Answer :
To determine the enthalpy change for the reaction, we will use the given equation and the formula provided for the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{rxn}} = \sum \left( \Delta H_{\text{f, products}} \right) - \sum \left( \Delta H_{\text{f, reactants}} \right) \][/tex]
First, let's write down the reaction:
[tex]\[ S ( s ) + O_2 ( g ) \rightarrow SO_2 ( g ) \][/tex]
Next, let's identify the enthalpy of formation for each substance involved in the reaction:
- Standard enthalpy of formation for \( S(s) \) is \( 0 \) kJ/mol because it is in its elemental form.
- Standard enthalpy of formation for \( O_2(g) \) is \( 0 \) kJ/mol because it is also in its elemental form.
- Standard enthalpy of formation for \( SO_2(g) \) is given as \( -296.8 \) kJ/mol.
We now substitute these values into the formula. The enthalpy change for the reaction is:
[tex]\[ \Delta H_{\text{rxn}} = \sum \left( \Delta H_{\text{f, products}} \right) - \sum \left( \Delta H_{\text{f, reactants}} \right) \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = \left( \Delta H_{\text{f, SO}_2} \right) - \left( \Delta H_{\text{f, S}} + \Delta H_{\text{f, O}_2} \right) \][/tex]
Substitute the given values:
[tex]\[ \Delta H_{\text{rxn}} = \left( -296.8 \, \text{kJ/mol} \right) - \left( 0 \, \text{kJ/mol} + 0 \, \text{kJ/mol} \right) \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -296.8 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change for the reaction \( S(s) + O_2(g) \rightarrow SO_2(g) \) is \(-296.8\) kJ/mol.
The correct answer is:
[tex]\[ \boxed{-296.8 \text{ kJ}} \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = \sum \left( \Delta H_{\text{f, products}} \right) - \sum \left( \Delta H_{\text{f, reactants}} \right) \][/tex]
First, let's write down the reaction:
[tex]\[ S ( s ) + O_2 ( g ) \rightarrow SO_2 ( g ) \][/tex]
Next, let's identify the enthalpy of formation for each substance involved in the reaction:
- Standard enthalpy of formation for \( S(s) \) is \( 0 \) kJ/mol because it is in its elemental form.
- Standard enthalpy of formation for \( O_2(g) \) is \( 0 \) kJ/mol because it is also in its elemental form.
- Standard enthalpy of formation for \( SO_2(g) \) is given as \( -296.8 \) kJ/mol.
We now substitute these values into the formula. The enthalpy change for the reaction is:
[tex]\[ \Delta H_{\text{rxn}} = \sum \left( \Delta H_{\text{f, products}} \right) - \sum \left( \Delta H_{\text{f, reactants}} \right) \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = \left( \Delta H_{\text{f, SO}_2} \right) - \left( \Delta H_{\text{f, S}} + \Delta H_{\text{f, O}_2} \right) \][/tex]
Substitute the given values:
[tex]\[ \Delta H_{\text{rxn}} = \left( -296.8 \, \text{kJ/mol} \right) - \left( 0 \, \text{kJ/mol} + 0 \, \text{kJ/mol} \right) \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -296.8 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change for the reaction \( S(s) + O_2(g) \rightarrow SO_2(g) \) is \(-296.8\) kJ/mol.
The correct answer is:
[tex]\[ \boxed{-296.8 \text{ kJ}} \][/tex]
