If `triangleH_(f) (C_(2)H_(5)-S-C_(2)H_(5))=-147` kJ/mole and
`triangleH_(f) (C_(2)H_(5)-S-S-C_(2)H_(5))=-202` Kj/mole and
`triangleH_(f) (S) (g)=+ 223 kJ/mole then the S-S bond energy will be

Options

A. 168 kJ
B. 126 kJ
C. 278 kJ
D. 572 kJ

Answer

To find the S-S bond energy based on the given enthalpy of formation values, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- \(\Delta H_f (C_2H_5-S-C_2H_5) = -147 \, \text{kJ/mol}\)
- \(\Delta H_f (C_2H_5-S-S-C_2H_5) = -202 \, \text{kJ/mol}\)
- \(\Delta H_f (S) = +223 \, \text{kJ/mol}\)

2. **Write the Formation Reactions:**
- For the first compound \(C_2H_5-S-C_2H_5\):
\[ C_2H_5 + C_2H_5 + \frac{1}{2} S_2 \rightarrow C_2H_5-S-C_2H_5 \quad \Delta H = -147 \, \text{kJ} \]
- For the second compound \(C_2H_5-S-S-C_2H_5\):
\[ C_2H_5 + C_2H_5 + S_2 \rightarrow C_2H_5-S-S-C_2H_5 \quad \Delta H = -202 \, \text{kJ} \]

3. **Set Up the Equations:**
- For the first reaction:
\[ -147 = \text{(Bond energy of reactants)} - \text{(Bond energy of products)} \]
- For the second reaction:
\[ -202 = \text{(Bond energy of reactants)} - \text{(Bond energy of products with S-S bond)} \]

4. **Assume Bond Energies:**
- Let \(E_{S-S}\) be the bond energy of the S-S bond.
- The bond energies for the carbon and hydrogen bonds remain constant in both reactions.

5. **Subtract the Two Equations:**
- By subtracting the first equation from the second:
\[ -202 - (-147) = \text{(Bond energy terms cancel out)} - E_{S-S} \]
- This simplifies to:
\[ -55 = -E_{S-S} \]
- Thus, we find:
\[ E_{S-S} = 278 \, \text{kJ/mol} \]

6. **Conclusion:**
- The bond energy of the S-S bond is \(278 \, \text{kJ/mol}\).

### Final Answer:
The S-S bond energy is **278 kJ/mol**.