\[N{{H}_{3}}(g)+3C{{l}_{2}}(g)\overset{{}}{leftrightarrows}NC{{l}_{3}}(g)+3HCl(g);-\Delta {{H}_{1}}\] |
\[{{N}_{2}}(g)+3{{H}_{2}}(g)\overset{{}}{leftrightarrows}2N{{H}_{3}}(g);-\Delta {{H}_{2}}\] |
\[{{H}_{2}}(g)+C{{l}_{2}}(g)\overset{{}}{leftrightarrows}2HCl(g);-\Delta {{H}_{3}}\] |
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) \[\Delta {{H}_{f}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{2}}\]
Correct Answer: B
Solution :
\[-\Delta {{H}_{1}}=-\Delta {{H}_{f\,\,NC{{l}_{3}}}}+3\times \left( -\frac{\Delta {{H}_{3}}}{2} \right)-\left( -\frac{\Delta {{H}_{2}}}{2} \right)\]You need to login to perform this action.
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