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

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 G at 298 K if the products of the combustion are CO2(g) and H2O(l) . (G f(C6H6(g))= 129.8 kJmol1, G f(CO2(g))=-394.4 kJmol1 , Gf(H2O(l))= -237.1 kJmol1).

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

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All attempts used; correct answer displayed

Part C

Determine G at 298 K if the products of the combustion are CO2(g) and H2O(g). (Gf(C6H6(g))= 129.8 kJmol1, Gf(CO2(g))=-394.4 kJmol1 , Gf(H2O(g))= -228.6 kJmol1).

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

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

2C6H6(g)+15O2(g)12CO2(g)+6H2O(l)

Explanation / Answer

The reaction taking place is:

2C6H6(g)+15O2(g)12CO2(g)+6H2O(g)

Using relation:

dGrxn = dGprod + dGreac

dGprod = 12*dGCO2 + 6dGH2O = 12*(-394.4) + 6*(-228.6) = -6104.4 kJ

dGreac = 2*dGC6H6 + 15*dGO2 = 2*129.8 + 15*0 = 259.6 kJ

So,

dGrxn = -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 !

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