Question

Consider the following reaction: A2 + B2 → 2AB ΔH = –321 kJ Bond energy (A2) = 1/2AB Bond energy (B2) =393 kJ/mol. What is the numerical value for Bond energy (A2) ?

Answer 1

Answer: The numerical value for Bond energy of $A_2$ is -238 kJ/mol

Explanation:

The balanced chemical reaction is,

$A_2+B_2(g)\rightarrow 2AB$    $\Delta H=-321kJ$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$

$\Delta H=[(n_{A_2}\times B.E_{A_2})+(n_{B_2}\times B.E_{B_2}) ]-[(n_{AB}\times B.E_{AB})]$

$\Delta H=[(1\times x)+(1\times B.E_{B_2}) ]-[(2\times 2x)]$

where,

n = number of moles

If $B.E_{A_2}=x$    $B.E_{AB}=2x$

Now put all the given values in this expression, we get

$-321=[(1\times x)+(1\times 393)]-[(2\times 2x)]$

$x=-238kJ/mol$

Therefore, the bond energy of $A_2$ is -238 kJ/mol