Answer:
The answer is a. -277.6 kJ/mol
Explanation:
We have to arrange and sum the given equations to obtain the desired reaction as follows:
3 H₂O(l) + 2 CO₂(g) → 3 O₂(g) + C₂H₅OH(l) ΔH1= (-1) x (-1367 KJ/mol)
2 C(s) + 2 O₂(g)→ 2 CO₂(g) ΔH2= 2 x (-393.5 KJ/mol)
3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l) ΔH3= 3 x (-285.8 KJ/mol)
In the first equation (ΔH1) , we multiply the enthalphy by (-1) because we use the reverse reaction. In the second and third equations (ΔH2 and ΔH3), we multiply by 2 and 3, respectively. Once we sum and cancel the terms that are repeated in both sides of the reactions, we obtain:
3H₂(g) + 2C(s) + ½O₂(g) → C₂H₅OH(l)
The total enthalphy is:
ΔH= ΔH1 + ΔH2 + ΔH3
ΔH= (-1) x (-1367 KJ/mol) + 2 x (-393.5 KJ/mol) + 3 x (-285.8 KJ/mol)
ΔH= -277.4 KJ/mol
The nearest answer between the given options is option a (-277.6 KJ/mol)
