Question

Part A

Write an equation for the combustion of one mole of benzene, C6H6(g).

Express your answer as a chemical equation. Identify all of the phases in your answer.

2C6H6(g)+15O2(g)→12CO2(g)+6H2O(l)

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Correct

Part B

Determine ΔG∘ at 298 K if the products of the combustion are CO2(g) and H2O(l) . (ΔG∘ f(C6H6(g))= 129.8 kJmol−1, ΔG∘ f(CO2(g))=-394.4 kJmol−1 , ΔG∘f(H2O(l))= -237.1 kJmol−1).

Express your answer to four significant figures and include the appropriate units.

ΔG∘ =

-3208 kJ

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All attempts used; correct answer displayed

Part C

Determine ΔG∘ at 298 K if the products of the combustion are CO2(g) and H2O(g). (ΔG∘f(C6H6(g))= 129.8 kJmol−1, ΔG∘f(CO2(g))=-394.4 kJmol−1 , ΔG∘f(H2O(g))= -228.6 kJmol−1).

Express your answer to four significant figures and include the appropriate units.

I got -6364 kJ and it says that's wrong.

Answer 1

The reaction taking place is:

2C₆H₆(g)+15O₂(g)→12CO₂(g)+6H₂O(g)

Using relation:

dG_rxn = dG_prod + dG_reac

dG_prod = 12*dG_CO2 + 6dG_H2O = 12*(-394.4) + 6*(-228.6) = -6104.4 kJ

dG_reac = 2*dG_C6H6 + 15*dG_O2 = 2*129.8 + 15*0 = 259.6 kJ

So,

dG_rxn = -6104.4-(259.6) = -6364 kJ

This is the answer when we write the equation for combustion of two moles of benzene. But the question has asked for combustion of one mole of benzene.

So the correct answer for this part is: -6364/2 = -3182 kJ

Hope this helps !