question_answer

Given:

\[N{{H}_{3}}(g)+3C{{l}_{2}}(g)\overset{{}}{leftrightarrows}NC{{l}_{3}}(g)+3HCl(g);-\Delta {{H}_{1}}\]

\[{{N}_{2}}(g)+3{{H}_{2}}(g)\overset{{}}{leftrightarrows}2N{{H}_{3}}(g);-\Delta {{H}_{2}}\]

\[{{H}_{2}}(g)+C{{l}_{2}}(g)\overset{{}}{leftrightarrows}2HCl(g);-\Delta {{H}_{3}}\]

The heat of formation of \[NC{{l}_{3}}(g)\] in terms of\[\Delta {{H}_{1}},\Delta {{H}_{2}}\], and \[\Delta {{H}_{3}}\] is :

A) \[\Delta {{H}_{1}}=\Delta {{H}_{f}}+\frac{\Delta {{H}_{2}}}{2}-\frac{3}{2}\Delta {{H}_{3}}\]

B) \[\Delta {{H}_{1}}=\Delta {{H}_{f}}-\frac{\Delta {{H}_{2}}}{2}+\frac{3}{2}\Delta {{H}_{3}}\]

C) \[\Delta {{H}_{f}}=\Delta {{H}_{1}}+\frac{\Delta {{H}_{2}}}{2}-\frac{3}{2}\Delta {{H}_{3}}\]

D) \[\Delta {{H}_{f}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{2}}\]