Answer:
-241.826 kJ·mol⁻¹; -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous
Explanation:
½O₂(g) + H₂(g) ⟶ H₂O(g)
ΔHf°/kJ·mol⁻¹: 0 0 -241.826
S°/J·K⁻¹mol⁻¹: 205.0 130.6 188.7
1. ΔᵣH
ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹
2. ΔᵣS
ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹
3. ΔS(univ)
4. Spontaneity
ΔG is negative, so the reaction is spontaneous.
Answer:
Butanoic acid.
Explanation:
Hello,
In this case, when a primary alcohol such as 1-butanol (OH is bonded to a primary carbon) is oxidized in the presence of a strong oxidizing media such as potassium dichromate (K2Cr2O7) and sulfuric acid, the stepwise oxidation goes to the corresponding aldehyde with a further oxidation to the corresponding carboxylic acid:
Therefore, on the attached picture you can find that the formed aldehyde is butanal and the inly organic product, due to the strong oxidizing media is finally butanoic acid.
Best regards.
