of \[NC{{l}_{3}}(g)\] in terms of x, y and z.
| Given: |
| I. \[N{{H}_{3(g)}}+3C{{l}_{2(g)}}\to NC{{l}_{3(g)}}+3HC{{l}_{(g),}}\Delta {{H}_{1}}=x\] |
| II. \[{{N}_{2(g)}}+3{{H}_{2(g)}}\to 2N{{H}_{3(g)}},\Delta {{H}_{2}}=y\] |
| III. \[{{H}_{2(g)}}+C{{l}_{2(g)}}\to 2HC{{l}_{(g)}},\Delta {{H}_{3}}=z\] |
A) \[x-\frac{y}{2}+\frac{3z}{2}\]
B) \[x+\frac{y}{2}-\frac{3z}{2}\]
C) \[x+(y-3z)\]
D) \[x-y+3z\]
Correct Answer: B
Solution :
| \[\frac{1}{2}{{N}_{2}}+\frac{3}{2}C{{l}_{2}}\to NC{{l}_{3}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
