A
If ΔH is negative, the reaction is ______, and if positive, _________
Exothermic, then endothermic
If ΔG is positive, the process is...
NONSPONTANEOUS
If ΔG is negative, equilibrium will shift to the...
A positive ΔS means that entropy or disorder changed in this way
INCREASED
The First Law of Thermodynamics
Energy cannot be created or destroyed, only transformed between heat, work, and internal energy
Mass also cannot be created nor destroyed
SURPRISE 500 POINT QUESTION, NO GOOGLING ALLOWED (15 second time limit here), WHAT TYPE OF CHEMISTRY DID DR. COOK GET HER PHD IN???
Theoretical Chemistry
Extra 250 points if you say where she got it from:
University of Calgary; Alberta, Canada
Gibb's Free Energy Formula, the one WITHOUT any relation to equilibrium
ΔG = ΔH - TΔS
If I set some gasoline on fire, it will completely burn out. However, I have to ignite it initially to begin the burning. This process could be said to be:
(exo/endo)thermic
(non)spontaneous
Exothermic and Spontaneous
Formula for Gibb's Free Energy in terms of equilibrium (NOT STANDARD)
ΔG = ΔGo + RTln(Q)
The Second Law of Thermodynamics
The net entropy of the universe is always increasing
I have a spontaneous reaction at a certain temperature. Both my ΔH and ΔS were positive. This reaction was _-driven
Entropy driven. A positive ΔH confers nonspontaneous processes, and ΔS confers spontaneous processes. Since we are told that the reaction is spontaneous, this means that the ΔS spontaneity outweighs the ΔH nonspontaneity, meaning the reaction is entropy-driven
There is 1.2 moles of gas in a vessel at 298 K and 1 atm. Constant pressure is maintained until volume doubles. How much work was done?
w = -PΔV
w = 29.3 L*atm
SEE B3 ON POWERPOINT
When ΔH and ΔS are both positive, this is the spontaneity of the reaction.
Spontaneous at high temperatures
OR
Nonspontaneous at low temperatures
ΔG = ΔH - TΔS
ΔG = (+) - T(+), get T high to make -TΔS negative, so ΔG is negative
A puddle is sitting on the sidewalk as the sun comes up. These 50 grams of water were originally at a cool 25 Co, but the sun brought the heat, and the water is now at a steamy 98 Co. Since water has a specific heat capacity of Cwater = 4.184 J/g*K, how much energy, IN JOULES, did it take to warm the water up?
q = mCΔT
q = (50 grams)(4.184 J/g*K)(98-25 Co)
q = 15271.6 JOULES
A vessel initially has 0.8 moles of gas at 350 K and 2 atm. A pressure of 1.5 atm is applied until volume reaches 10 L. How much work was done?
2.25 L*atm
SEE E3 ON POWERPOINT
The difference between ΔG and ΔGo
Should you answer, there is a follow-up opportunity for 100 extra points with NO INCORRECT PENALTY (but you have to get the above part of the question correct)
ΔG is for current/any conditions
ΔGo is for ideal/standard conditions
Bonus:
ΔGo "standard" conditions are:
1 atm of pressure
1 mol or 1 M
ANY temperature
ΔGo is not equilibrium. If you have 1 M or 1atm of everything, it tells you which way the reaction would shift, like comparing Q to K sort of
A reaction at equilibrium and at 298 K has the following concentrations:
C7H16(g) + 11O2(g) ⇌ 7CO2(g) + 8H2O(g)
C7H16 = 12 atm
O2 = 1.1 atm
CO2 = 1.4 atm
H2O = 2 atm
What is ΔGo?
ΔGo = -RTln(K)
ΔGo = -(8.314)(298)ln(K)
K = (H2O)8(CO2)7 / (C7H16)(O2)11
K = (28)(1.47) / (12)(1.111)
K = 78.8198
ΔGo = -(8.314)(298)ln(78.8198)
ΔGo = -10819.9644173 J/mol
OR
-10.819 kJ/mol
I have the following reaction:
N2O4 (g) ⇌ 2NO2 (g); ΔHo = 57.2 kJ/mol
At 298K, the K constant is K1 = 0.25. What is the K constant at a temperature of 350 K?
Hint: Van't Hoff Equation from the homework
K = 0.0080967...
OR
K = 8.1 x 10-3
SEE C4 ON POWERPOINT
20 grams of water is heated from 25 oC to 100 oC, all of it turned to steam. How much heat is required to do this in kilojoules?
Cwater = 4.184 J/g*K
ΔHvaporization = 40.7 kJ/mol
51.5 kJ
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I have some unknown substance, specifically 40 grams of it. I expelled 89 joules of energy warming it up from 20 degrees to 100 degrees celsius. Find its specific heat capacity, and provide the units for it.
q = mCΔT
89 = (40)(C)(100-20)
C = 89/3200
C = 0.0278 J/g*C
Solve for the ΔH for the combustion of methane (CH4) using the bond-enthalpies approach and the following bond energy information:
C-H has 414 kJ
O=O has 143 kJ
C=O has 1075 kJ
O-H has 464 kJ
ΔH = -1778 kJ/mol
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I toss my 800 gram ball of tungsten into the freezer until it reaches 10 oC. I then remove it and place it into 200 grams of warm, 80 oC water. The situation is insulated, so no heat is lost to the air around the iron and water. What is the final temperature when thermal equilibrium is reached?
Ctungsten = 0.449 J/g*C
Cwater = 4.184 J/g*C
Final Temperature = 58.98 oC
SEE B5 ON POWERPOINT
I have the following reaction:
2SO2(g) + O2(g) ⇌ 2SO3(g)
At 298 K and all standard conditions, the equilibrium constant is 3.5. I prepare a mixture of gases as follows:
P(SO2) = 0.8 atm
P(O2) = 0.5 atm
P(SO3) = 0.2 atm
Calculate ΔG at these conditions and 298 K
ΔG = -8255.8 kJ
SEE C5 ON POWERPOINT
I have this "fake-water" that is not like water at all. It has a melting point of 25 oC and a boiling point of 125 oC. There are other constants (heats of vaporization/fusion and heat capacities) given below. I took some unknown mass of it, starting at -25 oC where it was fake-ice, and heated it up to 200 oC where it was fake-steam. If the entire heating process cost me 2.5 x 106 J of energy, how much fake-water did I have in grams?
Cfake-ice = 5 J/g*C
ΔHfusion = 200 J/g
Cfake-water = 10 J/g*C
ΔHvaporization = 1000 J/g
Cfake-steam = 20 J/g*C
632.9 grams
SEE D5 ON POWERPOINT
What is ΔH for the overall reaction:
Target Reaction: C6H6O2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(l)
Reaction 1: C6H6O2 (aq) → C6H4O2(aq) + H2(g); ΔH1=+177.4 kJ
Reaction 2: H2(g) + O2(g) → H2O2(aq); ΔH2=-191.2 kJ
Reaction 3: H2(g) + ½O2(g) → H2O(g); ΔH3=-241.8 kJ
Reaction 4: H2O(g) → H2O(l); ΔH4=-43.8 kJ
ΔH = -202.6 kJ
SEE E5 ON POWERPOINT