Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
Answer:
Explanation:1. NaNH2 (1-Butene)
CH3CH2CH2CH2Cl --------------> CH3CH2CH=CH2 + HCl (elimination reaction)
2. Br2, CCl4
CH3CH2CH=CH2 ---------------> CH3CH2CH(Br)CH2Br (Simple addition Reaction)
3. NaNH2 (1-Butyne)
CH3CH2CH(Br)CH2Br ----------------> CH3CH2C≡CH + 2HBr
Sodamide (NaNH2) is a very strong base and generally results in Terminal Alkynes when treated with Vicinal Dihalides.
Alcoholic KOH on the other hand results in the formation of Alkynes with triple bonds in the middle of the molecule.
