Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual

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1 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) + 2H 2 O (l) 2NaOH(aq) + H 2 (g) c) 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 3. Order the following in order of increasing S, assuming the same temperature a) Ethane (C 2 H 6 ), Butane (C 4 H 10 ), Propane (C 3 H 8 ) b) ClO 4 - (aq), ClO 3 - (aq) ClO 2 - (aq) c) Heptane (C 7 H 16 ), Benzene (C 6 H 6 ), Cycloheptane (C 7 H 14 ) d) PF 2 Cl 3 (g), PF 3 (g), PF 5 (g) 4. Define what it means for a reaction to be entropy driven.

2 5. Consider the combustion of acetlyne gas: C 2 H 2 (g) + O 2 (g) CO 2 (g) + H 2 O(g) (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 (C 2 H 5 OH). 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.

3 8. Use values given in appendix II to calculate the theoretical boiling point of methanol (CH 3 OH). 9. The boiling point of acetone (CH 3 ) 2 CO is 56.2 C. Given that the reaction (CH 3 ) 2 CO(l) (CH 3 ) 2 CO(g) has a H = 31.9 kj/mol, calculate the S for this reaction. 10. For the reaction I 2 (s) 2 I(g), H = kj and S = 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 K p or K c? 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?

4 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, Cl 2 = 0.75 atm and NOCl = 0.65 atm. 2 NO(g) + Cl 2 (g) 2 NOCl(g) 12. For the reaction Ni 2+ (aq) + 6 NH 3 (aq) Ni(NH 3 ) 6 2+ (aq), K c = 5.6 x 10 8 at 25 C. a. What does the magnitude of K c 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(NH 3 ) 6 2+ ] = M, [Ni 2+ ] = M, and [NH 3 ] = M. In which direction will the reaction proceed to achieve equilibrium?

5 13. Calculate the maximum amount of work that could be obtained under standard conditions from the combustion of 5.00 g of sucrose (C 12 H 22 O 11 ). 14. Calculate the G at 95 C for the reaction given below when the partial pressure of each gas is atm. 2 NO(g) + Cl 2 (g) 2 NOCl(g)

6 Answer Key 1. a) S b) S c) S d) + S 2. a) S b) + S c) S 3. a) C 2 H 6 < C 3 H 8 < C 4 H 10 b) ClO 2 (aq) < ClO 3 (aq) < ClO 4 (aq) c) C 6 H 6 < C 7 H 14 < C 7 H 16 d) PF 3 (g) < PF 5 (g) < PF 2 Cl 3 (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 kj for 1 mole of ethanol (Determine this by calculating Gº from the Gº f values and max work = Gº) 8. At the boiling point G = 0 for the reaction CH 3 OH(l) CH 3 OH(g), so 0 = H T S. The answer is 59 C J/molK 10. a kj not spontaneous b kj not spontaneous c. higher than 598 C d x This would be K p because this reaction has a gas and not aqueous. e. P I = 4.85 x atm 11. G is negative, so reaction will proceed to the RIGHT 12. a. Since K c is large, G must be negative. b. 50 kj c. Spontaneous in the FORWARD direction because G is negative d x 10 4 J Since G is positive, this reaction will proceed LEFT to reach equilibrium kj kj (note: Use the equation G = G + RTlnQ, but the G isn t really G, it is G o 95 C which is the standard value at 95 C instead of 25 C. Calculate this by using G 95 H T S.

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