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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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The standard enthalpy of formation of $NH_3$ is $-46\: kJ/mol.$ If the enthalpy of formation of $H_2$ from its atoms is $-436\: kJ/mol$ and that of $N_2$ is $-712\: kJ/mol$, the average bond enthalpy of $N-H$ bond in $NH_3$ is

$\begin {array} {1 1} (A)\;-964\: kJ/mol & \quad (B)\;352\: kJ/mol \\ (C)\; 1056\: kJ/mol & \quad (D)\;-1102 \: kJ/mol \end {array}$

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1 Answer

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$\large\frac{1}{2}$$ N_2(g) + H_2(g) \rightarrow NH_3(g); \Delta H_f^{\circ} = -46 \: kJ/mol $
$2H(g) \rightarrow H_2(g);\: \: \: \: \: \Delta H_f^{\circ} = -436\: kJ/mol $
$2N(g) \rightarrow N_2(g);\: \: \: \: \: \Delta H_f^{\circ} = -712\: kJ/mol $
Let $x$ is bond energy of $N-H$ bond
$ \large\frac{1}{2}$$ \times(-712) + 3(-436)/2 – 3x = -46$
$ \Rightarrow 3x = 1056$
$\Rightarrow x = 352\: kJ/mol$
Ans : (B)
answered Mar 17, 2014 by thanvigandhi_1
 

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