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Kris Quinlan

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LECTURE 12 – Thermodynamics
Slide No. Notes
3-q represents heat, or the transfer of thermal energy, which is another
type of kinetic energy
-restating delta E = -PdeltaV plus q, which q being equal to delta E
plus P delta V
-common kinds of reactions
7- the addition sign is circled because the negative sign for work has already
been included
8-Ex. A)
-A change in volume from 0.274 L to 448L
-W = -p times delta V = -(1atm)(448L-0.274L) = -447.7 Latm times
(101J/Latm)(kJ/1000J) = -45.2kJ
-Negative sign means energy lost, leaving system, so work is done on
-Ex. B)
-W = -p times delta V = -1atm(5.6L)(101J/Latm) = -566J
-Ex. B) part 2
-Delta E = q + w = -(484kJ) + (-566J) = -484.6kJ
9 -What is absolutely true is B
-What could be true is all of them, so E
10 -heat capacity depends on amount of substance
-specific heat capacity – the amount is specified
-for water, specific heat capacity is around 2 – so it takes 2J to raise 1g
of water by 1 degree C
-metals have low heat capacities
11 -Ans: A
-Al transfers more energy, therefore melts more wax
12 -calorimeter is the surroundings; reaction is the system
-energy is conserved, so energy that is lost to the system goes to the
1-bomb calorimeter – how much water warms up or cools down
-in an insulated box so that any heat lost goes into the water
2-any energy goes into water
-EX. q of surr = q of water = C of water times mass times change in
temperature = 4.18J/gC (250g)(7.48C)(kJ/1000J) = 7.82 kJ
-Q of rxn = - q of surr = -7.82kJ
5- extensive – because enthalpy differs for 1 mole of reactants compared to 5
6- enthalpy is a state function
-A) times 2
-B) times –1
-C) times 2
-Delta H changes for each of those three equations as well
-Total H is then 52.3kJ
-This is an endothermic reaction, absorbs energy
-52.3kJ/1mol of graphite times 10.0g(mol/12.01g) = 21.8kJ
9- carbon most stable in form of graphite
10 -EX.
-Ca + 1/2O2 -> CaO
-Delta H is –635.1kJ/mol
-C + O2 -> CO2
-Delta H is –393.5kJ/mol
-CaCO3 -> Ca + C + 3/2O2
-Delta H is –(-1206. 9 kJ/mol
-So total H is 178.3kJ
-Therefore the reaction is endothermic
11 -breaking bonds require energy
-Delta H = (1mol)(243kJ/mol) + (1mol)(436kJ/mol) – (2mol)
(432kJ/mol) = -185kJ
-Exothermic reaction
12 - three ways of calculating enthalpy of formation
13 -Ans1: B
-Ans2: A
-Ans3: E