Question

A₂+B₂→2AB. \(ΔH^0_f\)= -200 kJ mol^-1 AB, A2 and B2 are diatomic molecule. If the bond enthalpies of A₂, B₂ and AB are in the ratio 1:0.5:1, then the bond enthalpy of A₂ is ___ kJmol^-1 (Nearest integer)

Answer 1

Correct Answer: 800

Solution:
The given reaction is: \[ A_2 + B_2 \rightarrow 2AB. \] The bond enthalpy change \( \Delta H \) for this reaction is given as \( -200 \, \text{kJ/mol} \). Let the bond enthalpies of \( A_2 \), \( B_2 \), and \( AB \) be \( x \), \( y \), and \( z \) respectively. We are told that the bond enthalpies are in the ratio: \[ \frac{x}{y} = 1:0.5:1 \quad \Rightarrow \quad x = 2y \quad \text{and} \quad z = y. \] Now, applying the concept of bond enthalpy for the reaction: \[ \Delta H = \text{Bonds broken} - \text{Bonds formed}. \] The bonds broken are the bonds in \( A_2 \) and \( B_2 \), and the bonds formed are the bonds in \( AB \). Hence, we have: \[ \Delta H = (x + y) - 2z. \] Substitute \( x = 2y \) and \( z = y \) into the equation: \[ \Delta H = (2y + y) - 2y = y. \] We are given that \( \Delta H = -200 \, \text{kJ/mol} \), so: \[ y = -200 \, \text{kJ/mol}. \] Finally, since \( x = 2y \), the bond enthalpy of \( A_2 \) is: \[ x = 2(-200) = -400 \, \text{kJ/mol}. \] Thus, the bond enthalpy of \( A_2 \) is -400 kJ/mol.