Hydrogen [tex]$\left( H_2 \right)$[/tex] reacts with bromine [tex]$\left( Br_2 \right)$[/tex] to produce hydrogen bromide [tex]$( HBr )$[/tex] gas as shown in the equation below.

[tex]\[ H_2( g ) + Br_2( g ) \rightarrow 2 HBr ( g ) \][/tex]

Standard Enthalpies of Formation:

\begin{tabular}{|c|c|}
\hline
Substance & [tex]$\Delta H_f ( kJ / mol )$[/tex] \\
\hline
[tex]$Br_2$[/tex] (l) & 0.0 \\
\hline
[tex]$Br_2( g )$[/tex] & 30.907 \\
\hline
[tex]$HBr ( g )$[/tex] & -36.4 \\
\hline
[tex]$HC \left( g \right)$[/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?



Answer :

To determine the enthalpy change for the reaction, we need to use the standard enthalpies of formation provided in the table. The reaction given is:

[tex]\[ \text{H}_2(\text{g}) + \text{Br}_2(\text{g}) \rightarrow 2 \text{HBr}(\text{g}) \][/tex]

The change in enthalpy for the reaction, \(\Delta H_{\text{reaction}}\), can be calculated using the relation:

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

### Step-by-Step Solution

1. List the given standard enthalpies of formation \(\Delta H_f\):
- \(\Delta H_f(\text{H}_2(\text{g})) = 0.0 \, \text{kJ/mol}\)
- \(\Delta H_f(\text{Br}_2(\text{g})) = 30.907 \, \text{kJ/mol}\)
- \(\Delta H_f(\text{HBr}(\text{g})) = -36.4 \, \text{kJ/mol}\)

2. Write the balanced reaction:
[tex]\[ \text{H}_2(\text{g}) + \text{Br}_2(\text{g}) \rightarrow 2 \text{HBr}(\text{g}) \][/tex]

3. Calculate the total enthalpy of the reactants:
- For \(\text{H}_2(\text{g})\): \(\Delta H_f(\text{H}_2(\text{g})) = 0.0 \, \text{kJ/mol}\)
- For \(\text{Br}_2(\text{g}))\): \(\Delta H_f(\text{Br}_2(\text{g})) = 30.907 \, \text{kJ/mol}\)

Total enthalpy of reactants:
[tex]\[ \text{Enthalpy of reactants} = 0.0 + 30.907 = 30.907 \, \text{kJ/mol} \][/tex]

4. Calculate the total enthalpy of the products:
- For \(2 \text{HBr}(\text{g})\), since we have 2 moles:
[tex]\[ 2 \times \Delta H_f(\text{HBr}(\text{g})) = 2 \times (-36.4) = -72.8 \, \text{kJ/mol} \][/tex]

Total enthalpy of products:
[tex]\[ \text{Enthalpy of products} = -72.8 \, \text{kJ/mol} \][/tex]

5. Calculate the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \text{Enthalpy of products} - \text{Enthalpy of reactants} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -72.8 - 30.907 = -103.707 \, \text{kJ/mol} \][/tex]

Thus, the enthalpy change for the reaction is [tex]\(-103.707 \, \text{kJ/mol}\)[/tex].