Chapter 16 Thermodynamics

Name________________________

 

Use the thermodynamics table in your book to look up any enthalpy, entropy ro free energy values needed.

 

1. Predict the sign of the entropy change for the following reaction:

solid ammonium dichromate is burned to give solid chromium oxide, water vapor and nitrogen gas.

 

2. Given the following data, calculate the free energy of the reaction and state if it is spontaneous or non spontaneous.

D H= -16kJ/mol, D S = 50J/K, T= 300K

 

3. The heat of fusion for actinium is 10.50kJ/mol. The entropy of fusion is 9.6 J/mol K. Calculate the melting point of actinium. (Hint) the melt point is an equilibrium between the solid and liquid phase, Therefore D G = ??????

 

4. Calculate D H for the following reaction 25°C:

2Ag2O ---> 4 Ag + O2

5. Calculate D S for the reaction in question #4

 

6. Using the answers from problem #4 and #5, calculate the free energy of the reaction.

 

7. Calculate the equilibrium constant for the reaction in question #4

 

8. Given the following reactions

N2 + 2O2 ---> 2NO2 D G = -51kJ

2NO + O2 ---> 2NO2 D G = -70kJ

Calculate D G for N2 + O2---> 2NO

 

9. Given Kb for ammonia at 298K is 1.8x10-5 calculate D G° for the following reaction:

NH3(g) + H2O(l) <--> NH4+(aq) + OH-(aq)

 

9b. What is the value of D G at equilibrium?

 

9c. What is the value of D G when [NH3] = 0.10 M, [NH4+] = 0.10M and [OH-] =0.050M

 

The following reaction is one that contributes significantly to the formation of photochemical smog.

2NO + O2 --> 2NO2 D H = -114.1 kJ, D S° = -146.5J/K

 

10a. Calculate the heat released when 73.1 g of NO is converted to NO2.

 

10b. For the reaction at 25°C, the value of the standard free energy change, D G°, is -70.4kJ. Calculate the value of the equilibrium constant, Kc, for the reaction at 25°C.

 

 

10c. Indicate whether the value of D G° would become more negative, less negative, or remain unchanged as the temperature is increased. Justify your answer.

 

 

10d. Give the standard molar entropy, S° for NO = 210.8J/mol K and NO2 = 240.1 J/mol K calculate S° for O2 at 25°C

 

 Answers:

1) D is positive

2) -31.0kJ/mol

3) 820.7 C

4) +62 kJ/mol

5) +133 J/mol k

6)22.7kJ/mol

7) 1.05x10 -4

8) +19 kJ/mol

9) +27.1 kJ/mol

9b zero

9c) 19.7 kJ/mol

10a) -139kJ

10b) 2.2x1012

10c) less negative, more positive

10d +205j/mol-k