4.66 - Thermodynamics study questions - 3

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.
show

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)

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
show
 Reactants formation enthalpy 1 x C3H4(g) = 1 x 186 = 186 kJ Total = 186 kJ Products formation enthalpy 3 x CO2(g) = 3 x -394 = -1182 kJ 2 x H2O(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.

show
 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:

 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
show
 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 Product entropy 1 x C3H8(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.

show

Standard free energy change, ΔG is given by the equation:

 ΔG = ΔH - TΔS

at 25ºC (298K) ΔG = -287 - 298 (-0.24)

∴ ΔG = -287 + 71.5 = -215.5 kJ