If `DeltaH_(f(C_(2)H_(6)))^(0)(g) = -85 KJH mol^(-1) , DeltaH_(f(C_(3)H_(8)))^(0) (g) = -104 KJ mol^(-1), DeltaH^(0)` for `C (s) rarr C(g)` is `718 KJ mol^(-1)` and heat of formation of H-atom is `218 KJmol^(-1)` then :

Options

A. `DeltaH_(C-C) = 345` KJ
B. `DeltaH_(C-H) = 414` KJ
C. `DeltaH_(H-H)=436 KJ`
D. `DeltaH_(H-H) = 456` KJ

Explanation

To solve the problem, we need to calculate the bond enthalpies of C-H and C-C bonds using the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Formation Reactions
The formation of ethane (C₂H₆) can be represented as:
\[ 2C(s) + 3H_2(g) \rightarrow C_2H_6(g) \]
The formation of propane (C₃H₈) can be represented as:
\[ 3C(s) + 4H_2(g) \rightarrow C_3H_8(g) \]

### Step 2: Write the Enthalpy Change for C₂H₆
The enthalpy change for the formation of C₂H₆ can be expressed as:
\[ \Delta H_f^{\circ}(C_2H_6) = \Delta H_{sublimation} + \Delta H_{formation \, of \, H_2} - \text{(bond energies)} \]

Where:
- \(\Delta H_{sublimation} = 2 \times 718 \, \text{kJ/mol}\) (for converting 2C(s) to 2C(g))
- \(\Delta H_{formation \, of \, H_2} = 3 \times 218 \, \text{kJ/mol}\) (for forming 3H₂ from H atoms)
- Bond energies: \(6A\) (for 6 C-H bonds) and \(B\) (for 1 C-C bond)

### Step 3: Set Up the Equation for C₂H₆
From the information given:
\[ -85 = 2 \times 718 + 3 \times 218 - (6A + B) \]
Simplifying this:
\[ -85 = 1436 + 654 - (6A + B) \]
\[ -85 = 2090 - (6A + B) \]
Rearranging gives:
\[ 6A + B = 2090 + 85 \]
\[ 6A + B = 2175 \] (Equation 1)

### Step 4: Write the Enthalpy Change for C₃H₈
Similarly, for the formation of C₃H₈:
\[ \Delta H_f^{\circ}(C_3H_8) = \Delta H_{sublimation} + \Delta H_{formation \, of \, H_2} - \text{(bond energies)} \]

Where:
- \(\Delta H_{sublimation} = 3 \times 718 \, \text{kJ/mol}\)
- \(\Delta H_{formation \, of \, H_2} = 4 \times 218 \, \text{kJ/mol}\)
- Bond energies: \(8A\) (for 8 C-H bonds) and \(2B\) (for 2 C-C bonds)

### Step 5: Set Up the Equation for C₃H₈
From the information given:
\[ -104 = 3 \times 718 + 4 \times 218 - (8A + 2B) \]
Simplifying this:
\[ -104 = 2154 + 872 - (8A + 2B) \]
\[ -104 = 3026 - (8A + 2B) \]
Rearranging gives:
\[ 8A + 2B = 3026 + 104 \]
\[ 8A + 2B = 3130 \] (Equation 2)

### Step 6: Solve the System of Equations
Now we have two equations:
1. \(6A + B = 2175\)
2. \(8A + 2B = 3130\)

From Equation 1, we can express \(B\) in terms of \(A\):
\[ B = 2175 - 6A \]

Substituting \(B\) into Equation 2:
\[ 8A + 2(2175 - 6A) = 3130 \]
\[ 8A + 4350 - 12A = 3130 \]
\[ -4A + 4350 = 3130 \]
\[ -4A = 3130 - 4350 \]
\[ -4A = -1220 \]
\[ A = 305 \, \text{kJ/mol} \]

Now substituting \(A\) back to find \(B\):
\[ B = 2175 - 6(305) \]
\[ B = 2175 - 1830 \]
\[ B = 345 \, \text{kJ/mol} \]

### Final Values
- \(A\) (C-H bond enthalpy) = 305 kJ/mol
- \(B\) (C-C bond enthalpy) = 345 kJ/mol