Be sure to answer all parts. consider the combustion of butane gas: c4h10(g) + 13 2 o2(g) → 4co2(g) + 5h2o(g) (a) predict the signs of δs o and δh o . δs o : negative positive δh o : negative positive (b) calculate δg o at 298 k. kj

Answers are:
1) ΔH is negative because that is combustion reaction and heat is released, enthalpy of combustion is -2877.5 kJ/mol.
2) ΔS is positive, because there more molecules on the right side of balanced chemical reaction, standard molar entropy is 310.23 J/mol·K.
3) ΔG = ΔH - TΔS.
ΔG = -2877.5 kJ/mol - 298 K · 310.23 J/mol·K.
ΔG = -2969.95 kJ/mol.

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