In the reaction `CS_(2)(l)+3O_(2)(g)rarrCO_(2)(g)+2SO_(2)(g)DeltaH= -265 kcal`
The enthalpies of formation of `CO_(2)` and `SO_(2)` are both negative and are in the ratio `4:3`. The enthalpy of formation of `CS_(2)` is `+26 kcal//mol`. Calculate the enthalpy of formation of `SO_(2)`.

Options

A. `-90 kcal//mol`
B. `-52 kcal//mol`
C. `-78 kcal//mol`
D. `-71.7 kcal//mol`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the reaction and the given data
The reaction is:
\[ \text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \]
We are given:
- \(\Delta H = -265 \, \text{kcal}\) (for the reaction)
- Enthalpy of formation of \(\text{CS}_2\) = \(+26 \, \text{kcal/mol}\)
- The enthalpies of formation of \(\text{CO}_2\) and \(\text{SO}_2\) are in the ratio \(4:3\).

### Step 2: Define variables for enthalpies of formation
Let:
- The enthalpy of formation of \(\text{CO}_2\) be \(4X\)
- The enthalpy of formation of \(\text{SO}_2\) be \(3X\)

### Step 3: Write the equation for the enthalpy of reaction
Using the formula for the enthalpy change of a reaction:
\[ \Delta H_{\text{reaction}} = \sum (\Delta H_{\text{products}}) - \sum (\Delta H_{\text{reactants}}) \]
We can write:
\[ -265 = [4X + 2(3X)] - (+26) \]

### Step 4: Simplify the equation
This simplifies to:
\[ -265 = 4X + 6X - 26 \]
\[ -265 = 10X - 26 \]

### Step 5: Solve for \(X\)
Rearranging gives:
\[ 10X = -265 + 26 \]
\[ 10X = -239 \]
\[ X = -23.9 \, \text{kcal} \]

### Step 6: Calculate the enthalpy of formation of \(\text{SO}_2\)
Now, substituting \(X\) back to find the enthalpy of formation of \(\text{SO}_2\):
\[ \Delta H_f(\text{SO}_2) = 3X = 3(-23.9) = -71.7 \, \text{kcal} \]

### Final Answer
The enthalpy of formation of \(\text{SO}_2\) is:
\[ \Delta H_f(\text{SO}_2) = -71.7 \, \text{kcal} \]

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