Answer:
The given reaction is:
\[{{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g);{{\Delta
}_{r}}{{H}^{{}^\circ }}=-92.4\,kJmo{{l}^{-1}}\]
\[{{\Delta
}_{r}}H_{{{N}_{2}}(g)}^{{}^\circ }=0;\,\,\,\,\,{{\Delta
}_{f}}H_{{{H}_{2}}(g)}^{{}^\circ }=0;\,\,\,{{\Delta }_{f}}H_{N{{H}_{3}}(g)}^{{}^\circ
}=x\]
\[{{\Delta
}_{r}}H=\Sigma \,\,{{\Delta }_{f}}H_{(products)}^{{}^\circ }-\Sigma {{\Delta
}_{f}}H_{(reactants)}^{{}^\circ }\]
\[=2{{\Delta
}_{f}}H_{(N{{H}_{3}})}^{{}^\circ }-[{{\Delta }_{f}}H_{{{N}_{2}}(g)}^{{}^\circ
}+3{{\Delta }_{f}}H_{{{H}_{2}}(g)}^{{}^\circ }]\]
\[-92.4=2x-[0+3\times
0]\]
\[x=-\frac{1}{2}\times
92.4\]
\[=-46.2kJ\,mo{{l}^{-1}}\]
\[i.e.,\,{{\Delta
}_{f}}H_{N{{H}_{3}}(g)}^{{}^\circ }=-46.2kJ\,mo{{l}^{-1}}\]
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