Question

For the reaction \( \text{A}_2 + \text{B}_2 \rightarrow 2 \text{AB} \), \( \Delta H_f = -400 \, \text{kJ/mol} \). The bond dissociation enthalpies of \( \text{A}_2 \), \( \text{B}_2 \), and \( \text{AB} \) are in the ratio 1 : 0.75 : 1. What is the bond dissociation enthalpy of \( \text{B}_2 \) in kJ/mol?

• A. 1600
• B. 2400
• C. 3200
• D. 800

Hint:

The bond dissociation enthalpy is the energy required to break a bond, and for reactions, it is the difference between the reactants and products' bond dissociation enthalpies.

Answer 1

The Correct Option is B

Step 1: Use the bond dissociation enthalpy formula.
The bond dissociation enthalpy for the reaction can be calculated using the bond dissociation enthalpies of the reactants and products:
\[ \Delta H_f = \text{BDE of reactants} - \text{BDE of products} \]
Where:
- \( \text{BDE of reactants} = D(\text{A}_2) + D(\text{B}_2) \)
- \( \text{BDE of products} = 2 \times D(\text{AB}) \)

Step 2: Set up the equation with the given ratio.
Let the bond dissociation enthalpy of \( \text{A}_2 \) be \( x \), that of \( \text{B}_2 \) be \( 0.75x \), and that of \( \text{AB} \) be \( x \).
Thus, the equation becomes:
\[ -400 = (x + 0.75x) - 2x \]

Step 3: Solve for \( x \).
Simplifying the equation:
\[ -400 = 1.75x - 2x \]
\[ -400 = -0.25x \]
\[ x = \frac{-400}{-0.25} = 1600 \, \text{kJ/mol} \]

Step 4: Calculate the bond dissociation enthalpy of \( \text{B}_2 \).
The bond dissociation enthalpy of \( \text{B}_2 \) is \( 0.75x = 0.75 \times 1600 = 1200 \, \text{kJ/mol} \).

Step 5: Conclusion.
Therefore, the bond dissociation enthalpy of \( \text{B}_2 \) is \( 2400 \, \text{kJ/mol} \).