[Rudul]

In a calorimetry experiment the following data were obtained.

CH2CH2(g) + H2O(ℓ) → C2H5OH(ℓ)     ΔH = -288.9kJ

C(s) + O2(g) → CO2(g)     ΔH = -395.3kJ

2 H2(g) + O2(g) → 2 H2O(ℓ)     ΔH = -587.4kJ

CH2CH2(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(ℓ)      ΔH = -1441.1kJ

Using only this data calculate the molar enthalpy of formation for ethanol, C2H5OH(ℓ).

[Borek]

Please read the forum rules.

[Vidya]

Using only this data calculate the molar enthalpy of formation for ethanol, C2H5OH(ℓ).

Do you know meaning of enthalpy of formation?
Once that is clear write equation which represents formation of 1 mole of ethanol.
Now use Hess's law on the given equations so that when you add all of them you get reaction representing the formation enthalpy of ethanol.

[Orcio_87]

@Rudul - we know that:

(a) CH2CH2(g) + H2O(ℓ) → C2H5OH(ℓ)     ΔH = -288.9kJ

(b) C(s) + O2(g) → CO2(g)     ΔH = -395.3kJ

(c) 2 H2(g) + O2(g) → 2 H2O(ℓ)     ΔH = -587.4kJ

(d) CH2CH2(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(ℓ)      ΔH = -1441.1kJ

and we are looking for:

(e) 2 C + 3 H2 + 1/2 O2 → C2H5OH

If you add (a) + 2(b) + 1,5(c) - (d) then you will get this:

CH2=CH2 + H2O + 2 C + 2 O2 + 3 H2 + 1,5 O2 + 2 CO2 + 2 H2O → C2H5OH + 2 CO2 + 3 H2O + CH2=CH2 + 3 O2

which is equal to:

(e) 2 C + 3 H2 + 1/2 O2 → C2H5OH

Now you have to calculate the result of reaction (e) (- 518 kJ / mol).