Chemistry Problems
Thermodynamics
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Blase Ferraris (did these problems
for Gangluff)
Final Project
1996B and 1997D
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
ΔS (J/molK)
ΔHf (kJ/mol)
C2H2(g)
H2(g)
C2H6(g)
200.9
130.7
-----
226.7
0
-84.7
Bond
Bond Energy (kJ/mol
C-C
C=C
C-H
H-H
347
611
414
436
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 freeenergy 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.
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
Calculate the value of the standard free-energy
change, ΔG, for the reaction. What does the sign
of ΔG indicate about the reaction above?
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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
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Negative ΔG° therefore reaction is spontaneous
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Calculate the value of the equilibrium constant, K,
for the reaction at 298 K.
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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
Calculate the value of the C=C bond energy in
C2H2 in kilojuoles per mole.
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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)
ITS DONE…
wait no, one more