To determine the enthalpy of the reaction given in the equation:
[tex]\[ 2 C (s, \text{graphite}) + H_2 (g) \rightarrow C_2H_2 (g) \][/tex]
we use the standard enthalpies of formation from the table given. The standard 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.
From the table, we have:
- [tex]\(\Delta H_f \ \text{for} \ C_2 H_2 (g) = 226.73 \ \text{kJ/mol} \)[/tex]
- [tex]\(\Delta H_f \ \text{for} \ C (s, \text{graphite}) = 0.0 \ \text{kJ/mol} \)[/tex]
- [tex]\(\Delta H_f \ \text{for} \ H_2 (g) = 0.0 \ \text{kJ/mol} \)[/tex]
Using the given equation:
[tex]\[ \Delta H_{\text{reaction}} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) \][/tex]
First, we calculate the total enthalpy of the reactants:
For the reactants:
- The enthalpy contribution from [tex]\(C (s, \text{graphite})\)[/tex] is [tex]\(2 \times \Delta H_f (C (s, \text{graphite})) = 2 \times 0.0 \ \text{kJ/mol} = 0.0 \ \text{kJ}\)[/tex]
- The enthalpy contribution from [tex]\(H_2 (g)\)[/tex] is [tex]\(1 \times \Delta H_f (H_2 (g)) = 1 \times 0.0 \ \text{kJ/mol} = 0.0 \ \text{kJ}\)[/tex]
Total enthalpy of reactants:
[tex]\[ \Delta H_{\text{reactants}} = 0.0 \ \text{kJ} + 0.0 \ \text{kJ} = 0.0 \ \text{kJ} \][/tex]
Next, we calculate the total enthalpy of the products:
For the products:
- The enthalpy contribution from [tex]\(C_2 H_2 (g)\)[/tex] is [tex]\(1 \times \Delta H_f (C_2 H_2 (g)) = 1 \times 226.73 \ \text{kJ/mol} = 226.73 \ \text{kJ}\)[/tex]
Total enthalpy of products:
[tex]\[ \Delta H_{\text{products}} = 226.73 \ \text{kJ} \][/tex]
Finally, we calculate the enthalpy of the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = 226.73 \ \text{kJ} - 0.0 \ \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = 226.73 \ \text{kJ} \][/tex]
Therefore, the enthalpy of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is [tex]\(226.73 \ \text{kJ}\)[/tex].
