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 = -296 kJ mol–1C2H5OH +3O2 →  3H2O (g) + 2CO2(g)       ΔHo =–1369 kJ mol–1 Use-full formula ΔrxnHo=  ∑ ΔfHoproducts -  ∑ ΔfHoreactants

The enthalpy of formation ΔfHo for C2H5OH(l) can be calculated using the given thermochemical data and the formula ΔrxnHo= ∑ ΔfHoproducts - ∑ ΔfHoreactants.

First, let's write down the enthalpy changes for the given reactions:

1. C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1
2. H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -296 kJ mol–1
3. C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1

We can calculate the enthalpy of formation for C2H5OH(l) by rearranging the formula to solve for ΔfHo of C2H5OH(l):

ΔfHo(C2H5OH) = ∑ ΔfHoproducts - ΔrxnHo

The enthalpy of formation for the products of the third reaction (3H2O and 2CO2) can be calculated using the enthalpy changes from the first two reactions:

ΔfHo(3H2O) = 3 * ΔHo(H2O) = 3 * -296 kJ mol–1 = -888 kJ mol–1 ΔfHo(2CO2) = 2 * ΔHo(CO2) = 2 * -393 kJ mol–1 = -786 kJ mol–1

Substituting these values into the formula gives:

ΔfHo(C2H5OH) = (-888 kJ mol–1 + -786 kJ mol–1) - (-1369 kJ mol–1) ΔfHo(C2H5OH) = -1674 kJ mol–1 + 1369 kJ mol–1 ΔfHo(C2H5OH) = -305 kJ mol–1

So, the enthalpy of formation for C2H5OH(l) is -305 kJ mol–1.

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 of a reaction is the sum of the enthalpy changes of the individual steps into which a reaction can be divided.

First, let's write down the reaction for which we want to find the enthalpy of formation:

C2H5OH(l) = 2C(s) + 3H2(g) + 1/2 O2(g)

Now, let's rearrange the given reactions:

1. C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1 We need 2 moles of CO2 for our target reaction, so we multiply the entire reaction by 2: 2C(s) + 2O2(g) → 2CO2(g) ΔHo = –2*393 kJ mol–1 = -786 kJ mol–1

2. H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -296 kJ mol–1 We need 3 moles of H2O for our target reaction, so we multiply the entire reaction by 3: 3H2(g) + 3/2 O2(g) → 3H2O (g) ΔHo = 3*-296 kJ mol–1 = -888 kJ mol–1

3. C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1 This reaction needs to be reversed to match our target reaction: 3H2O (g) + 2CO2(g) → C2H5OH +3O2 ΔHo = 1369 kJ mol–1

Now, we add up all the reactions:

2C(s) + 2O2(g) + 3H2(g) + 3/2 O2(g) + 3H2O (g) + 2CO2(g) → 2CO2(g) + 3H2O (g) + C2H5OH +3O2

This simplifies to:

C2H5OH(l) = 2C(s) + 3H2(g) + 1/2 O2(g)

The total enthalpy change ΔHo for this reaction is the sum of the enthalpy changes of the individual reactions:

ΔHo = -786 kJ mol–1 - 888 kJ mol–1 + 1369 kJ mol–1 = -305 kJ mol–1

So, the enthalpy of formation ΔfHo for C2H5OH(l) is -305 kJ mol–1.