1. Chemistry Problems Thermodynamics Blase Ferraris (did these problems for Gangluff) Blase Ferraris (did these problems for Gangluff) Final Project Final Project 1996B and 1997D 1996B and 1997D

2. 1996 B C2H2(g) + 2H2(g)  C2H6(g) Information about the substances involved in the reaction represented above is summarized in the following tables. Substance Δ S (J/molK) Δ H f (kJ/mol) C 2 H 2(g) H 2(g) C 2 H 6(g) 200.9130.7-----226.70-84.7 Bond Bond Energy (kJ/mol C-CC=CC-HH-H347611414436

3. A. If the value of the standard entropy change ΔS, for the reaction is -232.7 joules per mole Kelvin, calculate the standard molar entropy, ΔS, of C2H6 gas. B. Calculate the value of the standard free- energy change, ΔG, for the reaction. What does the sign of ΔG indicate about the reaction above? C. Calculate the value of the equilibrium constant, K, for the reaction at 298 K. D. Calculate the value of the C C bond energy in C2H2 in kilojuoles per mole.

4. If the value of the standard entropy change S, for the reaction is -232.7 joules per mole Kelvin, calculate the standard molar entropy, S, of C2H6 gas. If the value of the standard entropy change ΔS, for the reaction is -232.7 joules per mole Kelvin, calculate the standard molar entropy, ΔS, of C2H6 gas. Use this equation to solve for Entropy ΔS°rxn= ΔS°products-ΔS°reactants ΔS°rxn=ΔS°(C2H6)- [ΔS°(C2H2)+2*ΔS°(H2)] Just plug and chug -232.7= ΔS°(C2H6)- [200.9+2*130.7] Solve for variable ΔS°(C2H6)= 229.6J/K

5. Calculate the value of the standard free-energy change, ΔG, for the reaction. What does the sign of Δ G indicate about the reaction above? ΔG=ΔH-TΔS ΔG=ΔH-TΔS ΔH= ΔHproducts-ΔHreactants ΔH= ΔHproducts-ΔHreactants ΔH=(- 84.7 kJ) - (226.7 kJ) = -311.4kJ ΔH=(- 84.7 kJ) - (226.7 kJ) = -311.4kJ kJ units are needed for ΔS kJ units are needed for ΔS -232.7J/K(from part A) * 1kJ/1000J = - 0.2327kJ/K -232.7J/K(from part A) * 1kJ/1000J = - 0.2327kJ/K ΔG= -311.4kJ-T*(- 0.2327kJ/K) ΔG= -311.4kJ-T*(- 0.2327kJ/K) *note*(use standard temperature (25 C)) *note*(use standard temperature (25 C)) ΔG= -311.4kJ-(298K)*(- 0.2327kJ/K)= -242.1 kJ ΔG= -311.4kJ-(298K)*(- 0.2327kJ/K)= -242.1 kJ Negative ΔG° therefore reaction is spontaneous Negative ΔG° therefore reaction is spontaneous

6. Calculate the value of the equilibrium constant, K, for the reaction at 298 K. Use the following equation: Use the following equation: ΔG= -RTlnK ΔG= -RTlnK Just divide by RT Just divide by RT -lnk= ΔG/(RT) -lnk= ΔG/(RT) Plug and Chug Plug and Chug -ln K = -242.1 ÷ [(8.31 x 10 -3 ) (298)] = 97.7 -ln K = -242.1 ÷ [(8.31 x 10 -3 ) (298)] = 97.7 e 97.7 =k e 97.7 =k K=3 x 10 42 K=3 x 10 42

7. Calculate the value of the C=C bond energy in C2H2 in kilojuoles per mole. ΔH = bonds broken- bonds formed ΔH = bonds broken- bonds formed Plug and chug all over again Plug and chug all over again - 311.4 kJ = [(2) (436) + ΔH (C=C) + (2) (414)] - [(347) + (6) (414)] - 311.4 kJ = [(2) (436) + ΔH (C=C) + (2) (414)] - [(347) + (6) (414)] Solve for variable Solve for variable 820 kJ= ΔH (C=C) 820 kJ= ΔH (C=C)

8. ITS DONE… wait no, one more wait no, one more