The enthalpy of formation for [tex]C_6H_6(l)[/tex] is [tex]49.0 \, \text{kJ/mol}[/tex]. Consider the following reaction:

[tex]
6 \, C \text{ (s, graphite) } + 3 \, H_2(g) \rightarrow C_6H_6(l)
[/tex]

Is the reaction endothermic or exothermic, and what is the enthalpy of reaction? Use [tex]\Delta H_{r \times n} = \sum\left(\Delta H_{f, \text{products}}\right) - \sum\left(\Delta H_{f, \text{reactants}}\right)[/tex].

A. Exothermic; [tex]\Delta H_{r \times n} = 49.0 \, \text{kJ}[/tex]
B. Exothermic; [tex]\Delta H_{r \times n} = -49.0 \, \text{kJ}[/tex]
C. Endothermic; [tex]\Delta H_{r \times n} = 49.0 \, \text{kJ}[/tex]
D. Endothermic; [tex]\Delta H_{r \times n} = -49.0 \, \text{kJ}[/tex]



Answer :

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].