Date post: | 19-Dec-2015 |
Category: |
Documents |
View: | 241 times |
Download: | 0 times |
2
K
What is K at T = 298.15 K What is K at T = 773 K
How to evalute K?
Question? Is cyclobutane more stable than 2 moles of ethylene at T = 298 K?What about at T = 773 K?
Definition of Thermodynamic Symbols
E is the intrinsic energy; (J/mol)H is the enthalpy (= E + PV); (J/mol)G is the Gibbs energy (= H – TS); (J/mol)Cp is the heat capacity at constant temperature; (J/mol K)
The superscript o denotes the value given is for the standard reference state.For gases this is the ideal gas state at 1 atm. For example Hv refers to theenthalpy of vaporization of the liquid phase to the real gas at the saturationpressure, while Hov refers to the vaporization enthalpy of the pure liquid tothe ideal gas at 1 atm. For liquids and solids the standard state is the puresubstance at 1 atm pressure.
(GoT - H
o298)/T is the Gibbs energy function and is equal to (Ho
T - Ho
298)/T -So
T. This function is tabulated because it shows greater linearity than GoT
thus facilitating interpolation between temperatures.
SoT is the practical entropy in the standard state at temperature T omitting
contributions from isotopic mixing and nuclear spins.
refers to the difference between the final state and the initial one. Forexample Hf is the enthalpy of formation of substance A relative to theenthalpy of the elements of this substance with each substance in aspecified state. Ho
f is the enthalpy of formation of substance A relative tothe elements with each substance in its standard state at the specifiedtemperature.
Kp is the equilibrium constant expressed in terms of pressure Kp =(Pa)m(Pb)n/(Pc)o(Pd)p.
GoT = -RT lnKp
Elements: Compounds:
Hof = 0 at all temperatures Ho
f = Hoproduct - Ho
elementsGo
T = 0 at all temperatures Gof = Ho
f - TSof
Elements and Compounds:
Cp increases slowly with temperatureHo
T - Ho298 = Cp dT Cp [T2 – T1]
SoT not zero except at 0 K
Cp: increases with increasing temperature; Cp liquid > Cp solid> Cp gas
1. How do we evaluate Hf ?
C4H8 + 6 O2 = 4 CO2 + 4 H2O Hc(298) = -2721 kJ mol-1
Bomb calorimetry: Bomb calorimetry actually measures Ef
since the reaction starts off with 7 moles of gas and end up withonly 4 mol at T = 288.15.
C + O2 = CO2
1/2 H2 + O2 = H2O
Hf (298.15 K
kcal mol-1
Hf (298.15 K)kJ mol-1
H2O (l) -68.315 -285.83CO2 (g) -94.049 -393.5
C4H8 + 6 O2 = 4 CO2 + 4 H2O Hc(298) = -2721 kJ mol-1
4 CO2 = 4 C + 4 O2 Hf (298) = 4(393.5) = 1574.0
4 H2O = 4 H2 + 2 O2 Hf (298) = 4(285.83) = 1143.4
___________________________________________________
C4H8 = 4 C + 4 H2
[-2721 +1574.0+ 1143.4] = -3.6 kJ mol-1
C4H8 Hf (298 K) = 3.6 kJ mol-1
Cyclobutane is a gas at 298.15 K and at 1 atmosphere. Howeverthe sample was weighed as a liquid and was burned as a liquid.Therefore it is necessary to add the vaporization enthalpy ofcyclobutane to the enthalpy of formation at T = 298.15 K.
Hf (g, 298 K) = H
v (298 K) + Hf (l, 298 K)
Hf (g, 298 K) = 24.7 + 3.6 = 28.3 kJ mol-1
A more complete list can be found atwww.umsl.edu/~jscumsl/
Enthalpy of Formation of Some Small Molecules
Hf (298.15 K
kcal mol-1
Hf (298.15 K
kJ mol-1
H2O -68.315 -285.83CO2 -94.049 -393.5C2H4 ethylene 12.5 52.3C2H6 ethane -20.0 -83.7C3H4 cyclopropene 66.23 277.1C3H6 propene 4.8 20.1C3H6 cyclopropane 12.74 53.3C3H8 propane -25 -104.6C4H4O furan -8.34 -34.9C4H6 cyclobutene 37.5 156.9C4H6 butadiene 26.3 110C4H6 methylenecyclopropane 47.9 200.4C4H6O divinyl ether -3.25 -13.6C4H8 cyclobutane 6.79 28.4C4H8 methylcyclopropane 5.6 23.4C4H8 isobutene -4 -16.7C4H8 cis 2-butene -1.7 -7.1C4H8 trans 2-butene -2.7 -11.4C4H8 1-butene 0.2 0.8
cyclobutane Hf (g, 298 K) = 28.4 kJ mol-1;
ethylene Hf (g, 298 K) = 52.3 kJ mol-1
C4H8 2 C2H4
Hr (g, 298 K) = 2(52.3) - 28.4 = 76.2 kJ mol-1
What is the enthalpy of reaction at 773 K ?
