\[NC{{l}_{3}}(g)+3HCl(g);-\Delta {{H}_{1}}\] (1) |
\[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g);-\Delta {{H}_{2}}\] (2) |
\[{{H}_{2}}(g)+C{{l}_{2}}(g)\rightleftharpoons 2HCl(g);\Delta {{H }_{3}}\] (3) |
A) \[\Delta {{H}_{f}}=\Delta {{H}_{1}}+\frac{\Delta {{H}_{2}}}{2}-\frac{3}{2}\Delta {{H}_{3}}\]
B) \[\Delta {{H}_{f}}=-\Delta {{H}_{1}}+\frac{\Delta {{H}_{2}}}{2}-\frac{3}{2}\Delta {{H}_{3}}\]
C) \[\Delta {{H}_{f}}=\Delta {{H}_{1}}-\frac{\Delta {{H}_{2}}}{2}-\frac{3}{2}\Delta {{H}_{3}}\]
D) None of the above
Correct Answer: B
Solution :
To get\[\frac{1}{2}{{N}_{2}}+\frac{3}{2}C{{l}_{2}}\to NC{{l}_{3}};\,\Delta {{\Eta }_{f}}\] Multiply equation 2 by \[\frac{1}{2},\] equation 3 by \[\frac{3}{2}\] and subtract from the equation 1 and \[\Delta {{H}_{f}}=\Delta {{H}_{1}}-\left( -\frac{\Delta {{H}_{2}}}{2}+\frac{3}{2}\Delta {{\Eta }_{3}} \right)\] \[=-\Delta {{H}_{1}}+\frac{\Delta {{\Eta }_{2}}}{2}-\frac{3}{2}\Delta {{\Eta }_{3}}\] Hence, the correct option is (b).You need to login to perform this action.
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