Given the values of ΔGfo given below in kJ/mol, calculate the value of ΔGo in kJ for the combustion of 1 mole of butane to form carbon dioxide and liquid water. ΔGfo (C4H10(g)) = -17 ΔGfo (CO2(g)) = -395 ΔGfo (H2O(l)) = -238

Now the change in the Gibbs free energy for this reaction is computed via (don't forget that oxygen's Gibbs free energy of formation is 0 since it is an element):

Δ[tex]_rG=[/tex]4Δ[tex]_fG_{CO_2}+[/tex]5Δ[tex]_fG_{H_2O}[/tex]-Δ[tex]_fG_{C_4H_{10}}[/tex]

Δ[tex]_rG=4(-395kJ/mol)+5(-238kJ/mol)-(-17kJ/mol)[/tex]

Δ[tex]_rG=-2753kJ/mol[/tex]

Now, since the calculation is done for 1 mol of butane, the change in the Gibbs free energy for this fuel is:

Δ[tex]_rG=-2753kJ[/tex]

Best regards.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-07T08:08:50+01:00

The change in free energy of the reaction is -5506 KJ/mol

Given the equation of the reaction;

2C4H10(g) + 13O2(g) -------> 8CO2(g) + 10H2O(l)

We have the following information from the question;

ΔGf C4H10(g)= -17kJ/mol

ΔGf O2(g) = 0 KJ/mol

ΔGfCO2(g) = -395 KJ/mol

ΔGf H2O(l) = -238 KJ/mol

Now, we can obtain the ΔGrxn as follows;

[8(-395 KJ/mol) + 10(-238 KJ/mol)] - [2( -17kJ/mol) + 13 (0 KJ/mol)]

ΔGrxn =[(-3160) + (-2380)] - [(-34) + 0]

ΔGrxn = (-5540) + 34

ΔGrxn = -5506 KJ/mol

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