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.