Question

Find entropy, enthalpy, Gibbs free energy of reaction at 298K. Find if it is exothermic and whether it is spontaneous? C3H8 + 502 -> 3CO2 + 4H2O

Hf (kJ/mol)
C3H8 = -103.85
O2 = 0
C02 = -393.5
H20 = -286

S (J/Mol K)
C3H8 = 269.9
O2 = 205
C02 = 214
H20 = 70

Answer 1

Here's what I get.

Explanation:

You have the following information.

`color(white)(mmmmmmmmm)"C"_3"H"_8"(g)" + "5O"_2"(g)" → "3CO"_2"(g)" + "4H"_2"O(l)"`
`Δ_fH"/kJ·mol"^"-1":color(white)(ml) "-103.85" color(white)(mmml)0 color(white)(mmmll)"-393.5" color(white)(mmm)"-286"`
`S"/J·K"^"-1""mol"^"-1": color(white)(mml)269.9color(white)(mmm) 205 color(white)(mmmm)214color(white)(mmmmm) 70`

Calculate the enthalpy of reaction

The formula for enthalpy of reaction is

`color(blue)(|bar(ul(color(white)(a/a) Δ_rH = sumΔ_fH_text(products) - sumΔ_fH_text(reactants)color(white)(a/a)|)))" "`

∴ `Δ_rH = [3("-393.5") + 4("-286") - 1("-103.85")]color(white)(l) "kJ" = "-2221 kJ"`

`ΔH < 0`, so the reaction is exothermic.

Calculate the entropy of reaction

The formula for the entropy of reaction is

`color(blue)(|bar(ul(color(white)(a/a)Δ_rS = sumS_text(products) - sumS_text(reactants)color(white)(a/a)|)))" "`

∴ `Δ_rS = (3×214 + 4×70 - 1×269.9 - 5× 205)color(white)(l) "J/K" = "-373 J/K"`

Calculate the Gibbs free energy change for the reaction

The formula for free energy change is

`color(blue)(|bar(ul(color(white)(a/a)ΔG = ΔH - TΔScolor(white)(a/a)|)))" "`

∴ `Δ_rG = "-2220.65 kJ" - 298.15 color(red)(cancel(color(black)("K"))) × ("-0.3729 kJ"·color(red)(cancel(color(black)("K"^"-1")))) = "-2220.65 kJ" + "111.18 kJ" = "-2109 kJ"`

`ΔG < 0`, so the reaction is spontaneous.

Summary

`Δ_rH = "-2221 kJ"`; exothermic
`Δ_rS = "-373 J/K"`
`Δ_rG = "-2109 kJ"`; spontaneous