\begin{tabular}{|c|r|}
\hline
[tex]$CO (g)$[/tex] & -393.50 \\
\hline
[tex]$CO_2 (g)$[/tex] & -285.8 \\
\hline
[tex]$H_2O (l)$[/tex] & -241.8 \\
\hline
[tex]$H_2O (g)$[/tex] & 1.895 \\
\hline
[tex]$C (s)$[/tex], diamond & 0.0 \\
\hline
[tex]$C (s)$[/tex], graphite & \\
\hline
\end{tabular}

Based on the equation and the information in the table, what is the enthalpy of the reaction?

Use [tex]$\Delta H_{\text {reaction}} = \sum \left(\Delta H_{f, \text {products}}\right) - \sum \left(\Delta H_{f, \text {reactants}}\right)$[/tex].

A. [tex]$-453.46 \, \text{kJ}$[/tex]
B. [tex]$-226.73 \, \text{kJ}$[/tex]
C. [tex]$226.73 \, \text{kJ}$[/tex]
D. [tex]$453.46 \, \text{kJ}$[/tex]



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.