Chemistry ProblemsChemistry Problems

ThermodynamicsThermodynamics

Blase Ferraris (did these problems for Blase Ferraris (did these problems for Gangluff)Gangluff)

Final ProjectFinal Project 1996B and 1997D1996B and 1997D

1996 B1996 BC2H2(g) + 2H2(g) C2H2(g) + 2H2(g) C2H6(g) C2H6(g)

Information about the substances involved in the Information about the substances involved in the reaction represented above is summarized in the reaction represented above is summarized in the

following tables.following tables.

SubstanceSubstance ΔΔS (J/molK)S (J/molK) ΔΔHHff (kJ/mol) (kJ/mol)

CC22HH2(g)2(g)

HH2(g)2(g)

CC22HH6(g)6(g)

200.9200.9130.7130.7----------

226.7226.700-84.7-84.7

BondBond Bond Energy (kJ/molBond Energy (kJ/mol

C-CC-CC=CC=CC-HC-HH-HH-H

347347611611414414436436

A. If the value of the standard entropy A. If the value of the standard entropy change ΔS, for the reaction is -232.7 change ΔS, for the reaction is -232.7 joules per mole Kelvin, calculate the joules per mole Kelvin, calculate the standard molar entropy, ΔS, of C2H6 gas.standard molar entropy, ΔS, of C2H6 gas.

B. Calculate the value of the standard free-B. Calculate the value of the standard free-energy change, ΔG, for the reaction. energy change, ΔG, for the reaction. What does the sign of ΔG indicate about What does the sign of ΔG indicate about the reaction above?the reaction above?

C. Calculate the value of the equilibrium C. Calculate the value of the equilibrium constant, K, for the reaction at 298 K. constant, K, for the reaction at 298 K.

D. Calculate the value of the C C bond D. Calculate the value of the C C bond energy in C2H2 in kilojuoles per mole.energy in C2H2 in kilojuoles per mole.

If the value of the standard entropy If the value of the standard entropy change change ΔS, for the reaction is -232.7 joules S, for the reaction is -232.7 joules per mole Kelvin, calculate the standard per mole Kelvin, calculate the standard molar entropy, molar entropy, ΔS, of C2H6 gas.S, 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

Calculate the value of the standard free-energy Calculate the value of the standard free-energy change, ΔG, for the reaction. What does the sign change, ΔG, for the reaction. What does the sign

of of ΔΔG indicate about the reaction above?G indicate about the reaction above?

ΔG=ΔH-TΔSΔG=ΔH-TΔS ΔH= ΔHproducts-ΔHreactantsΔH= ΔHproducts-ΔHreactants ΔH=ΔH=(- 84.7 kJ) - (226.7 kJ) = -311.4kJ(- 84.7 kJ) - (226.7 kJ) = -311.4kJ kJ units are needed for kJ units are needed for ΔΔSS -232.7J/K(from part A) * 1kJ/1000J = - 0.2327kJ/K-232.7J/K(from part A) * 1kJ/1000J = - 0.2327kJ/K ΔΔG= -311.4kJ-T*(G= -311.4kJ-T*(- 0.2327kJ/K) - 0.2327kJ/K) *note*(use standard temperature (25 C))*note*(use standard temperature (25 C)) ΔΔG= -311.4kJ-(298K)*(G= -311.4kJ-(298K)*(- 0.2327kJ/K)= -242.1 kJ- 0.2327kJ/K)= -242.1 kJ

Negative ΔG° therefore reaction is spontaneousNegative ΔG° therefore reaction is spontaneous

Calculate the value of the equilibrium constant, K, Calculate the value of the equilibrium constant, K, for the reaction at 298 K.for the reaction at 298 K.

Use the following equation:Use the following equation: ΔG= -RTlnKΔG= -RTlnK

Just divide by RTJust divide by RT -lnk= ΔG/(RT)-lnk= ΔG/(RT)

Plug and ChugPlug and Chug -ln K = -242.1 ÷ [(8.31 x 10-ln K = -242.1 ÷ [(8.31 x 10-3-3) (298)] = 97.7) (298)] = 97.7

ee97.797.7=k=k K=K=3 x 103 x 104242

Calculate the value of the CCalculate the value of the C==C bond energy in C bond energy in C2H2 in kilojuoles per mole.C2H2 in kilojuoles per mole.

ΔH = bonds broken- bonds formedΔH = bonds broken- bonds formed

Plug and chug all over againPlug and chug all over again - 311.4 kJ = [(2) (436) + ΔH (C- 311.4 kJ = [(2) (436) + ΔH (C==C) + (2) (414)] - C) + (2) (414)] -

[(347) + (6) (414)][(347) + (6) (414)]

Solve for variableSolve for variable 820 kJ= ΔH (C820 kJ= ΔH (C==C) C)

ITS DONE…ITS DONE…

wait no, one morewait no, one more