What is the standard entropy change for the formation of butyric acid at 25 degrees C
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What is the standard entropy change for the formation of butyric acid at 25 degrees C

[From: ] [author: ] [Date: 11-05-04] [Hit: ]
for butyric acid.(b) Write a correctly balanced equation for the formation of butyric acid from its elements.(c) Calculate the standard entropy change, DSf°, for the formation of butyric acid at 25°C. The entropy change,......
I need help solving this little thermodynamics issue.

Standard Heat of Absolute
Formation, DHf°, Entropy, S°,
Substance in kJ mol-1 in J mol-1 K-1
------------------ ----------------------------- ------------------------
C(s) 0.00 5.69
CO2(g) -393.5 213.6
H2(g) 0.00 130.6
H2O(l) -285.85 69.91
O2(g) 0.00 205.0
C3H7COOH(l) ? 226.3

The enthalpy change for the combustion of butyric acid at 25°C, DH°comb, is -2,183.5 kilojoules per mole. The combustion reaction is

C3H7COOH(l) + 5 O2(g) ® 4 CO2(g) + 4 H2O(l)

(a) From the above data, calculate the standard heat of formation, DHf°, for butyric acid.

(b) Write a correctly balanced equation for the formation of butyric acid from its elements.

(c) Calculate the standard entropy change, DSf°, for the formation of butyric acid at 25°C. The entropy change, DS°, for the combustion reaction above is -117.1 J K-1 at 25°C.

(d) Calculate the standard free energy of formation, DG°f, for butyric acid at 25°C.

I already have the answers, and am mostly interested solely in parts C and D, but i need to see the process and how it is done. Please show the steps taken to calculate entropy and free energy of formation, any help is appreciated.

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c) Calculate the standard entropy change, ∆Sf°, for the formation of butyric acid at 25° C. The entropy change, ∆S°, for the combustion reaction above is - 117.1 J K-1 at 25° C.

∆S°f (butyric acid) = S°(butyric acid) - [4 S°C + 4 S°H2 + S°O2]
226.3 - [4(5.69) + 4(130.6) + 205] = - 523.9 J/K

d) Calculate the standard free energy of formation, ∆Gf°, for butyric acid at 25° C.

∆G°f = ∆H°f - T∆S°f = 533.8 - (298)(-0.5239) kJ = - 377.7 kJ
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