Answer:
The given reaction:
\[2N{{H}_{3}}(g)\to
{{N}_{2}}(g)+2{{H}_{2}}(g)\]
\[{{\Delta }_{r}}{{H}^{{}^\circ
}}=[{{\Delta }_{f}}H_{{{N}_{2}}}^{{}^\circ }+3{{\Delta
}_{f}}H_{{{H}_{2}}}^{{}^\circ }]-[2{{\Delta }_{f}}H_{N{{H}_{3}}}^{{}^\circ
}]\]\[=0+3\times 0-2\times (-91.8)\]
\[=+183.6kJ\]
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