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

[tex]6C(s, \text{graphite}) + 3H_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_{\text{reaction}} = \sum\left(\Delta H_{\text{f, products}}\right) - \sum\left(\Delta H_{\text{f, reactants}}\right)[/tex].

A. exothermic; [tex]\Delta H_{\text{rxn}} = 49.0 \, \text{kJ}[/tex]
B. exothermic; [tex]\Delta H_{\text{rxn}} = -49.0 \, \text{kJ}[/tex]
C. endothermic; [tex]\Delta H_{\text{rxn}} = 49.0 \, \text{kJ}[/tex]
D. endothermic; [tex]\Delta H_{\text{rxn}} = -49.0 \, \text{kJ}[/tex]



Answer :

To determine if the reaction is endothermic or exothermic and to find the enthalpy of reaction, let's follow these steps:

1. Identify the enthalpy of formation for the reactants and products:
- For the given reaction:
[tex]\[ 6 \, \text{C (graphite)} + 3 \, \text{H}_2 \, \text{(g)} \rightarrow \text{C}_6\text{H}_6 \, \text{(l)} \][/tex]
- The enthalpy of formation for C (graphite) and H\(_2\) (g) in their standard states is zero because they are elements in their standard states.

2. Write down the enthalpy of formation for the product:
- The enthalpy of formation (\(\Delta H_f\)) of benzene, C\(_6\)H\(_6\) (l), is given as \(49.0 \, \text{kJ/mol}\).

3. Calculate the enthalpy of the reactants:
[tex]\[ \sum \left( \Delta H_{\text{f, reactants}} \right) = 0 \, \text{kJ/mol} \][/tex]
since the enthalpy of formation of elements in their standard state is zero.

4. Calculate the enthalpy of the products:
[tex]\[ \sum \left( \Delta H_{\text{f, products}} \right) = 49.0 \, \text{kJ/mol} \][/tex]
given that this is the enthalpy of formation of C\(_6\)H\(_6\) (l).

5. Apply the enthalpy of reaction formula:
[tex]\[ \Delta H_{\text{reaction}} = \sum \left( \Delta H_{\text{f, products}} \right) - \sum \left( \Delta H_{\text{f, reactants}} \right) \][/tex]
Substitute the known values:
[tex]\[ \Delta H_{\text{reaction}} = 49.0 \, \text{kJ/mol} - 0 \, \text{kJ/mol} = 49.0 \, \text{kJ/mol} \][/tex]

6. Determine whether the reaction is endothermic or exothermic:
- Since \(\Delta H_{\text{reaction}}\) is positive (+49.0 kJ/mol), this indicates that the reaction absorbs energy.
- Thus, the reaction is endothermic.

So, the enthalpy of reaction (\(\Delta H_{\text{reaction}}\)) is \(49.0 \, \text{kJ/mol}\) and the reaction is endothermic.

Therefore, the correct answer is:

endothermic; [tex]\(\Delta H_{\text{reaction}} = 49.0 \, \text{kJ/mol}\)[/tex]