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Thermodynamic Analysis of the C2H2 and H2 Reaction to C2H6

This project explores the thermodynamics of the reaction C2H2(g) + 2H2(g) → C2H6(g), focusing on standard entropy change (ΔS), standard free-energy change (ΔG), and equilibrium constant (K). Using the provided standard entropy change of -232.7 J/mol·K, we calculate the standard molar entropy of C2H6, the free-energy change, and equilibrium constant at 298 K. Additionally, the bond energy of the C=C bond in C2H2 is determined. The results confirm that the reaction is spontaneous based on the negative ΔG value.

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Thermodynamic Analysis of the C2H2 and H2 Reaction to C2H6

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  1. Chemistry ProblemsThermodynamics • Blase Ferraris (did these problems for Gangluff) • Final Project • 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.

  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 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 ΔG indicate about the reaction above? • ΔG=ΔH-TΔS • ΔH= ΔHproducts-ΔHreactants • ΔH=(- 84.7 kJ) - (226.7 kJ) = -311.4kJ • kJ units are needed for ΔS • -232.7J/K(from part A) * 1kJ/1000J = - 0.2327kJ/K • ΔG= -311.4kJ-T*(- 0.2327kJ/K) • *note*(use standard temperature (25 C)) • ΔG= -311.4kJ-(298K)*(- 0.2327kJ/K)= -242.1 kJ • 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: • ΔG= -RTlnK • Just divide by RT • -lnk= ΔG/(RT) • Plug and Chug • -ln K = -242.1 ÷ [(8.31 x 10-3) (298)] = 97.7 • e97.7=k • K=3 x 1042

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

  8. ITS DONE… wait no, one more

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