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1st Law of Thermodynamics
Law of conservation of energy Energy in the universe is conserved
System vs. surroundings vs. universe
Internal energy
Specific heat capacity (C)
Quantity of energy to increase the temperature of 1 gram of a substance by 1 degree C (liquid water) = 4.184 J/g·K C (ice) = 2.06 J/g·K C (steam) = 1.84 J/g·K C (aluminum)= 0.902 J/g·K
Molar heat capacity Quantity of energy that must be transferred to
increase the temperature of 1 mole of a substance by 1 °C
Heat transfer
qgained + qlost = 0
qgained = - qlost
55.0 g of iron at 99.8°C is plunged into 225 g of water at 21°C. What is the final temperature? C (iron) = 0.451 J/g-K
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?
Changes in state
Temperature stays the same during changes of state
Gas/Vapor
Liquid
Solid
ENERGY
q = mass x constant
q = moles x constant
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)
What is the minimum amount of ice at 0 °C that must be added to a 340 mL of water to cool it from 20.5°C to 0°C?
qwater + qice = 0
Cwatermwater∆Twater + mice∆Hfus = 0
(4.184 J/K-g)(340 g)(0°C - 20.5°C) + (333 J/g)mice = 0
mice = 87.6 g
A rainstorm deposits 2.5 x 1010 kg of rain. Calculate the quantity of thermal energy in joules transferred when this much rain forms. (∆Hvap = - 44 KJ/mol)
Exothermic or endothermic?
q = 2.5 x 1010 Kg x (103g/kg) x (1 mol/18 g) x -44 KJ/mol q = -6.1 x 1013 KJ
Exothermic
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 (aw) 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(cofee 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 cofee 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
Often combine qbomb + qwater into 1 calorimeter term with qcal = Ccal∆T
combustion chamber
Bomb calorimeter math
K & T: 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
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 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 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