Download Dr. Baxley`s Thermodynamics Worksheet

Chem 201B
Dr. Baxley
Thermodynamics Worksheet
I also highly recommend Worksheets 13 and 14 in the Lab Manual
1. Predict the sign of entropy change in the following processes
a) The process of carbonating water to make a soda drink
b) Condensation of water
c) Precipitation of Calcium Carbonate
d) Spray Painting
2. In the following equation indicate if ∆S is positive or negative.
a) HCl(g) → HCl(aq)
b) 2Na(s) + 2H2O (l) → 2NaOH(aq) + H2(g)
c) 2 H2(g) + O2(g) → 2 H2O(g)
3. Order the following in order of increasing S°, assuming the same temperature
a) Ethane (C2H6), Butane (C4H10), Propane (C3H8)
b) ClO4- (aq), ClO3- (aq) ClO2- (aq)
c) Heptane (C7H16), Benzene (C6H6), Cycloheptane (C7H14)
d) PF2Cl3 (g) , PF3 (g) , PF5 (g)
4. Define what it means for a reaction to be entropy driven.
Chem 201B
5. Consider the combustion of acetlyne gas:
C2H2(g) + O2(g) → CO2(g) + H2O(g)
Dr. Baxley
(unbalanced)
a) Predict the signs of ∆H, and ∆S
b) Calculate ∆G° for the combustion of one mole of acetylene by two different methods.
6. (from Brady, Russell and Holum) Considering the fact that the formation of a bond between two
atoms is exothermic and is accompanied by an entropy decrease, explain why all chemical
compounds decompose into individual atoms if heated to a high enough temperature.
7. (from Brady, Russell and Holum) Gasohol is a mixture of gasoline and ethanol (C2H5OH).
Calculate the maximum work that could be obtained at 25 °C and 1 atm by burning 1 mol of
liquid ethanol. You will need to write and balance the chemical equation. Remember that
when a hydrocarbon burns, it reacts with oxygen to produce carbon dioxide and gaseous
water.
Chem 201B
Dr. Baxley
8. Use values given in appendix II to calculate the theoretical boiling point of methanol
(CH3OH).
9. The boiling point of acetone (CH3)2CO is 56.2 °C. Given that the reaction
(CH3)2CO(l) → (CH3)2CO(g) has a ∆H° = 31.9 kJ/mol, calculate the ∆S° for this reaction.
10. For the reaction I2(s) → 2 I(g), ∆H° = +213.6 kJ and ∆S° = +245.2 J/K.
a. Calculate ∆G° for this reaction. Is the reaction spontaneous at this temperature?
b. Calculate ∆G at 250 °C. Is the reaction spontaneous at this temperature?
c. At what Celsius temperatures will this reaction be spontaneous?
d. Calculate the thermodynamic equilibrium constant for this reaction at 25 °C. Will this be
Kp or Kc?
e. If this reaction occurred in a closed vessel at 25 °C until it reached equilibrium, what
would the equilibrium pressure of I(g) be?
Chem 201B
Dr. Baxley
11. For the reaction given below, calculate ∆G° using the ∆G°f values given in Appendix II.
Without calculating K, determine whether this reaction will proceed to the right or to the left
if the partial pressures are NO = 0.50 atm, Cl2 = 0.75 atm and NOCl = 0.65 atm.
2 NO(g) + Cl2(g) → 2 NOCl(g)
12. For the reaction Ni2+(aq) + 6 NH3(aq) ⇌ Ni(NH3)62+(aq), Kc = 5.6 x 108 at 25 °C.
a. What does the magnitude of Kc tell you about the value of ∆G°?
b. Calculate ∆G°.
c. Is this reaction spontaneous in the forward or reverse direction?
d. Calculate ∆G when [Ni(NH3)62+] = 0.020 M, [Ni2+] = 0.0020 M, and [NH3] = 0.0010 M.
In which direction will the reaction proceed to achieve equilibrium?
Chem 201B
Dr. Baxley
13. Calculate the maximum amount of work that could be obtained under standard conditions
from the combustion of 5.00 g of sucrose (C12H22O11).
14. Calculate the ∆G at 95 °C for the reaction given below when the partial pressure of each gas
is 0.120 atm.
2 NO(g) + Cl2(g) → 2 NOCl(g)
Chem 201B
Dr. Baxley
Answer Key
1. a) −∆S
b) −∆S
c) −∆S
d) +∆S
2. a) −∆S
b) +∆S
c) −∆S
3. a) C2H6 < C3H8 < C4H10
b) ClO2−(aq) < ClO3−(aq) < ClO4−(aq)
c) C6H6 < C7H14 < C7H16
d) PF3(g) < PF5(g) < PF2Cl3(g)
4. This is when a spontaneous reaction has a + ∆H and a + ∆S. The reaction is spontaneous
because of its ∆S, so ∆S drives the reaction.
5. a. balance it first! ∆S is probably − because there are more gaseous reactants than products.
∆H is − because this is a combustion reaction and they are always exothermic (otherwise
you would look at how many bonds are formed vs how many break)
b. Using ∆G°f, I get −1226 kJ. Using ∆H°f and S°f, then ∆G° = ∆H°–T∆S°, I get –1227 kJ
6. Since formation of a bond has − ∆H° and − ∆S°, breaking of bonds has + ∆H° and + ∆S°.
Putting this into the equation ∆G° = ∆H° − T∆S°, you get sign of ∆G° = (+) − [T(+)]. When
temperatures are low, the product of T∆S° < ∆H°, so ∆G° is + and the breakdown of the
compound is nonspontaneous. At high temperatures, T∆S° > ∆H°, so ∆G° is − and the
breakdown of the compound is spontaneous. Therefore, the breakdown of compounds is
spontaneous at higher temperatures. The actual temperature depends on the compound.
7. The maximum amount of work would be 1299.8 kJ for 1 mole of ethanol (Determine this by
calculating ∆Gº from the ∆Gºf values and max work = –∆Gº)
8. At the boiling point ∆G° = 0 for the reaction CH3OH(l) → CH3OH(g), so 0 = ∆H° − T∆S°.
The answer is 59 °C
9. 96.8 J/molK
10. a. 140.5 kJ not spontaneous
b. 85.4 kJ not spontaneous
c. higher than 598 °C
d. 2.35 x 10−25 This would be Kp because this reaction has a gas and not aqueous.
e. PI = 4.85 x 10−13 atm
11. ∆G is negative, so reaction will proceed to the RIGHT
12. a. Since Kc is large, ∆G° must be negative.
b. −50 kJ
c. Spontaneous in the FORWARD direction because ∆G° is negative
d. +5.8 x 104 J
Since ∆G is positive, this reaction will proceed LEFT to reach equilibrium.
13. 83.3 kJ
14. –26.10 kJ
(note: Use the equation ∆G = ∆G° + RTlnQ, but the ∆G° isn’t really ∆G°, it is ∆Go95°C
which is the standard value at 95 °C instead of 25 °C. Calculate this by using ∆G95 ≈ ∆H° –
T∆S°.