Question

Bond enthalpies of A₂, B₂ and AB are in the ratio 2 : 1 : 2. If bond enthalpy of formation of AB is -100 KJ mol^-1. The bond enthalpy of B₂ is

• A. 100 KJ mol^-1
• B. 50 KJ mol^-1
• C. 200 KJ mol^-1
• D. 150 KJ mol^-1

Answer 1

The Correct Option is C

To solve the problem, we need to determine the bond enthalpy of $ B_2 $ given the ratio of bond enthalpies and the bond enthalpy of formation of $ AB $. Let us proceed step by step.

Step 1: Define the variables
Let:
- The bond enthalpy of $ A_2 $ be $ x \, \text{kJ mol}^{-1} $.
- The bond enthalpy of $ B_2 $ be $ y \, \text{kJ mol}^{-1} $.
- The bond enthalpy of $ AB $ be $ z \, \text{kJ mol}^{-1} $.

From the problem, we are given:
1. The ratio of bond enthalpies is $ x : y : z = 2 : 1 : 2 $.
2. The bond enthalpy of formation of $ AB $ is $ -100 \, \text{kJ mol}^{-1} $.

Step 2: Express the bond enthalpies in terms of a common variable
Using the ratio $ x : y : z = 2 : 1 : 2 $, we can write:
$$ x = 2k, \quad y = k, \quad z = 2k $$
where $ k $ is a proportionality constant.

Step 3: Use the bond enthalpy of formation of $ AB $
The bond enthalpy of formation of $ AB $ is given by the equation:
$$ \Delta H_f = \frac{1}{2}x + \frac{1}{2}y - z $$
Substituting the values of $ x $, $ y $, and $ z $ in terms of $ k $:
$$ \Delta H_f = \frac{1}{2}(2k) + \frac{1}{2}(k) - (2k) $$
Simplify:
$$ \Delta H_f = k + \frac{k}{2} - 2k $$
Combine like terms:
$$ \Delta H_f = \frac{2k}{2} + \frac{k}{2} - \frac{4k}{2} = \frac{2k + k - 4k}{2} = \frac{-k}{2} $$

We are given that the bond enthalpy of formation of $ AB $ is $ -100 \, \text{kJ mol}^{-1} $. Therefore:
$$ \frac{-k}{2} = -100 $$

Step 4: Solve for $ k $
Multiply both sides by $ -2 $:
$$ k = 200 $$

Step 5: Determine the bond enthalpy of $ B_2 $
The bond enthalpy of $ B_2 $ is $ y = k $. From the previous step, we found $ k = 200 $. Therefore:
$$ y = 200 \, \text{kJ mol}^{-1} $$

Final Answer:
$$ {200 \, \text{kJ mol}^{-1}} $$

Thus, the correct option is:
$$ \boxed{\text{(C)}} $$

Answer 2

The bond enthalpies are in the ratio of 2 : 1 : 2 for A₂, B₂, and AB respectively. The bond enthalpy of AB formation is given as -100 KJ mol^-1. Based on the given ratio, the bond enthalpy of B₂ can be calculated by considering the relationship between bond enthalpies of the reactants and products. Using the given ratio, we find that the bond enthalpy of B₂ is 200 KJ mol^-1.

The correct answer is (C) : 200 KJ mol^-1.