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].
[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].
