Answer :
To determine the enthalpy of the reaction using the given enthalpies of formation from the table, we'll follow a systematic process. We'll use the formula for the change in enthalpy (ΔH_reaction):
[tex]\[ \Delta H_{\text {reaction}} = \sum \left( \Delta H_{f, \text {products}} \right) - \sum \left( \Delta H_{f, \text {reactants}} \right) \][/tex]
Let's consider a typical chemical reaction involving the formation of [tex]\( CO_2 \)[/tex] from graphite:
[tex]\[ \text{C (s, graphite)} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \][/tex]
From the given table:
- ΔH_f for [tex]\(CO_2 (\text{g})\)[/tex] = -285.8 kJ/mol
- ΔH_f for [tex]\(\text{C (s, graphite)}\)[/tex] is missing but usually assumed as 0 kJ/mol (it's a standard state).
For [tex]\(O_2 (\text{g})\)[/tex], ΔH_f = 0 kJ/mol (by convention as a diatomic molecule in its elemental form).
Applying these values to the reaction:
Reactants:
[tex]\[ \Delta H_f (\text{C (s, graphite)}) + \Delta H_f (\text{O}_2 (\text{g})) = 0 + 0 = 0 \text{ kJ/mol} \][/tex]
Products:
[tex]\[ \Delta H_f (\text{CO}_2 (\text{g})) = -285.8 \text{ kJ/mol} \][/tex]
Now, plug these values into the enthalpy change formula:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \][/tex]
[tex]\[ \Delta H_{\text {reaction}} = -285.8 - 0 = -285.8 \text{ kJ/mol} \][/tex]
Thus, the enthalpy of the reaction is:
[tex]\[ -285.8 \text{ kJ/mol} \][/tex]
Even though the given set of answer options doesn't include [tex]\(-285.8 \text{ kJ/mol}\)[/tex] directly, this computed enthalpy value best represents the scenario described in the problem and calculations. If the context of the problem had more details about other reactants/products or specific reactions required to match the given options, then that would need to be revisited. But based on the current information, -285.8 kJ/mol is the enthalpy change, leading me to understand that there might be a mismatch between the option sets provided and the correct answer.
[tex]\[ \Delta H_{\text {reaction}} = \sum \left( \Delta H_{f, \text {products}} \right) - \sum \left( \Delta H_{f, \text {reactants}} \right) \][/tex]
Let's consider a typical chemical reaction involving the formation of [tex]\( CO_2 \)[/tex] from graphite:
[tex]\[ \text{C (s, graphite)} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \][/tex]
From the given table:
- ΔH_f for [tex]\(CO_2 (\text{g})\)[/tex] = -285.8 kJ/mol
- ΔH_f for [tex]\(\text{C (s, graphite)}\)[/tex] is missing but usually assumed as 0 kJ/mol (it's a standard state).
For [tex]\(O_2 (\text{g})\)[/tex], ΔH_f = 0 kJ/mol (by convention as a diatomic molecule in its elemental form).
Applying these values to the reaction:
Reactants:
[tex]\[ \Delta H_f (\text{C (s, graphite)}) + \Delta H_f (\text{O}_2 (\text{g})) = 0 + 0 = 0 \text{ kJ/mol} \][/tex]
Products:
[tex]\[ \Delta H_f (\text{CO}_2 (\text{g})) = -285.8 \text{ kJ/mol} \][/tex]
Now, plug these values into the enthalpy change formula:
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \][/tex]
[tex]\[ \Delta H_{\text {reaction}} = -285.8 - 0 = -285.8 \text{ kJ/mol} \][/tex]
Thus, the enthalpy of the reaction is:
[tex]\[ -285.8 \text{ kJ/mol} \][/tex]
Even though the given set of answer options doesn't include [tex]\(-285.8 \text{ kJ/mol}\)[/tex] directly, this computed enthalpy value best represents the scenario described in the problem and calculations. If the context of the problem had more details about other reactants/products or specific reactions required to match the given options, then that would need to be revisited. But based on the current information, -285.8 kJ/mol is the enthalpy change, leading me to understand that there might be a mismatch between the option sets provided and the correct answer.