Certainly! Let's walk through the process of determining whether the given reaction is endothermic or exothermic, and calculating the enthalpy of reaction.
### Given Information:
1. The enthalpy of formation (\(\Delta H_f\)) of \(C_6H_6(l)\) is \(49.0 \, \text{kJ/mol}\).
2. The reaction is:
[tex]\[
6 \, C \text{ (s, graphite)} + 3 \, H_2 \text{ (g)} \rightarrow C_6H_6 \text{ (l)}
\][/tex]
### Step-by-Step Solution:
1. Enthalpy of Formation for Reactants:
- The enthalpy of formation (\(\Delta H_f\)) for elements in their standard state (like graphite carbon and hydrogen gas) is zero.
- Therefore, for the reactants \(6 \, C \text{ (s, graphite)}\) and \(3 \, H_2 \text{ (g)}\):
[tex]\[
\sum(\Delta H_{f, \text{reactants}}) = 0 \, \text{kJ/mol}
\][/tex]
2. Enthalpy of Formation for the Product:
- Given the enthalpy of formation for \(C_6H_6(l)\) is:
[tex]\[
\Delta H_{f, \text{product}} = 49.0 \, \text{kJ/mol}
\][/tex]
3. Calculate the Enthalpy of Reaction:
Using the formula for the enthalpy of reaction:
[tex]\[
\Delta H_{\text{rxn}} = \sum \left( \Delta H_{f, \text{products}} \right) - \sum \left( \Delta H_{f, \text{reactants}} \right)
\][/tex]
We substitute the values:
[tex]\[
\Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol} - 0 \, \text{kJ/mol} = 49.0 \, \text{kJ/mol}
\][/tex]
4. Determine if the Reaction is Endothermic or Exothermic:
- A reaction is endothermic if it absorbs energy from the surroundings (i.e., \(\Delta H_{\text{rxn}} > 0\)).
- A reaction is exothermic if it releases energy to the surroundings (i.e., \(\Delta H_{\text{rxn}} < 0\)).
Since \(\Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol}\) is positive, the reaction is endothermic.
### Conclusion:
The reaction is endothermic with an enthalpy of reaction \(\Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol}\).
- Answer: Endothermic; \(\Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol}\).
Hence, the correct selection from the given options is:
endothermic; [tex]\(\Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol}\)[/tex].
