To determine whether the given reaction is endothermic or exothermic and to find the enthalpy of reaction, we will use the enthalpy of formation values and apply the given formula:
Equation:
[tex]\[
\Delta H_{\text{reaction}} = \sum \left( \Delta H_{\text{f,products}} \right) - \sum \left( \Delta H_{\text{f,reactants}} \right)
\][/tex]
Given:
- The enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) for [tex]\(C_6H_6(l)\)[/tex] is [tex]\(49.0 \, \text{kJ/mol}\)[/tex].
- The reactants are elemental carbon (graphite) and hydrogen gas [tex]\(H_2(g)\)[/tex]. Since these are elements in their standard states, their enthalpies of formation are [tex]\(0 \, \text{kJ/mol}\)[/tex].
Let’s calculate the enthalpy change for the reaction step-by-step:
### Step 1: Identify the enthalpies of formation of products and reactants
- [tex]\( \Delta H_f \)[/tex] for [tex]\( C_6H_6(l) \)[/tex] = [tex]\( 49.0 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] for [tex]\( C (s, \text{graphite}) \)[/tex] = [tex]\( 0 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] for [tex]\( H_2(g) \)[/tex] = [tex]\( 0 \, \text{kJ/mol} \)[/tex]
### Step 2: Apply the formula
The enthalpy of reaction [tex]\(\Delta H_{\text{reaction}}\)[/tex] is given by:
[tex]\[
\Delta H_{\text{reaction}} = \Delta H_f(C_6H_6(l)) - [6 \cdot \Delta H_f(C(s, \text{graphite})) + 3 \cdot \Delta H_f(H_2(g))]
\][/tex]
Plugging in the values:
[tex]\[
\Delta H_{\text{reaction}} = 49.0 \, \text{kJ/mol} - [6 \cdot 0 + 3 \cdot 0] \\
\Delta H_{\text{reaction}} = 49.0 \, \text{kJ/mol} - 0 \\
\Delta H_{\text{reaction}} = 49.0 \, \text{kJ/mol}
\][/tex]
### Step 3: Determine the type of reaction
Since the enthalpy of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is positive ([tex]\(49.0 \, \text{kJ/mol}\)[/tex]), it indicates that the reaction absorbs heat.
Therefore, the reaction is endothermic.
### Conclusion
- The reaction is endothermic.
- The enthalpy of reaction [tex]\(\Delta H_{\text{reaction}}\)[/tex] is [tex]\(49.0 \, \text{kJ/mol}\)[/tex].
Thus, the correct choice is:
endothermic; [tex]$\Delta H_{\text{rxn}} = 49.0 \, \text{kJ}$[/tex].
