1. Lecture 8: Thermochemistry
Lecture 8 Topics Brown chapter 5
8.1: Kinetic vs. potential energy
8.2: Transferring energy as heat & work
Thermal energy
8.3: System vs. surroundings
Closed systems
8.4: First Law of Thermodynamics
Internal energy of chemical reactions
Energy diagrams
E, system & surroundings
8.5: Enthalpy
Exothermic vs. endothermic
Guidelines thermochemical equations 5.4
Hess’s Law
8.6: Calorimetry
Constant pressure calorimetry
8.7: Enthlapy of formation

2. Calorimetry is the lab study of heat flow.
Constant pressure calorimetry for reactions that don’t change volume (do work).
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Calorimetry: Is heat lost or gained?
The amount of heat lost or gained by any material can be determined using calorimetry and specific heat values for that material.
Calorimetry
determination of enthalphy by experimental measurement of heat flow
Specific heat
the amount of energy required to raise the temp of 1 g of material by 1C
J/C-g
Heat capacity
the amount of energy required to raise the temp of an object by 1C
J/C
Molar heat capacity
the amount of energy required to raise the temp of 1 mole by 1C
J/C-mol
H = q = (specific heat)(grams)(T) = J
How much heat is needed to warm 250 g of water from 22°C to 98°C?
H = (4.18 J/g-K)(250 g)( C) = 7.9x104
Sh of water = J/g-K
What is the molar heat capacity of water?
Molar heat capacity = (4.18 J/g-K)(18.0g/1 mole) = 75.2 J/mol-K p

4. Examples: calorimetry
What is the specific heat of a substance that takes 547 J to raise 43 g from 25C to 53C?
q = (sh)(mass)(T) --> sh = q/(mass)(T)
sh = +547 J/(43 g)( C) = J/g-C
How much heat is given off when 75.0 g of water is cooled from 88 to 25C?
q = (sh)(mass)(T)
q = (4.184 J/g-C)(75.0 g)( C) = kJ
Some solar homes use large beds of rock to absorb heat.
Calculate the amount of heat absorbed by 50.0 kg of rocks if their temperature increases by 12°C. Assume that rock has a specific heat of 0.82 J/g-C.
q = (0.82 J/g-C)(5x104 g)(12°C)
= 4.92x105 J

5. Constant-pressure calorimetry
Remember that at constant pressure H = qp. So the most common form of calorimetry is constant pressure.
The calorimeter serves to prevent loss of heat from the system to the surroundings.
Thus, all heat generated by the system is absorbed by the system.
System = water + calorimeter
exothermic  heat produced causes the temperature within the calorimeter to increase
endothermic  heat absorbed by the system causes the temperature within the calorimeter to decrease
qrxn = - ((mass sol’n)(sh sol’n)(T) (mass cal)(sh cal)(T))
T for both is change in water temp. p

6. Measuring enthalpy with calorimeters
Hsol’n = - Hrxn
Why?
An exothermic reaction produces heat (-), and that heat is absorbed (+) by the solution contained within the calorimeter.
In a calorimeter, 50 mL of 0.65 M HCl is reacted with 50 mL of 1.00 M NaOH (excess base). Temperature of the solution increases from to 38.4C. Calculate Hrxn.
qrxn = -qsoln = -((4.184 J/g-C)(100 g)(13.4C) = -5.6 kJ
Molar enthalpy of rxn?
(0.050 L)(0.65 mol/L) = mol HCl or H2O kJ/0.033 mole = -170 kJ/mol p

7. Example: calorimetry
When 50.0 mL of M AgNO3 and 50.0 mL of M HCl are mixed in a constant pressure calorimeter, the temperature increases from 22.3C to 23.11C. The reaction is as follows:
AgNO3 + HCl --> AgCl (insoluble) + HNO3
Calculate H of this reaction, assuming that the combined solution has a mass of g and a specific heat of 4.18 J/g-C.
AgNO3  (0.050 L)(0.100 M) = 5x10-3 moles
HCl --> (0.050 L)(0.100 M) = 5x10-3 moles
T = 0.81C
qrxn = - (4.18 J/g-C)(100.0 g)(0.81C)
= x102 J
H= x102 J/5x10-3 moles
= x104 J/moles p