Question

Given that \(\Delta H_f(H) = 218\,kJ/mol\), express the \(H-H\) bond energy in \(kcal/mol\).

• A. \(52.15\)
• B. \(911\)
• C. \(104\)
• D. \(52153\)

Hint:

If enthalpy is given for \(\frac{1}{2}H_2 \rightarrow H\), multiply by \(2\) to get \(H_2 \rightarrow 2H\) bond energy.

Answer 1

The Correct Option is C

Step 1: Understand the meaning of \(\Delta H_f(H)\).
\(\Delta H_f(H)\) represents the enthalpy required to form \(1\) mole of hydrogen atoms from hydrogen molecules.
Reaction:
\[ \frac{1}{2}H_2(g) \rightarrow H(g) \]
Given:
\[ \Delta H = 218\,kJ/mol \]
Step 2: Convert this to bond dissociation energy of \(H_2\).
Bond energy is for:
\[ H_2(g) \rightarrow 2H(g) \]
So bond dissociation energy is:
\[ D(H-H) = 2 \times 218 = 436\,kJ/mol \]
Step 3: Convert \(kJ/mol\) into \(kcal/mol\).
\[ 1\,kcal = 4.184\,kJ \Rightarrow 436\,kJ = \frac{436}{4.184}\,kcal \]
\[ D(H-H) \approx 104\,kcal/mol \]
Final Answer:
\[ \boxed{104\,kcal/mol} \]