∆H2 = 2x∆H°f = -571.6 KJ
∆H°rxn = (103.85 KJ) + (-1180.5 KJ) + (-571.6 KJ)
∆H°rxn = -1648.25 KJ
Using ∆H°f to get ∆H°rxn2nd way
C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)
∆H°rxn = Σn ∆H°f (products) - Σn ∆H°f (reactants)
∆H°rxn = [3x(-393.5 KJ/mol) + 2x(-285.8 KJ/mol)] – [(-103.85 KJ/mol) + 0]
∆H°rxn = [-1752.1 KJ] – [-103.85 KJ]
∆H°rxn = -1648.25 KJ
Spontaneity
Some thought that ∆H could predict spontaneity
Sounds great BUT . . . . .
Spontaneity
Some thought that ∆H could predict spontaneity Exothermic – spontaneous Endothermic – non-spontaneous
Sounds great BUT some things spontaneous at ∆H > 0 Melting Dissolution Expansion of a gas into a vacuum Heat transfer
Clearly enthalpy not the whole story
Entropy (Measurement of disorder)
Related to number of microstates
∆Suniverse = ∆Ssystem + ∆Ssurroundings
2nd Law of Thermodynamics Entropy of the universe increases with spontaneous
reactions
Reversible reactions vs. Irreversible reaction
Standard heat of formation (∆H°f)
Enthalpy change for the formation of a substance from its elements at standard state
Na(s) + ½ Cl2 (g) NaCl (s) ∆H°f = -411.1 kJ
Key points
Entropy(Measurement of disorder)
Related to number of microstates S = klnW
∆Suniverse = ∆Ssystem + ∆Ssurroundings
2nd Law of Thermodynamics Entropy of the universe increases with spontaneous reactions
Reversible reactions ∆Suniverse = ∆Ssystem + ∆Ssurroundings = 0 Can be restored to the original state by exactly reversing the
change Each step is at equilibrium
Irreversible reaction ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0 Original state can not be restored by reversing path spontaneous
3rd Law of thermodynamicsS = O at O K
S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions
3rd Law of thermodynamicsS = O at O K
T
qS rev
S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions
∆S° = ΣS°(products) - ΣS°(reactants)
Example
Sulfur (2.56 g) was burned in a bomb calorimeter with excess O2. The temperature increased from 21.25 ºC to 26.72 ºC. The bomb had a heat capacity of 923 J/ºC and the calorimeter contained 815 g of water. Calculate the heat evolved per mole of SO2 formed.
S(s) + O2 (g) SO2 (g)
Standard heat of reaction (∆H°rxn)
Same standard conditions as before:
Using ∆H°f to get ∆H°rxn
2 ways to look at the problem
Calculate ∆H°rxn for:
C3H8 (g) + 5 O2 3 CO2 (g) + 4 H2O (l)
Given:
3 C(s) + 4 H2 (g) C3H8 (g) ∆H°f = -103.85 KJ/mol
C(s) + O2 (g) CO2 (g) ∆H°f = -393.5 KJ/mol
O2 (g) + 2 H2 (g) 2H2O (l) ∆H°f = -285.8 KJ/mol
Degrees of freedom
translational motion molecules in gas > liquid > solid
vibrational motion movement of a atom inside a molecule
rotational motion rotation of a molecule
Entropy trends
Entropy increases: with more complex molecules with dissolution of pure gases/liquids/solids with increasing temperature with increasing volume with increasing # moles of gases
Which has higher entropy?
dry ice or CO2
liquid water at 25°C or liquid water at 50°C
pure Al2O3(s) or Al2O3 with some Al2+ replaced with Cr3+
1 mole of N2 at 1 atm or 1 mol of N2 at 10 atm
CH3CH2CH2CH3 (g) or CH3CH3 (g)
Is the reaction spontaneous?
Gibbs Free Energy (∆G)
∆G° = ∆H° - T∆S°
∆G = ∆H - T∆S
∆G° = Σn∆Gf°(products) - Σn∆Gf°(reactants)
Gibbs Free Energy ∆G = ∆H - T∆S
∆H ∆S -T∆S ∆G spontaneous
? example
- + 2O3 (g) 3O2 (g)
+ - 3O2(g) 2O3 (g)
- - H2O (l) H2O (s)
+ + H2O (s) H2O (l)
Gibbs Free Energy (∆G) and equilibrium
QlnRTGG
R = 8.314 J/mol-K
Example
A hot plate is used to heat two 50-mL beakers at the same constant rate. One beaker contains 20.0 grams of graphite (C=0.79 J/g-K) and one contains 10 grams of ethanol (2.46 J/g-K).
Which has a higher temperature after 3 minutes of heating?
Example
59.8 J are required to change the temperature of 25.0 g of ethylene glycol by 1 K.
What is the specific heat capacity of ethylene glycol?