To calculate the $\Delta H^\circ_f$ for $SO_3(g)$, let's analyze the given data and solve step by step:

Given Information:

1. Reaction: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$
2. $\Delta H^\circ_{rxn} = -198 \, \text{kJ}$
3. Standard enthalpy of formation ($\Delta H^\circ_f$) of $SO_2(g)$: $-297 \, \text{kJ/mol}$
4. We are solving for $\Delta H^\circ_f$ of $SO_3(g)$.

Enthalpy Formula for Reaction:

The enthalpy of the reaction ($\Delta H^\circ_{rxn}$) can be expressed in terms of the standard enthalpy of formation of products and reactants: $\Delta H^\circ_{rxn} = \Sigma \Delta H^\circ_f(\text{products}) - \Sigma \Delta H^\circ_f(\text{reactants})$

Substituting the terms from the reaction: [ \Delta H^\circ_{rxn} = [2 \cdot \Delta H^\circ_f(SO_3)] - [2 \cdot \Delta H^\circ_f(SO_2)] ]

Substitution and Calculation:

1. Substitute $\Delta H^\circ_{rxn} = -198 \, \text{kJ}$ and $\Delta H^\circ_f(SO_2) = -297 \, \text{kJ/mol}$: [ -198 = [2 \cdot \Delta H^\circ_f(SO_3)] - [2 \cdot (-297)] ]

2. Simplify: $-198 = [2 \cdot \Delta H^\circ_f(SO_3)] + 594$

3. Solve for $[2 \cdot \Delta H^\circ_f(SO_3)]$: $2 \cdot \Delta H^\circ_f(SO_3) = -198 - 594$ $2 \cdot \Delta H^\circ_f(SO_3) = -792 \, \text{kJ}$

4. Divide by 2 to find $\Delta H^\circ_f(SO_3)$: $\Delta H^\circ_f(SO_3) = \frac{-792}{2} = -396 \, \text{kJ/mol}$

Final Answer:

$\Delta H^\circ_f(SO_3) = -396 \, \text{kJ/mol}$

Would you like further clarification or more details?

5 Related Questions:

1. How is the enthalpy of formation different from the reaction enthalpy?
2. What is the significance of negative $\Delta H^\circ_f$ values for compounds?
3. How do you apply Hess's Law in reaction enthalpy calculations?
4. Can this approach be extended to calculate $\Delta H^\circ_f$ of compounds in multi-step reactions?
5. What role does oxygen ($O_2$) play in the reaction's thermodynamics?

Tip:

Always check that all the units match (e.g., $\text{kJ}$ vs. $\text{kJ/mol}$) to avoid errors in thermodynamic calculations!