Sure! Let's solve the problem step by step.
1. Write Down the Given Information:
• The given reaction is:
[tex]\[
6 \, C(\text{s, graphite}) + 3 \, H_2(\text{g}) \rightarrow C_6H_6(\text{l})
\][/tex]
• The enthalpy of formation for [tex]\( C_6H_6(\text{l}) \)[/tex] is [tex]\( \Delta H_f = 49.0 \, \text{kJ/mol} \)[/tex].
2. Understand Enthalpy of Formation:
• The enthalpy of formation ([tex]\( \Delta H_f \)[/tex]) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
• For elements in their standard states, the enthalpy of formation is zero:
[tex]\[
\Delta H_f(C(\text{s, graphite})) = 0 \, \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H_f(H_2(\text{g})) = 0 \, \text{kJ/mol}
\][/tex]
3. Apply the Enthalpy of Reaction Formula:
The enthalpy change of the reaction ([tex]\( \Delta H_{\text{rxn}} \)[/tex]) can be found using the formula:
[tex]\[
\Delta H_{\text{rxn}} = \sum \left( \Delta H_{f, \text{products}} \right) - \sum \left( \Delta H_{f, \text{reactants}} \right)
\][/tex]
4. Calculate the Enthalpy of the Products:
Since benzene ([tex]\(C_6H_6(\text{l})\)[/tex]) is the only product and its formation enthalpy is given as [tex]\(49.0 \, \text{kJ/mol}\)[/tex]:
[tex]\[
\sum \left( \Delta H_{f, \text{products}} \right) = \Delta H_f (C_6H_6(\text{l})) = 49.0 \, \text{kJ/mol}
\][/tex]
5. Calculate the Enthalpy of the Reactants:
Since carbon in its graphite form and hydrogen gas are in their standard states:
[tex]\[
\sum \left( \Delta H_{f, \text{reactants}} \right) = 6 \cdot \Delta H_f(C(\text{s})) + 3 \cdot \Delta H_f(H_2(\text{g})) = 6 \cdot 0 + 3 \cdot 0 = 0 \, \text{kJ/mol}
\][/tex]
6. Determine the Enthalpy Change of the Reaction:
Using the enthalpy of reaction formula:
[tex]\[
\Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol} - 0 \, \text{kJ/mol} = 49.0 \, \text{kJ/mol}
\][/tex]
7. Identify if the Reaction is Endothermic or Exothermic:
• A positive enthalpy change ([tex]\(\Delta H_{\text{rxn}} > 0\)[/tex]) indicates that the reaction is endothermic since energy is absorbed from the surroundings.
8. Final Answer:
The reaction is endothermic with an enthalpy change of [tex]\( \Delta H_{\text{rxn}} = 49.0 \, \text{kJ/mol} \)[/tex].
Therefore, the correct answer is:
[tex]\[
\text{endothermic; } \Delta H_{\text{rxn}} = 49.0 \, \text{kJ}
\][/tex]
