Q466-03
Define the term standard enthalpy of formation and illustrate your answer
with an equation including state symbols for the formation of nitric acid.
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The energy change when 1 mole of a substance is formed from
its constituent elements in their states under standard conditions.
The enthalpy of formation of nitric acid:
½H2(g) + ½N2(g) + 1½O2(g) 
HNO3(l) |
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The standard enthalpy of formation of propyne = +186 kJ. It undergoes complete
combustion as follows:
| C3H4(g) + 4O2(g)
3CO2(g) + 2H2O(l) |
Calculate the enthalpy change of this reaction given the following values:
ΔHf
of CO2(g) = -394 kJ
ΔHf of
H2O(l) = -286 kJ |
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Reactants formation enthalpy
1 x C3H4(g) = 1 x 186 = 186 kJ
Total = 186 kJ
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Products formation enthalpy
3 x CO2(g) = 3 x -394 = -1182 kJ
2 x H2O(g) = 2 x -286 = -572 kJ
Total = -1754 kJ
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| Reaction enthalpy ΔH = -1754 - 186
= -1940 kJ |
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Predict and explain whether the value of ΔS
for the combustion of propyne would be negative, close to zero or positive.
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| The combustion reaction has 5 moles of gas on the left hand
side and only 3 moles of gas on the right hand side. The expected
entropy change would be negative as the number of moles of gas
decreases. |
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Propyne reacts with hydrogen as follows:
C3H4(g) + 2H2(g)
C3H8(g) ΔH
= -287 kJ |
Calculate the standard entropy change of this reaction given the following
information:
S of H2
(g) = 131 J K-1 mol-1
S of C3H4
(g) = 248 J K-1 mol-1
S of C3H8
(g) = 270 J K-1 mol-1 |
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Reactant entropy
1 x C3H4(g) = 1 x 248 = 248 J
K-1
2 x H2(g) = 2 x 131 = 262 J K-1
Total = 510 J K-1
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Product entropy
1 x C3H8(g) = 1 x 270 = 270 J
K-1
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The entropy change, ΔS = 270 - 510 = -240
J K-1
The sign of the entropy change is negative as the overall
entropy decreases.
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Calculate the standard free energy change at 298K. ΔG
for the hydrogenation reaction above.
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Standard free energy change, ΔG is given by the
equation:
at 25ºC (298K) ΔG
= -287 - 298 (-0.24)
∴ ΔG
= -287 + 71.5 = -215.5 kJ
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