Propane (C_3 H_8) is a common fuel source for heating and cooking. The combustio

Question

Propane (C_3 H_8) is a common fuel source for heating and cooking. The combustion of propane gas in a typical cooking situation is governed by the following equation at standard temperature and pressure: C_3H_8 (g) + 5 O_2 (g) rightarrow 3 CO_2 (g) + 4 H_2 O (g) a) Knowing that the above is a combustion reaction, what can you predict (i.e., use logic, not calculations) about the sign of the change in enthalpy (Delta H) for the reaction? Briefly explain your answer. b) Predict the sign of the change in entropy (Delta S) for the reaction. Briefly explain your answer. c) Predict the sign of the change in gibbs free energy (Delta G) and briefly explain if the reaction is always spontaneous, always non-spontaneous, or if it depends on temperature (high/low). Now consider the same combustion of propane with the following thermodynamic data at standard temperature (25 degree C) and pressure: d) Calculate (i.e., use numbers and formulas) the theoretical change in enthalpy (Delta H) for the reaction. e) Calculate the theoretical change in entropy (Delta S) for the reaction. f) Calculate the theoretical change in gibbs free energy (Delta G) for the reaction.

Explanation / Answer

C3H8(g) + 5O2 = 3CO2(g) + 4H2O(g)
Ans d-)
using standard heats of formation:
Horeaction=Hof(products)Hof(Reactants)

dH = [(3mol CO2)(-393.5kj/mol) + (4mol H2O)(-285.8kJ/mol)] - (-103.9)

dH = [-1180.5 - 1143.2] - (-103.9)

dH = - 2323.7 + 103.9

dH = -2219.8 kJ = 2219800 J

Ans e)

Soreaction=Sof(products)Sof(Reactants)

dS = [(3mol CO2)(213.6j/mol-K) + (4mol H2O)(69.9J/mol-K)] - [(1molC3H8)(269.9) + (5 mol O2) (-205)]

dS = [640.8 + 279.6] - [269.9 - 1025]

dS = 1675.5 J/K

Ans f) dG = dH - TdS =( - 2219800 J) - (298 K)*(1675.5 J/K)

dG = 2719099 J = 2719.099 kJ

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