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AIEEE 2010 Solved Paper

Section: Chemistry

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Question : 5 of 89

Marks: +1, -0

The standard enthalpy of formation of

\mathrm{NH}_3

-46.0\ \text{kJ mol}^{-1}

. If the enthalpy of formation of

\mathrm{H}_2

from its atoms is

-436\ \text{kJ mol}^{-1}

\mathrm{N}_2

-712\ \text{kJ mol}^{-1}

, the average bond enthalpy of

\mathrm{N}-\mathrm{H}

\mathrm{NH}_3

$-964\ \text{kJ mol}^{-1}$
$+352\ \text{kJ mol}^{-1}$
$+1056\ \text{kJ mol}^{-1}$
$-1102\ \text{kJ mol}^{-1}$

Solution:

\mathrm{N}_2+3\ \mathrm{H}_2 \rightarrow 2\ \mathrm{NH}_3

\Delta H = 2 \times -46.0\ \text{kJ mol}^{-1}

Let

x

be the bond enthalpy of

\mathrm{N}-\mathrm{H}

bond then

[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive. ]

\Delta H = \sum

Bond energies of products

-\sum

Bond energies of reactants

2 \times -46 = 712 + 3 \times (436) - 6 x

-92 = 2020 - 6 x

6 x = 2020 + 92

\Rightarrow 6 x = 2112

\Rightarrow x = +352\ \text{kJ}/\text{mol}

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