Standard Enthalpies of Formation

\begin{tabular}{|c|c|}
\hline Substance & [tex]$\Delta H_f(kJ / mol )$[/tex] \\
\hline [tex]$Br _2( l )$[/tex] & 0.0 \\
\hline [tex]$Br _2(g)$[/tex] & 30.907 \\
\hline [tex]$HBr ( g )$[/tex] & -36.4 \\
\hline [tex]$HCl ( g )$[/tex] & -92.307 \\
\hline [tex]$HI ( g )$[/tex] & 26.48 \\
\hline [tex]$H _2(g)$[/tex] & 0.0 \\
\hline [tex]$I _2(g)$[/tex] & 62.438 \\
\hline
\end{tabular}

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

A. [tex]$-134.6 kJ$[/tex]
B. [tex]$-103.7 kJ$[/tex]
C. [tex]$103.7 kJ$[/tex]
D. [tex]$134.6 kJ$[/tex]



Answer :

To determine the enthalpy change of the reaction, we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the steps that make up the reaction. The general formula for the enthalpy change of a reaction, ΔH_reaction, can be written as:

[tex]\[ \Delta H_{\text{reaction}} = \Sigma (\Delta H_{\text{formation of products}}) - \Sigma (\Delta H_{\text{formation of reactants}}) \][/tex]

Using the provided standard enthalpies of formation from the table, let's walk through the calculations step-by-step:

1. Identify the Reactants and Products.

Suppose the reaction is:
[tex]\[ Br_2(g) + H_2(g) + I_2(g) \rightarrow HBr(g) + HI(g) \][/tex]

2. Write the Enthalpy Changes for Each Component

- For reactants:
- [tex]\( \Delta H_{\text{Br}_2(g)} = 30.907 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_{\text{H}_2(g)} = 0.0 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_{\text{I}_2(g)} = 62.438 \, \text{kJ/mol} \)[/tex]

- For products:
- [tex]\( \Delta H_{\text{HBr(g)}} = -36.4 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_{\text{HI(g)}} = 26.48 \, \text{kJ/mol} \)[/tex]

3. Sum the Enthalpy Changes for the Reactants and Products

- Sum for the reactants:
[tex]\[ \Sigma (\Delta H_{\text{reactants}}) = \Delta H_{\text{Br}_2(g)} + \Delta H_{\text{H}_2(g)} + \Delta H_{\text{I}_2(g)} \][/tex]
[tex]\[ \Sigma (\Delta H_{\text{reactants}}) = 30.907 + 0.0 + 62.438 = 93.345 \, \text{kJ/mol} \][/tex]

- Sum for the products:
[tex]\[ \Sigma (\Delta H_{\text{products}}) = \Delta H_{\text{HBr(g)}} + \Delta H_{\text{HI(g)}} \][/tex]
[tex]\[ \Sigma (\Delta H_{\text{products}}) = -36.4 + 26.48 = -9.92 \, \text{kJ/mol} \][/tex]

4. Calculate the Enthalpy Change of the Reaction

[tex]\[ \Delta H_{\text{reaction}} = \Sigma (\Delta H_{\text{products}}) - \Sigma (\Delta H_{\text{reactants}}) \][/tex]

[tex]\[ \Delta H_{\text{reaction}} = -9.92 - 93.345 = -103.265 \, \text{kJ/mol} \][/tex]

Based on this detailed calculation, the enthalpy change of the reaction is -103.265 kJ/mol. From the given options, the closest value to our calculated result is:

[tex]\(-103.7 \, \text{kJ}\)[/tex]

Hence, the answer is:

[tex]\[ -103.7 \, \text{kJ} \][/tex]