JEE MAIN - Chemistry (2010 - No. 9)
The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N–H bond in NH3 is :
–964 kJ mol–1
+352 kJ mol–1
+ 1056 kJ mol–1
–1102 kJ mol–1
Explanation
$${N_2} + 3{H_2} \to 2N{H_3}\,$$
$$\,\Delta H = 2 \times - 46.0\,\,kJ\,mo{l^{ - 1}}$$
Let $$x$$ be the bond enthalpy of $$N-H$$ 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
$$2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x;$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\, - 92 = 2020 - 6x$$
$$6x = 2020 + 92$$
$$ \Rightarrow 6x = 2112$$
$$ \Rightarrow x = + 352\,kJ/mol$$
$$\,\Delta H = 2 \times - 46.0\,\,kJ\,mo{l^{ - 1}}$$
Let $$x$$ be the bond enthalpy of $$N-H$$ 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
$$2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x;$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\, - 92 = 2020 - 6x$$
$$6x = 2020 + 92$$
$$ \Rightarrow 6x = 2112$$
$$ \Rightarrow x = + 352\,kJ/mol$$
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