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# 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?

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Solution

## Step 1: Given dataBond Enthalpies Of ${A}_{2},AbAnd{B}_{2}$ Are In The Ratio $1:1:0.5$$∆\mathrm{H}=-100\mathrm{KJ}$Step 2: Deriving formula for determining bond energyLet bond energy of ${A}_{2}beX$, then bond energy of $AB$ is also $\mathrm{x}$Let bond energy of ${B}_{2}$ be $\frac{x}{2}$Enthalpy of formation of$AB$ is $-100KJ/mol.$$={A}_{2}+{B}_{2}\phantom{\rule{0ex}{0ex}}=2AB$ $=\frac{1}{2}{A}_{2}+\frac{1}{2}{B}_{2}$ $=AB$Step 3: Calculating Bond energy $-100=\left(\frac{x}{2}+\frac{x}{4}\right)-x$$-100=\frac{2x+x+4x}{4}$$=400KJ$Therefore, the bond energy of ${A}_{2}is400KJ$

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