Silanes are silicon hydrides of general formula `Si_(n)Hn_(2n+2)` and have several applications. From the data given below, the bond dissociation enthalpy of `Si-Si` bond `("in kJ mol"^(-1))` is
Given:
`DeltaH` of the reaction
`2Si(s)+3H_(2)(g)rarrSi_(2)H_(6)(g)" is "80.3 kJ mol"^(-1)`
Bond dissociation enthalpy for `H-H=436" kJ mol"^(-1)`
Bond dissociation enthalpy for `Si-H=304" kJ mol"^(-1)`
`Delta_(f)H[Si(g)]="450 kJ mol"^(-1)`

Answer

To find the bond dissociation enthalpy of the Si-Si bond, we can follow these steps:

### Step 1: Write the Reaction and Given Data
The reaction given is:
\[ 2 \text{Si}(s) + 3 \text{H}_2(g) \rightarrow \text{Si}_2\text{H}_6(g) \]
The enthalpy change for this reaction (\(\Delta H\)) is given as \(80.3 \, \text{kJ/mol}\).

Other given data:
- Bond dissociation enthalpy for \( \text{H-H} = 436 \, \text{kJ/mol} \)
- Bond dissociation enthalpy for \( \text{Si-H} = 304 \, \text{kJ/mol} \)
- Enthalpy of sublimation for \(\text{Si}(g) = 450 \, \text{kJ/mol}\)

### Step 2: Identify the Components of the Reaction
In the reaction, we have:
- 2 moles of solid silicon (\( \text{Si}(s) \))
- 3 moles of hydrogen gas (\( \text{H}_2(g) \))
- 1 mole of silane (\( \text{Si}_2\text{H}_6(g) \))

### Step 3: Write the Enthalpy Change Equation
The enthalpy change for the reaction can be expressed as:
\[ \Delta H_{\text{reaction}} = \Delta H_{\text{reactants}} - \Delta H_{\text{products}} \]

### Step 4: Calculate the Enthalpy of Reactants
The enthalpy of the reactants includes:
- The sublimation of silicon: \(2 \times 450 \, \text{kJ/mol}\)
- The bond dissociation of hydrogen: \(3 \times 436 \, \text{kJ/mol}\)

Thus, the total enthalpy of the reactants is:
\[ \Delta H_{\text{reactants}} = 2 \times 450 + 3 \times 436 \]

### Step 5: Calculate the Enthalpy of Products
The products consist of:
- The bond dissociation enthalpy of silane (\( \text{Si}_2\text{H}_6 \)):
- There are 6 Si-H bonds and 1 Si-Si bond.

Thus, the total enthalpy of the products is:
\[ \Delta H_{\text{products}} = 1 \times \Delta H_{\text{Si-Si}} + 6 \times 304 \]

### Step 6: Set Up the Equation
Now we can set up the equation:
\[ 80.3 = \left(2 \times 450 + 3 \times 436\right) - \left(\Delta H_{\text{Si-Si}} + 6 \times 304\right) \]

### Step 7: Calculate the Values
Calculating the left side:
\[ 2 \times 450 = 900 \, \text{kJ} \]
\[ 3 \times 436 = 1308 \, \text{kJ} \]
Thus:
\[ \Delta H_{\text{reactants}} = 900 + 1308 = 2208 \, \text{kJ} \]

Calculating the right side:
\[ 6 \times 304 = 1824 \, \text{kJ} \]

Now substituting back into the equation:
\[ 80.3 = 2208 - \left(\Delta H_{\text{Si-Si}} + 1824\right) \]

### Step 8: Solve for \(\Delta H_{\text{Si-Si}}\)
Rearranging gives:
\[ \Delta H_{\text{Si-Si}} = 2208 - 1824 - 80.3 \]
\[ \Delta H_{\text{Si-Si}} = 2208 - 1904.3 = 303.7 \, \text{kJ/mol} \]

### Final Answer
Thus, the bond dissociation enthalpy of the Si-Si bond is approximately:
\[ \Delta H_{\text{Si-Si}} \approx 304 \, \text{kJ/mol} \]

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