`4S(s)+60_2to4SO_(3)(g),DeltaH` f or this reaction is - 1583.2 kJ . The enthalpy of formation of sulphur trioxide is -

Options

A. `-395.8 kJ`
B. `395 .8 kJ`
C. `49.5 kJ`
D. `595.5kJ`

Answer

To find the enthalpy of formation of sulfur trioxide (SO₃) from the given reaction, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ 4S(s) + 6O_2(g) \rightarrow 4SO_3(g) \]
The enthalpy change (ΔH) for this reaction is provided as -1583.2 kJ.

### Step 2: Determine the Enthalpy Change per Mole
The given ΔH value corresponds to the formation of 4 moles of SO₃. To find the enthalpy of formation for 1 mole of SO₃, we need to divide the total ΔH by the number of moles produced.

### Step 3: Calculate the Enthalpy of Formation for 1 Mole
\[ \Delta H_{f} \text{ (for 1 mole of SO₃)} = \frac{\Delta H \text{ (for 4 moles of SO₃)}}{4} \]
Substituting the values:
\[ \Delta H_{f} = \frac{-1583.2 \text{ kJ}}{4} = -395.8 \text{ kJ} \]

### Step 4: Conclusion
The enthalpy of formation of sulfur trioxide (SO₃) is -395.8 kJ/mol.

### Final Answer
The enthalpy of formation of sulfur trioxide (SO₃) is -395.8 kJ/mol.

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