Consider the combustion of butane gas and predict the signs of ΔS, ΔH and ∆G.
C₄H₁₀ (g) + 13/2 O₂ (g)  ⟶  4 CO₂ (g) + 5 H₂O (g)

• A. ΔS = (−), ΔH = (+), ∆G = (+)
• B. ΔS = (+), ΔH = (−), ∆G = (+)
• C. ΔS = (+), ΔH = (+), ∆G = (−)
• D. ΔS = (+), ΔH = (−), ∆G = (−)

Answer: (D)

Verified step by step guidance

1. Start by understanding the chemical reaction: the combustion of butane (C4H10) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). This is an exothermic reaction, meaning it releases heat.
2. Consider the change in entropy (ΔS). Combustion reactions typically result in an increase in the number of gas molecules, which leads to an increase in disorder or entropy. Here, the reaction goes from 1 molecule of butane and 6.5 molecules of oxygen to 4 molecules of carbon dioxide and 5 molecules of water, totaling 9 molecules. Therefore, ΔS is positive.
3. Evaluate the change in enthalpy (ΔH). Combustion reactions are exothermic, meaning they release heat. This implies that the enthalpy change is negative, as energy is released to the surroundings.
4. Consider the change in Gibbs free energy (ΔG). For a spontaneous reaction, ΔG is negative. Since combustion is a spontaneous process under standard conditions, ΔG will be negative.
5. Combine these insights: ΔS is positive due to increased disorder, ΔH is negative due to heat release, and ΔG is negative indicating spontaneity. Therefore, the correct signs are ΔS = (+), ΔH = (−), ∆G = (−).