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 use the given reaction and the bond dissociation enthalpies provided. Let's break down the steps to solve this problem.

### Step 1: Write the Reaction
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 is given as \( \Delta H = 80.3 \, \text{kJ mol}^{-1} \).

### Step 2: Convert Solid Silicon to Gaseous Silicon
To convert silicon from solid to gaseous form, we need to consider the enthalpy of sublimation. The enthalpy change for converting 2 moles of solid silicon to gaseous silicon is:
\[ 2 \times \Delta H_{\text{sublimation}} = 2 \times 450 \, \text{kJ mol}^{-1} = 900 \, \text{kJ} \]

### Step 3: Break the H-H Bonds
We need to break 3 moles of H-H bonds. The bond dissociation enthalpy for H-H is given as \( 436 \, \text{kJ mol}^{-1} \). Thus, the energy required to break these bonds is:
\[ 3 \times 436 \, \text{kJ} = 1308 \, \text{kJ} \]

### Step 4: Form the Bonds in Si2H6
In the product Si2H6, we have:
- 1 Si-Si bond
- 6 Si-H bonds

The bond dissociation enthalpy for Si-H is given as \( 304 \, \text{kJ mol}^{-1} \). Therefore, the energy released when forming these bonds is:
- For Si-H bonds: \( 6 \times 304 \, \text{kJ} = 1824 \, \text{kJ} \)
- For Si-Si bond: Let the bond dissociation enthalpy of Si-Si be \( x \).

### Step 5: Set Up the Enthalpy Equation
The overall enthalpy change for the reaction can be expressed as:
\[ \Delta H = \text{Energy input} - \text{Energy output} \]
Substituting the values we have:
\[ 80.3 = (900 + 1308) - (1824 + x) \]

### Step 6: Solve for x
Rearranging the equation gives:
\[ 80.3 = 2208 - 1824 - x \]
\[ 80.3 = 384 - x \]
\[ x = 384 - 80.3 = 303.7 \, \text{kJ mol}^{-1} \]
Thus, the bond dissociation enthalpy of the Si-Si bond is approximately:
\[ \boxed{304 \, \text{kJ mol}^{-1}} \]