The `Delta_(r)H^(@)` for the reaction,
`4S(s) + 6O_(2)(g) to 4SO_(3)(g)`
is -1583.2 kJ.Calculate `DeltaH_(f)^(@)` of sulphur trioxide.

Explanation

To calculate the standard enthalpy of formation (ΔH_f^(@)) of sulfur trioxide (SO₃) from the given reaction:

**Given Reaction:**
\[ 4S(s) + 6O_2(g) \rightarrow 4SO_3(g) \]
**Given ΔH_r^(@):**
\[ ΔH_r^(@) = -1583.2 \text{ kJ} \]

### Step-by-Step Solution:

1. **Understand the Reaction:**
The reaction shows that 4 moles of sulfur (S) react with 6 moles of oxygen (O₂) to produce 4 moles of sulfur trioxide (SO₃). The enthalpy change for this reaction is given as -1583.2 kJ.

2. **Relate Enthalpy of Reaction to Enthalpy of Formation:**
The enthalpy of formation (ΔH_f^(@)) is defined for the formation of one mole of a compound from its elements in their standard states. In this case, we need to find the enthalpy of formation for 1 mole of SO₃.

3. **Adjust the Reaction for One Mole of Product:**
Since the given reaction produces 4 moles of SO₃, we can divide the entire reaction by 4 to find the enthalpy change for the formation of 1 mole of SO₃.

\[ \frac{1}{4} \times (4S(s) + 6O_2(g) \rightarrow 4SO_3(g)) \]
This simplifies to:
\[ S(s) + \frac{3}{2}O_2(g) \rightarrow SO_3(g) \]

4. **Calculate the Enthalpy of Formation:**
Since we divided the entire reaction by 4, we also need to divide the enthalpy change by 4.

\[ ΔH_f^(@) = \frac{-1583.2 \text{ kJ}}{4} = -395.8 \text{ kJ} \]

5. **Final Result:**
The standard enthalpy of formation of sulfur trioxide (SO₃) is:
\[ ΔH_f^(@) = -395.8 \text{ kJ/mol} \]