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system

The standard enthalpy of formation of a substance equals the heat change when 1 mole of the substance is formed from its constituent elements in their standard states (at a pressure of 1 atm and a specified temperature).

Determine the heat change of the reaction:

$$6 \mathrm{C}(\text{graphite}) + 3 \mathrm{H}_2(g) = \mathrm{C}_6\mathrm{H}_6(g) + Q_{\text{formation}}(\mathrm{C}_6\mathrm{H}_6}).$$

Given:

$$
\mathrm{C}_2\mathrm{H}_2(g) = 2 \mathrm{C}(\text{graphite}) + \mathrm{H}_2(g) + 226.7 \text{ kJ;}
$$

$$
3 \mathrm{C}_2\mathrm{H}_2(g) = \mathrm{C}_6\mathrm{H}_6(g) + 631.1 \text{ kJ;}
$$

$$
\mathrm{C}_6\mathrm{H}_6(g) = \mathrm{C}_6\mathrm{H}_6(\text{l}) - 33.9 \text{ kJ.}
$$

The heat change of the reaction is:

$$
Q_{\text{formation}} (\mathrm{C}_6\mathrm{H}_6) = 226.7 \cdot (-3) + 631.1 - 33.9 = -82.9 \text{ kJ/mol}.
$$

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