
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

The standard enthalpy of formation (ΔH_f°) of a substance is equal to the heat effect of the formation reaction of 1 mole of the substance from simple substances in their standard states (at 1 atm pressure and a given temperature). Therefore, it is necessary to find the heat effect of the reaction:

$$
\underset{\text {graphite}}{6 \mathrm{C}(\kappa)} + 3 \mathrm{H}_{2}(g) = \mathrm{C}_{6} \mathrm{H}_{6} (l) + \mathrm{Q}_{\text{formation}}\left(\mathrm{C}_{6} \mathrm{H}_{6}(l)\right) (4)
$$

According to Hess's law, the heat effect of the reaction depends only on the types and states of the reactants and products and does not depend on the path of the transition.

Hess's law allows dealing with thermochemical equations like algebraic expressions, i.e., based on it, by combining equations of reactions with known heat effects, one can calculate the unknown heat effect of the overall reaction.

Thus, we obtain:

$$
\mathrm{C}_{2}\mathrm{H}_{2}(g) = \underset{\text {graphite}}{2 \mathrm{C}(\kappa)} + \mathrm{H}_{2}(g) + 226.7 \text { kJ; } \quad -3
$$

$$
\begin{array}{lll} 
3 \mathrm{C}_{2} \mathrm{H}_{2}(g) & = \mathrm{C}_{6} \mathrm{H}_{6}(l) + 631.1 \text { kJ;} & 1 \\
\mathrm{C}_{6} \mathrm{H}_{6}(l) & = \mathrm{C}_{6} \mathrm{H}_{6}(l) - 33.9 \text { kJ;} & 1 
\end{array}
$$

$$
\underset{\text {graphite}}{6 \mathrm{C}(\kappa)} + 3 \mathrm{H}_{2}(g) = \mathrm{C}_{6} \mathrm{H}_{6}(l) + \mathrm{Q}_{\text{formation}}\left(\mathrm{C}_{6} \mathrm{H}_{6}(l)\right);
$$

$$
\mathrm{Q}_{\text{formation}}\left(\mathrm{C}_{6} \mathrm{H}_{6}(l)\right) = 226.7 \cdot (-3) + 631.1 - 33.9 = -82.9 \text{ kJ/mol}.
$$

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