Answer :
To determine the enthalpy change (\(\Delta H_{\text{reaction}}\)) for the reaction \( S (s) + O_2 (g) \rightarrow SO_2 (g) \), we use the given enthalpy of formation (\(\Delta H_f\)) of sulfur dioxide (\(SO_2\)) and the standard enthalpy of formation values for the reactants.
The enthalpy change of the reaction can be calculated using the formula:
[tex]\[ \Delta H_{\text{reaction}} = \sum\left(\Delta H_{\text{f,products}}\right) - \sum\left(\Delta H_{\text{f,reactants}}\right) \][/tex]
Given data:
- \(\Delta H_f\) of \(SO_2 (g) = -296.8 \, \text{kJ/mol}\)
- The enthalpy of formation (\(\Delta H_f\)) for elements in their standard state (i.e., \(S (s)\) and \(O_2 (g)\)) is zero.
We need to calculate:
[tex]\[ \Delta H_{\text{reaction}} \][/tex]
Step-by-step solution:
1. Identify the enthalpy of formation of the products:
- For \(SO_2 (g)\), \(\Delta H_f = -296.8 \, \text{kJ/mol}\)
2. Identify the enthalpy of formation for the reactants:
- For \(S (s)\), \(\Delta H_f = 0 \, \text{kJ/mol}\) (since it's in its standard state)
- For \(O_2 (g)\), \(\Delta H_f = 0 \, \text{kJ/mol}\) (since it's in its standard state)
3. Apply the enthalpy change formula:
[tex]\[ \Delta H_{\text{reaction}} = \left(\Delta H_f \, \text{of} \, SO_2 (g)\right) - \left(\Delta H_f \, \text{of} \, S (s) + \Delta H_f \, \text{of} \, O_2 (g)\right) \][/tex]
Substitute the values into the equation:
[tex]\[ \Delta H_{\text{reaction}} = (-296.8 \, \text{kJ/mol}) - (0 \, \text{kJ/mol} + 0 \, \text{kJ/mol}) \][/tex]
4. Simplify the equation:
[tex]\[ \Delta H_{\text{reaction}} = -296.8 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change (\(\Delta H_{\text{reaction}}\)) for the reaction \( S (s) + O_2 (g) \rightarrow SO_2 (g) \) is \(-296.8 \, \text{kJ/mol}\).
The correct answer is:
[tex]\[ \boxed{-296.8 \, \text{kJ}} \][/tex]
The enthalpy change of the reaction can be calculated using the formula:
[tex]\[ \Delta H_{\text{reaction}} = \sum\left(\Delta H_{\text{f,products}}\right) - \sum\left(\Delta H_{\text{f,reactants}}\right) \][/tex]
Given data:
- \(\Delta H_f\) of \(SO_2 (g) = -296.8 \, \text{kJ/mol}\)
- The enthalpy of formation (\(\Delta H_f\)) for elements in their standard state (i.e., \(S (s)\) and \(O_2 (g)\)) is zero.
We need to calculate:
[tex]\[ \Delta H_{\text{reaction}} \][/tex]
Step-by-step solution:
1. Identify the enthalpy of formation of the products:
- For \(SO_2 (g)\), \(\Delta H_f = -296.8 \, \text{kJ/mol}\)
2. Identify the enthalpy of formation for the reactants:
- For \(S (s)\), \(\Delta H_f = 0 \, \text{kJ/mol}\) (since it's in its standard state)
- For \(O_2 (g)\), \(\Delta H_f = 0 \, \text{kJ/mol}\) (since it's in its standard state)
3. Apply the enthalpy change formula:
[tex]\[ \Delta H_{\text{reaction}} = \left(\Delta H_f \, \text{of} \, SO_2 (g)\right) - \left(\Delta H_f \, \text{of} \, S (s) + \Delta H_f \, \text{of} \, O_2 (g)\right) \][/tex]
Substitute the values into the equation:
[tex]\[ \Delta H_{\text{reaction}} = (-296.8 \, \text{kJ/mol}) - (0 \, \text{kJ/mol} + 0 \, \text{kJ/mol}) \][/tex]
4. Simplify the equation:
[tex]\[ \Delta H_{\text{reaction}} = -296.8 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change (\(\Delta H_{\text{reaction}}\)) for the reaction \( S (s) + O_2 (g) \rightarrow SO_2 (g) \) is \(-296.8 \, \text{kJ/mol}\).
The correct answer is:
[tex]\[ \boxed{-296.8 \, \text{kJ}} \][/tex]
