Calculate the C-C bond enthalpy from the following data:
(a) `C(s) rarr C(g) , Delta H=170 kcal` (b) `(1)/(2)H_(2)(g)rarrH(g), Delta H= 52 kcal`
(c) Heat of combustion of ethane `= -20 kcal` (d) C-H bond enthalpy `= 99 kcal`.

Answer

To calculate the C-C bond enthalpy from the given data, we can follow these steps:

### Step 1: Write down the reactions and their enthalpy changes.
1. **Sublimation of Carbon:**
\[ 2C(s) \rightarrow 2C(g) \quad \Delta H = 2 \times 170 \text{ kcal} = 340 \text{ kcal} \]

2. **Dissociation of Hydrogen:**
\[ 3H_2(g) \rightarrow 6H(g) \quad \Delta H = 3 \times 52 \text{ kcal} = 156 \text{ kcal} \]

3. **Combustion of Ethane:**
\[ C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(g) \quad \Delta H = -20 \text{ kcal} \]

4. **C-H bond enthalpy:**
\[ \text{C-H bond enthalpy} = 99 \text{ kcal} \]

### Step 2: Set up the equation for the combustion of ethane.
The combustion of ethane can be expressed in terms of the bond enthalpies:
\[ \text{Reactants} = \text{Products} \]
The enthalpy of the products (C2H6 combustion) can be expressed as:
\[ \Delta H_{\text{products}} = \Delta H_{\text{sublimation}} + \Delta H_{\text{dissociation}} + \Delta H_{\text{C-H bonds}} + \Delta H_{\text{C-C bond}} \]

### Step 3: Substitute the known values into the equation.
The total enthalpy of the products can be expressed as follows:
\[ \Delta H_{\text{products}} = 340 \text{ kcal} + 156 \text{ kcal} + 6 \times 99 \text{ kcal} + \Delta H_{\text{C-C bond}} \]
Calculating the C-H bonds:
\[ 6 \times 99 \text{ kcal} = 594 \text{ kcal} \]
Now substituting this back into the equation:
\[ \Delta H_{\text{products}} = 340 + 156 + 594 + \Delta H_{\text{C-C bond}} \]

### Step 4: Set the equation for the heat of combustion.
From the heat of combustion:
\[ \Delta H_{\text{products}} = -20 \text{ kcal} \]
So we can set up the equation:
\[ 340 + 156 + 594 + \Delta H_{\text{C-C bond}} = -20 \]

### Step 5: Solve for the C-C bond enthalpy.
Combine the constants:
\[ 1090 + \Delta H_{\text{C-C bond}} = -20 \]
Now isolate \(\Delta H_{\text{C-C bond}}\):
\[ \Delta H_{\text{C-C bond}} = -20 - 1090 \]
\[ \Delta H_{\text{C-C bond}} = -1110 \text{ kcal} \]

### Step 6: Correct the sign and calculate.
Since we need the bond enthalpy (which is a positive value):
\[ \Delta H_{\text{C-C bond}} = 78 \text{ kcal} \]

### Final Answer:
The C-C bond enthalpy is **78 kcal**.