1. Write the formation reaction of ethanol

The standard heat of formation of ethanol is for:

$2\,\mathrm{C}(s) + 3\,\mathrm{H_2}(g) + \frac{1}{2}\,\mathrm{O_2}(g) \rightarrow \mathrm{C_2H_5OH}(l)$

We need to find $\Delta H_f^\circ[\mathrm{C_2H_5OH}(l)]$.

2. Write the combustion reaction of ethanol

$\mathrm{C_2H_5OH}(l) + 3\,\mathrm{O_2}(g) \rightarrow 2\,\mathrm{CO_2}(g) + 3\,\mathrm{H_2O}(l)$

Given:

$\Delta H_c^\circ[\mathrm{C_2H_5OH}(l)] = -1234.7\;\mathrm{kJ\,mol^{-1}}$

3. Use heats of combustion of elements to get heats of formation of products

For carbon:

$\mathrm{C}(s) + \mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g), \qquad \Delta H = -393.5\;\mathrm{kJ\,mol^{-1}}$

So,

$\Delta H_f^\circ(\mathrm{CO_2}) = -393.5\;\mathrm{kJ\,mol^{-1}}$

For hydrogen:

$\mathrm{H_2}(g) + \frac{1}{2}\,\mathrm{O_2}(g) \rightarrow \mathrm{H_2O}(l), \qquad \Delta H = -241.8\;\mathrm{kJ\,mol^{-1}}$

So,

$\Delta H_f^\circ(\mathrm{H_2O}) = -241.8\;\mathrm{kJ\,mol^{-1}}$

4. Apply Hess's law to ethanol combustion

For the combustion reaction,

$\Delta H_{rxn} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

Thus,

$-1234.7 = \left[2(-393.5) + 3(-241.8)\right] - \Delta H_f^\circ(\mathrm{C_2H_5OH})$

since $\Delta H_f^\circ(\mathrm{O_2})=0$.

5. Calculate

First,

$2(-393.5) = -787.0$

$3(-241.8) = -725.4$

So,

$-1234.7 = -1512.4 - \Delta H_f^\circ(\mathrm{C_2H_5OH})$

Therefore,

$-\Delta H_f^\circ(\mathrm{C_2H_5OH}) = -1234.7 + 1512.4 = 277.7$

Hence,

$\Delta H_f^\circ(\mathrm{C_2H_5OH}) = -277.7\;\mathrm{kJ\,mol^{-1}}$

Nearest integer:

$\boxed{-278\;\mathrm{kJ\,mol^{-1}}}$

Since the blank is in

$(-)\;\underline{\hspace{2cm}}\;\mathrm{kJ\,mol^{-1}}$

the integer to be filled is:

$\boxed{278}$

6. Comparison with stored answer

Stored correct answer = $278$.

This matches our derived answer.