Question

D2 + ½O2 gives D2O, ΔH=-294 KJ. If bond enthalpies of D2 and O2 respectively are 440 and 498 KJ/mole, then bond enthalpy of D-o is (KJ)​

Answer 1

The bond enthalpy of D-O is 491.5 kJ

Explanation:

The balanced chemical reaction is,

$D_2+\frac{1}{2}O_2\rightarrow D_2O$

The expression for enthalpy change is,

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

$\Delta H=[(n_{D_2}\times B.E_{D_2})+(n_{O_2}\times B.E_{O_2}) ]-[(n_{D_2O}\times B.E_{D_2O})]$

$\Delta H=[(n_{D_2}\times B.E_{D=D})+(n_{O_2}\times B.E_{O=O}) ]-[(n_{D-2O}\times 2\times B.E_{D-O})]$

where,

n = number of moles

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

$-294=[(1\times 440)+(\frac{1}{2}\times 498)]-[(1\times 2\times B.E_{D-O})]$

$B.E_{D-O}=491.5kJ$

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