Calculate the standard free energy change at 25 C for 2CgH18 (I) + 2502 (g) 16CO

Question

Calculate the standard free energy change at 25 C for 2CgH18 (I) + 2502 (g) 16CO2(g) + 18H20() AG (J/mol-K) (kJ/mol) (kJ/mol) CgH18 358 -249.96.4 O2 3) 205.14 0 CO2(3) 213.74 -393.51 -394.36 H20 ( 69.91-285.83 -237.13 0 Is this a spontaneous reaction at the given temperature? C -10591 kJ/molrxn. It is spontaneous at the given temperature. C 10565 kJ/molrxn. It is spontaneous at the given temperature. C 10565 kJ/molrxn. It is nonspontaneous at the given temperature. C10591 kJ/molrxn. It is spontaneous at the given temperature. C -10565 kJ/molrxn. It is spontaneous at the given temperature C -10591 kJ/molrxn. It is nonspontaneous at the given temperature. C -10565 kJ/molrxn. It is nonspontaneous at the given temperature. 10591 kJ/molrxn. It is nonspontaneous at the given temperature.

Explanation / Answer

1)

we have:

Gof(C8H18(l)) = 6.4 KJ/mol

Gof(O2(g)) = 0.0 KJ/mol

Gof(CO2(g)) = -394.36 KJ/mol

Gof(H2O(l)) = -237.13 KJ/mol

we have the Balanced chemical equation as:

2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(l)

deltaGo rxn = 16*Gof(CO2(g)) + 18*Gof(H2O(l)) - 2*Gof( C8H18(l)) - 25*Gof(O2(g))

deltaGo rxn = 16*(-394.36) + 18*(-237.13) - 2*(6.4) - 25*(0.0)

deltaGo rxn = -10591 KJ/mol

since deltaGo rxn is negative, it is spontaneous

Answer: option 1

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