Question

The enthalpy of formation for C6 H6 (I) is 49.0 kJ/mol. Consider the following reaction.

6 upper C (s, graphite) plus 3 upper H subscript 2 (g) right arrow upper C subscript 6 upper H subscript g (l).

Is the reaction endothermic or exothermic, and what is the enthalpy of reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
exothermic; Delta.Hrxn = 49.0 kJ
exothermic; Delta.Hrxn = –49.0 kJ
endothermic; Delta.Hrxn = 49.0 kJ
endothermic; Delta.Hrxn = –49.0 kJ

Answer 1

To determine whether the reaction is endothermic or exothermic and to calculate the enthalpy change (\( \Delta H_{rxn} \)), we can use the provided formula:

\[
\Delta H_{rxn} = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)}
\]

For the reaction:

\[
6 \text{C (s, graphite)} + 3 \text{H}_2(g) \rightarrow \text{C}_6\text{H}_6(l)
\]

- The product is \( \text{C}_6\text{H}_6(l) \), for which the enthalpy of formation (\( \Delta H_f \)) is given as +49.0 kJ/mol.
- The reactants \( \text{C (s, graphite)} \) and \( \text{H}_2(g) \) are both in their standard states, so their enthalpy of formation values are 0 kJ/mol.

Thus, we can calculate \( \Delta H_{rxn} \):

\[
\Delta H_{rxn} = \Delta H_f \text{(products)} - \Delta H_f \text{(reactants)}
\]
\[
\Delta H_{rxn} = [1 \times 49.0 \text{ kJ/mol}] - [6 \times 0 \text{ kJ/mol} + 3 \times 0 \text{ kJ/mol}]
\]
\[
\Delta H_{rxn} = 49.0 \text{ kJ/mol} - 0
\]
\[
\Delta H_{rxn} = 49.0 \text{ kJ/mol}
\]

Since the calculated enthalpy change is positive (\( +49.0 \) kJ), the reaction is **endothermic**.

Therefore, the correct conclusion is:
- **endothermic; \( \Delta H_{rxn} = 49.0 \, \text{kJ}**.