Question

Ab, A₂ And B₂ Are Diatomic Molecules. If The Bond Enthalpies Of A₂, Ab And B₂ Are In The Ratio 1:1:0.5 And Enthalpy Of Formation Of Ab From A ₂ And B₂ Is −100 Kj Mol−1. What Is The Bond Energy Of A₂?

Answer 1

Step 1: Given data

Bond Enthalpies Of $A_2,AbAnd{B}_2$ Are In The Ratio $1:1:0.5$

$∆\mathrm{H}=-100\mathrm{KJ}$

Step 2: Deriving formula for determining bond energy

Let bond energy of $A_{2}beX$, then bond energy of $AB$ is also $\mathrm{x}$

Let bond energy of $B_2$ be $\frac{x}{2}$

Enthalpy of formation of$AB$ is $-100KJ/mol.$

$$\begin{gathered} =A_2+B_2 \\ =2AB \end{gathered}$$

$=\frac{1}{2}{A}_2+\frac{1}{2}{B}_2$

$=AB$

Step 3: Calculating Bond energy

$-100=(\frac{x}{2}+\frac{x}{4})-x$

$-100=\frac{2x+x+4x}{4}$

$=400KJ$

Therefore, the bond energy of $A_{2}is400KJ$