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. `-814.4kJ mol^(-1)`
B. `+814.4 kJ"mole"^(-1)`
C. `-650.3 kJ"mole"^(-1)`
D. `-433.7 kJ "mole"^(-1)`

Explanation

To calculate 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 reactions and their enthalpy changes

1. \( S + O_2 \rightarrow SO_2 \)
\( \Delta H = -298.2 \, \text{kJ/mol} \)

2. \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 \)
\( \Delta H = -98.7 \, \text{kJ/mol} \)

3. \( SO_3 + H_2O \rightarrow H_2SO_4 \)
\( \Delta H = -130.2 \, \text{kJ/mol} \)

4. \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \)
\( \Delta H = -287.3 \, \text{kJ/mol} \)

### Step 2: Write the desired reaction for the formation of \( H_2SO_4 \)

The formation reaction of \( H_2SO_4 \) can be represented as:
\[ H_2 + S + 2O_2 \rightarrow H_2SO_4 \]

### Step 3: Combine the given reactions to obtain the desired reaction

To obtain the desired reaction, we need to manipulate the given reactions:

- **Reaction 1**: \( S + O_2 \rightarrow SO_2 \) (Use as is)
- **Reaction 2**: \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 \) (Use as is)
- **Reaction 3**: \( SO_3 + H_2O \rightarrow H_2SO_4 \) (Use as is)
- **Reaction 4**: \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \) (Reverse this reaction)

When we reverse Reaction 4, we change the sign of \( \Delta H \):
\[ H_2O \rightarrow H_2 + \frac{1}{2} O_2 \]
\( \Delta H = +287.3 \, \text{kJ/mol} \)

### Step 4: Add the reactions

Now, we add the reactions together:

1. \( S + O_2 \rightarrow SO_2 \)
\( \Delta H = -298.2 \, \text{kJ/mol} \)

2. \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 \)
\( \Delta H = -98.7 \, \text{kJ/mol} \)

3. \( SO_3 + H_2O \rightarrow H_2SO_4 \)
\( \Delta H = -130.2 \, \text{kJ/mol} \)

4. \( H_2O \rightarrow H_2 + \frac{1}{2} O_2 \)
\( \Delta H = +287.3 \, \text{kJ/mol} \)

Now, when we add these reactions, the \( SO_2 \), \( SO_3 \), and \( H_2O \) will cancel out, leaving us with:
\[ S + 2O_2 + H_2 \rightarrow H_2SO_4 \]

### Step 5: Calculate the total enthalpy change

Now, we sum the enthalpy changes:
\[ \Delta H = (-298.2) + (-98.7) + (-130.2) + (+287.3) \]
\[ \Delta H = -298.2 - 98.7 - 130.2 + 287.3 \]
\[ \Delta H = -814.4 \, \text{kJ/mol} \]

### Final Answer

The enthalpy of formation of \( H_2SO_4 \) at 298 K is:
\[ \Delta H_f = -814.4 \, \text{kJ/mol} \]

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