Given the following thermochemical data, what is the enthalpy of formation ΔfHo for C2H5OH(l) C(s)  +  O2(g)  →  CO2(g)                               ΔHo = –393 kJ mol–1H2(g)  +  1/2 O2(g)    →  H2O (g)                  ΔHo = -295 kJ mol–1C2H5OH +3O2 →  3H2O (g) + 2CO2(g)       ΔHo =–1369 kJ mol–1 Use-full formula ΔrxnHo=  ∑ ΔfHoproducts -  ∑ ΔfHoreactants

To find the enthalpy of formation ΔfHo for C2H5OH(l), we need to rearrange the given reactions and use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction.

First, we need to write down the formation reactions for all the compounds involved:

1. C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1
2. H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -295 kJ mol–1
3. C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1

The formation reaction for C2H5OH(l) is what we're trying to find. We can write it as:

4. 2C(s) + 3H2(g) + 1/2 O2(g) → C2H5OH(l) ΔfHo = ?

Now, we can express reaction 4 as a combination of reactions 1, 2, and 3.

We need 2 moles of CO2 for reaction 4, so we multiply reaction 1 by 2:

5. 2C(s) + 2O2(g) → 2CO2(g) ΔHo = –2*393 kJ mol–1 = -786 kJ mol–1

We need 3 moles of H2O for reaction 4, so we multiply reaction 2 by 3:

6. 3H2(g) + 3/2 O2(g) → 3H2O (g) ΔHo = 3*-295 kJ mol–1 = -885 kJ mol–1

Adding reactions 5 and 6 gives us:

7. 2C(s) + 3H2(g) + 2O2(g) → 2CO2(g) + 3H2O (g) ΔHo = -786 - 885 = -1671 kJ mol–1

Now, we subtract reaction 3 from reaction 7 to get reaction 4:

8. 2C(s) + 3H2(g) + 1/2 O2(g) → C2H5OH(l) ΔfHo = -1671 - (-1369) = -302 kJ mol–1

So, the enthalpy of formation ΔfHo for C2H5OH(l) is -302 kJ mol–1.