The change in entropy for a reaction is 3.0 J/(mol·K) at 25°C. Calculate the change in free energy for the reaction when ΔH = –394 kJ/mol and determine whether it will occur spontaneously at this temperature.

Question

Is the answer -395 kJ/mol?

and second question: using this equation below to calculate the change in enthalpy for this reaction: C8H18(l) + 25/2O2(g) --> 8C02(g) + 9H2O(g)

H2(g) + 1/2O2(g) --> H2O(g) DeltaH = -241.8 kJ
C(s) + O2(g) --> CO2(g) DeltaH = -393.5 kJ
8C(s) +9H2(g) --> C8H18(l) DeltaH = -224.13 kJ

Is the answer -5100 kJ? Sorry for asking a lot I'm not good at these questions :(

Answer 1

1.
dG = dH - TdS
dG = -394 kJ - 298(0.003) = -394.89 kJ = -395 kJ.

2. Yes, I obtained -5100.2 kJ which rounds to -5100 kJ.