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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]
[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]
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