Find bond enthalpy of S-S bond from the following data:
`C_(2)H_(5)-S-C_(2)H_(5)," "DeltaH_(f)^(@)=-147.2kJ" "mol^(-1)`
`C_(2)H_(5)-S-S-C_(2)H_(5)," "DeltaH_(f)^(@)=-201.9kJ" "mol^(-1)`
`S(g)," "DeltaH_(f)^(@)=222.8kJ" "mol^(-1)`

Options

A. `-168.1 KJ//"mol"`
B. `+168.1 KJ//"mol"`
C. `-277.5 KJ//"mol"`
D. `+277.5 KJ//"mol"`

Explanation

To find the bond enthalpy of the S-S bond from the provided data, we can use the enthalpy of formation values given for the compounds involved in the reaction. Here is the step-by-step solution:

### Step 1: Write the reaction
We start with the reaction where diethyl sulfide reacts with sulfur gas to form diethyl disulfide. The reaction can be represented as:

\[ \text{C}_2\text{H}_5\text{-S-C}_2\text{H}_5 (g) + \text{S}(g) \rightarrow \text{C}_2\text{H}_5\text{-S-S-C}_2\text{H}_5 (g) \]

### Step 2: Identify the enthalpy changes
From the problem, we have the following enthalpy of formation values:
- For \(\text{C}_2\text{H}_5\text{-S-C}_2\text{H}_5\): \(\Delta H_f = -147.2 \, \text{kJ/mol}\)
- For \(\text{C}_2\text{H}_5\text{-S-S-C}_2\text{H}_5\): \(\Delta H_f = -201.9 \, \text{kJ/mol}\)
- For \(\text{S}(g)\): \(\Delta H_f = 222.8 \, \text{kJ/mol}\)

### Step 3: Calculate the enthalpy change of the reaction
Using Hess's law, the enthalpy change of the reaction (\(\Delta H_{reaction}\)) can be calculated as follows:

\[ \Delta H_{reaction} = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \]

Substituting the values:

\[ \Delta H_{reaction} = \Delta H_f (\text{C}_2\text{H}_5\text{-S-S-C}_2\text{H}_5) - \Delta H_f (\text{C}_2\text{H}_5\text{-S-C}_2\text{H}_5) - \Delta H_f (\text{S}) \]

\[ \Delta H_{reaction} = (-201.9 \, \text{kJ/mol}) - (-147.2 \, \text{kJ/mol}) - (222.8 \, \text{kJ/mol}) \]

### Step 4: Perform the calculations
Now, we simplify the equation:

\[ \Delta H_{reaction} = -201.9 + 147.2 - 222.8 \]

Calculating this step-by-step:

1. Calculate \(-201.9 + 147.2\):
\[ -201.9 + 147.2 = -54.7 \, \text{kJ/mol} \]

2. Now, subtract \(222.8\):
\[ -54.7 - 222.8 = -277.5 \, \text{kJ/mol} \]

### Step 5: Find the bond enthalpy of the S-S bond
The enthalpy change of the reaction corresponds to the bond enthalpy of the S-S bond. Since we are looking for the bond enthalpy, we take the negative of the calculated value:

\[ \text{Bond enthalpy of S-S} = 277.5 \, \text{kJ/mol} \]

### Final Answer
The bond enthalpy of the S-S bond is \( 277.5 \, \text{kJ/mol} \).

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