To determine the enthalpy change of the reaction [tex]\( H_2 (g) + Br_2 (g) \rightarrow 2 HBr (g) \)[/tex], we use the concept of standard enthalpies of formation. Here are the steps:
1. Identify the substances and their standard enthalpies of formation involved in the reaction:
From the table:
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( H_2 (g) = 0.0 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( Br_2 (g) = 30.907 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f^\circ \)[/tex] for [tex]\( HBr (g) = -36.4 \, \text{kJ/mol} \)[/tex]
2. Calculate the total enthalpy of the reactants:
The reactants are [tex]\( H_2 (g) \)[/tex] and [tex]\( Br_2 (g) \)[/tex].
- For [tex]\( H_2 (g) \)[/tex], the enthalpy is [tex]\( 0.0 \, \text{kJ/mol} \)[/tex].
- For [tex]\( Br_2 (g) \)[/tex], the enthalpy is [tex]\( 30.907 \, \text{kJ/mol} \)[/tex].
Total enthalpy of reactants:
[tex]\[
\Delta H_{\text{reactants}} = 0.0 + 30.907 = 30.907 \, \text{kJ}
\][/tex]
3. Calculate the total enthalpy of the products:
The product is [tex]\( 2 \, \text{mol} \)[/tex] of [tex]\( HBr (g) \)[/tex].
- For [tex]\( HBr (g) \)[/tex], the enthalpy is [tex]\( -36.4 \, \text{kJ/mol} \)[/tex].
Total enthalpy of products:
[tex]\[
\Delta H_{\text{products}} = 2 \times (-36.4) = -72.8 \, \text{kJ}
\][/tex]
4. Determine the enthalpy change of the reaction (ΔH):
[tex]\[
\Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}}
\][/tex]
Plugging in the values:
[tex]\[
\Delta H_{\text{reaction}} = -72.8 - 30.907 = -103.707 \, \text{kJ}
\][/tex]
Summary:
- The total enthalpy of reactants is [tex]\( 30.907 \, \text{kJ} \)[/tex].
- The total enthalpy of products is [tex]\( -72.8 \, \text{kJ} \)[/tex].
- The enthalpy change of the reaction is [tex]\( \Delta H = -103.707 \, \text{kJ} \)[/tex].
Therefore, the enthalpy of the reaction is [tex]\( -103.707 \, \text{kJ} \)[/tex].
