Question

Consider the reaction: \(C_2H_5Cl(g) \rightarrow 2C(g) + 5H(g) + Cl(g)\), where \(\Delta H = 3047 \, {kJ mol}^{-1}\). The bond dissociation enthalpy (\(\Delta_{{bond}} H\)) values of C–Cl and C=C bonds are 431 and 414 kJ mol\(^{-1}\), respectively. What is the average bond enthalpy of C–H in kJ mol\(^{-1}\)?

• A. \(436\)
• B. \(431\)
• C. \(357.6\)
• D. \(446\)

Hint:

When calculating bond energies in a reaction, remember to account for all types of bonds broken and formed, using their respective bond energy values.

Answer 1

The Correct Option is C

Step 1: Identify the bonds involved in the reaction
In \( C_2H_5Cl \), the bonds present are:
\( 5 \) C–H bonds
\( 1 \) C–Cl bond
\( 1 \) C–C bond
After decomposition, all these bonds break, and we get atoms as products. The bond dissociation enthalpy formula is: \[ \Delta H = \sum \text{{Bond energies of reactants}} - \sum \text{{Bond energies of products}} \] Step 2: Formulating the equation \[ 3047 = (5E + 431 + 414) - (0) \] Where \( E \) is the bond energy of C–H.
Step 3: Solving for \( E \) \[ 5E + 431 + 414 = 3047 \] \[ 5E = 3047 - 845 \] \[ 5E = 2202 \] \[ E = \frac{2202}{5} = 440.4 \, \text{kJ/mol} \] Thus, the average bond energy of C–H is \( 440.4 \, \text{kJ/mol} \).