If `S+O_(2)toSO_(2),DeltaH=-298.2 " kJ" " mole"^(-1)`
`SO_(2)+(1)/(2)O_(2)toSO_(3)DeltaH=-98.7 " kJ" " mole"^(-1)`
`SO_(3)+H_(2)OtoH_(2)SO_(4),DeltaH=-130.2 " kJ" " mole"^(-1)`
`H_(2)+(1)/(2)O_(2)toH_(2)SO_(4),DeltaH=-287.3 " kJ" " mole"^(-1)`
the enthlapy of formation of `H_(2)SO_(4)` at 298 K will be

Options

A. `-814.4 kJ`
B. `-650.3 kJ`
C. `-320.5 kJ`
D. `-433.5 kJ`

Answer

To find the enthalpy of formation of \( H_2SO_4 \) at 298 K using the given reactions, we will follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes
1. \( S + O_2 \rightarrow SO_2 \) \(\Delta H_1 = -298.2 \, \text{kJ/mol}\)
2. \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 \) \(\Delta H_2 = -98.7 \, \text{kJ/mol}\)
3. \( SO_3 + H_2O \rightarrow H_2SO_4 \) \(\Delta H_3 = -130.2 \, \text{kJ/mol}\)
4. \( H_2 + \frac{1}{2} O_2 \rightarrow H_2SO_4 \) \(\Delta H_4 = -287.3 \, \text{kJ/mol}\)

### Step 2: Identify the target reaction
The target reaction for the formation of \( H_2SO_4 \) from its elements is:
\[ H_2 + S + 2 O_2 \rightarrow H_2SO_4 \]

### Step 3: Manipulate the given reactions to achieve the target reaction
We will combine the reactions in such a way that we can cancel out the intermediate species and arrive at the target reaction.

1. From Reaction 1, we have \( S + O_2 \rightarrow SO_2 \) (keep as is).
2. From Reaction 2, we have \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 \) (keep as is).
3. From Reaction 3, we have \( SO_3 + H_2O \rightarrow H_2SO_4 \) (keep as is).
4. From Reaction 4, we need to reverse it to get \( H_2SO_4 \rightarrow H_2 + \frac{1}{2} O_2 \) (change sign of \(\Delta H\)).

### Step 4: Write the modified reactions and their enthalpy changes
1. \( S + O_2 \rightarrow SO_2 \) \(\Delta H_1 = -298.2 \, \text{kJ/mol}\)
2. \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 \) \(\Delta H_2 = -98.7 \, \text{kJ/mol}\)
3. \( SO_3 + H_2O \rightarrow H_2SO_4 \) \(\Delta H_3 = -130.2 \, \text{kJ/mol}\)
4. \( H_2SO_4 \rightarrow H_2 + \frac{1}{2} O_2 \) \(\Delta H_4 = +287.3 \, \text{kJ/mol}\) (reversed)

### Step 5: Combine the reactions
When we add these reactions together, we can cancel out \( SO_2 \), \( SO_3 \), and \( H_2SO_4 \):
- \( S + O_2 + SO_2 + \frac{1}{2} O_2 + H_2O + SO_3 \rightarrow H_2 + \frac{1}{2} O_2 + H_2SO_4 \)

This simplifies to:
\[ H_2 + S + 2 O_2 \rightarrow H_2SO_4 \]

### Step 6: Calculate the total enthalpy change
Now, we will sum the enthalpy changes:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 - \Delta H_4 \]
\[ \Delta H = (-298.2) + (-98.7) + (-130.2) + 287.3 \]
\[ \Delta H = -298.2 - 98.7 - 130.2 + 287.3 = -239.8 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of formation of \( H_2SO_4 \) at 298 K is \( -239.8 \, \text{kJ/mol} \).

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