How do we calculate it?
For each compound
Hf (T2 ) = Hf (298 K) + i Hi
(Ti)phase changes +
dTCT
p2
298
Figure. Heat capacity of ethylene and cyclobutane as a function oftemperature (triangles, cyclobutane; circles, ethylene).
Temperature /K
200 300 400 500 600 700 800 900
Cp ga
s / J
mol
-1
20
40
60
80
100
120
140
160
180
200
Since both ethylene and cyclobutane are gases at room temperature and above, all we need is the heat capacity of the gas phase.
How do we measure the Cp(g)(298) of benzoic acid (solid), or of methyl salicylate (liquid)?
At 800 K: cyclobutane, Cp(g) = 177.4 J mol-1K-1
At 550 K: cyclobutane, Cp(g) 135 J mol-1K-1
Hf (g, 773 K) / J mol-1 = H
f (g, 298 K) + Cp(g)[T2 -T1]
Hf (g, 773 K) / J mol-1 = 28400 + 135[773-298] = 92525
At 800 K: ethylene, Cp(g) = 84.3 J mol-1K-1
At 550 K: ethylene, Cp(g) 67 J mol-1K-1
Hf (g, 773 K)/ J mol-1 = 52300 + 67[773-298] = 84125
Therefore:
C4H8 2 C2H4
Hr (g, 298 K)/ kJ mol-1 = 2(52.3) - 28.4 = 76.2
Hr (g, 773 K)/ kJ mol-1 = [2(84125) - 92525]/1000 = 75.73
On the basis of enthalpy considerations alone, the reaction no morefavored at 773 than at 298 K.
Entropy
fusT
dTT
cCp
0
)(Sf (g, 298) = + Hfus
/Tfus
+ b
fus
T
T
dTT
lCp )( + Hvap
/Tb + 298 )(
BT
dTT
gCp
Ethylene at 298 K
Sf (298) = Sethylene - 2 Sf H2 -2 SfC
Sethylene(298) = 219.5 J mol-1K-1
2 Sf H2 = 2(4.184)(31.2) = 261.2 J mol-1K-1
2 SfC = 2(4.184)(1.361) = 11.4 J mol-1K-1
Sf (298) (C2H4) = 219.5-261.2-11.4 = -53.1 J mol-1K-1
Cyclobutane at 298 K
Sf (298) = Scyclobutane - 4 Sf H2 - 4 SfC
Scyclobutane(298) = 265.4 J mol-1K-1
4 Sf H2 = 4(4.184)(31.2) = 522.4 J mol-1K-1
2 SfC = 2(4.184)(1.361) = 22.8 J mol-1K-1
Sf (298) (C4H8) = 265.4-522.4-22.8 = -279.8 J mol-1K-1
Sf (298) (C4H8) = -279.8 J mol-1K-1
Sf (298) (C2H4) = -53.1 J mol-1K-1
Cyclobutane 2 ethylene
Sr (298) = 2(-53.1) - (-279.8) = 173.6 J mol-1K-1
Sr (298) = 2 Sethylene - Scyclobutane
Sr (298) = 2(219.5) - 265.4 = 173.6
Gf (298) = Hf (298) - T Sf (298)
Ethylene: Hf (298) = 52300; Sf (298) = -53.1
Gf (298) = 52300 - (298)(-53.1) = 68130 J mol-1
Cyclobutane: Hf (298) = 28400; Sf (298) = -279.8
Gf (298) = 28400 - (298)(-279.8) = 111780 J mol-1
Cyclobutane 2 ethylene
Gr (298) = 2(68.13) - 111.78 = 24.5 kJ mol-1
Gr (298) = 76200 - (298)(173.6) = 24.5 kJ mol-1
Sethylene(773) Sethylene(298) = 219.5 J mol-1K-1
Sethylene(773) = Sethylene(298) + 773
298
)(dT
T
gCp
Sethylene(773) = Sethylene(298) + Cp(g) ln(773/298)
Sethylene(773) = 219.5 + 63.9 = 283.4 J mol-1 K-1
Cp(g) = 67 J mol-1K-1Ethylene
Scyclobutane (298) = 265.4 J mol-1K-1
Scyclobutane(773) = Scyclobutane(298) + Cp(g)ln(773/298)
Scyclobutane(773) = 265.4 + 128.7 = 394 J mol-1K-1
Cyclobutane Cp(g) = 135 J mol-1K-1