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]
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]
