Q46603
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:
½H_{2}(g) + ½N_{2}(g) + 1½O_{2}(g)
HNO_{3}(l) 


The standard enthalpy of formation of propyne = +186 kJ. It undergoes complete
combustion as follows:
C_{3}H_{4}(g) + 4O_{2}(g)
3CO_{2}(g) + 2H_{2}O(l) 
Calculate the enthalpy change of this reaction given the following values:
ΔHf
of CO_{2}(g) = 394 kJ
ΔHf of
H_{2}O(l) = 286 kJ 
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Reactants formation enthalpy
1 x C_{3}H_{4}(g) = 1 x 186 = 186 kJ
Total = 186 kJ

Products formation enthalpy
3 x CO_{2}(g) = 3 x 394 = 1182 kJ
2 x H_{2}O(g) = 2 x 286 = 572 kJ
Total = 1754 kJ

Reaction enthalpy ΔH = 1754  186
= 1940 kJ 


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. 

Propyne reacts with hydrogen as follows:
C_{3}H_{4}(g) + 2H_{2}(g)
C_{3}H_{8}(g)ΔH
= 287 kJ 
Calculate the standard entropy change of this reaction given the following
information:
S of H_{2}
(g) = 131 J K^{1} mol^{1}
S of C_{3}H_{4}
(g) = 248 J K^{1} mol^{1}
S of C_{3}H_{8}
(g) = 270 J K^{1} mol^{1} 
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Reactant entropy
1 x C_{3}H_{4}(g) = 1 x 248 = 248 J
K^{1}
2 x H_{2}(g) = 2 x 131 = 262 J K^{1}
Total = 510 J K^{1}

Product entropy
1 x C_{3}H_{8}(g) = 1 x 270 = 270 J
K^{1}

The entropy change, ΔS = 270  510 = 240
J K^{1}
The sign of the entropy change is negative as the overall
entropy decreases.



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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