Q466-04
Hex-1-ene gas, C
6H
12, burns in oxygen to produce carbon dioxide and water
vapour. Write an equation to represent this reaction.
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C6H12(g) + 9O2(g)
6CO2(g) + 6H2O(g) |
|
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Use the data below to calculate the values of ΔHc
and ΔSc
for the combustion of hex-1-ene.
Substance
|
O2(g) |
C6H12(g) |
CO2(g) |
H2O(g) |
Standard enthalpy of formation, ΔHf /kJmol -1
|
0.0 |
-43 |
-394 |
-242 |
Entropy, S /JK -1
mol -1
|
205 |
385 |
214 |
189 |
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Reactants enthalpy of formation
1 x C6H12(g) = -43 kJ
|
Products enthalpy of formation
6 x CO2(g) = 6 x -394 = -2364 kJ
6 x H2O(g) = 6 x -242 = -1452 kJ
Total = -3816 kJ
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The reactants are 'unformed' so change the sign of
the reactants and add it to the products:
Reaction enthalpy = Products ΔHf - ΔReactants
Hf = -3816 - (-43) = -3773
kJ
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Reactants entropy
1 x C6H12(g) = 1 x 385 = 385
JK-1
9 x O2(g) = 9 x 205 = 1845 JK-1
Total = 2230 JK-1
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Products entropy
6 x CO2(g) = 6 x 214 = 1284 JK-1
6 x H2O(g) = 6 x 189 = 1134 JK-1
Total = 2418 JK-1
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There is an increase in entropy by 2418 - 2230 = +188
JK-1
ΔSc
= +188 JK-1
|
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Calculate the standard free energy change for the combustion of hex-1-ene.
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Using the Gibbs free energy equation:
At 298 K: ΔG = -3773 - 298(0.188) = -3773 - 56 = -3829
kJ
|
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State and explain whether or not the combustion of hex-1-ene is spontaneous
at 25ºC
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The final value for Gibbs free energy is negative at 298K,
so the reaction is spontaneous. |
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