Using the following thermochemical data, what is the change in enthalpy for the following reaction:
3h2(g) + 2c(s) + ½o2(g) → c2h5oh(l)
c2h5oh(l)+3o2(g)→2co2(g)+3h2o(l), δh = –1367 kj/mol
c(s)+o2(g)→co2(g), δh = –393.5 kj
h2(g)+½o2(g)→h2o(l), δh = –285.8 kj

a. -277.6 kj/mol
b. -194.7 kj/mol
c. 194.7 kj/mol
d. 486 kj/mol

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)

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