| \[C{{l}_{2}}(g)\xrightarrow{{}}2Cl(g),\,\,242.3\,kJ\,\,mo{{l}^{-1}}\] |
| \[{{I}_{2}}(g)\xrightarrow{{}}2I(g),\,\,151.0\,\,kJ\,\,mo{{l}^{-1}}\] |
| \[ICI(g)\xrightarrow{{}}I(g)+Cl(g),\,211.3\,\,kJ\,\,mo{{l}^{-1}}\] |
| \[{{I}_{2}}(s)\xrightarrow{{}}{{I}_{2}}(g),\,\,62.76\,\,kJ\,\,mo{{l}^{-1}}\] |
A) \[-14.6\,\,kJ\,\,mo{{l}^{-1}}\]
B) \[-16.8\,\,kJ\,\,mo{{l}^{-1}}\]
C) \[+16.8\,\,kJ\,\,mo{{l}^{-1}}\]
D) \[+244.8\,\,kJ\,\,mo{{l}^{-1}}\]
Correct Answer: C
Solution :
\[\frac{1}{2}{{l}_{2}}(s)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}lCl(g)\] \[\Delta H=\left[ \frac{1}{2}{{H}_{s\to g}}+\frac{1}{2}\Delta {{H}_{diss}}(C{{l}_{2}})+\frac{1}{2}\Delta {{H}_{diss}}({{l}_{2}}) \right]\] \[=\left( \frac{1}{2}\times 62.76+\frac{1}{2}\times 242.3+\frac{1}{2}\times 151.0 \right)-211.3\] = 228.03 - 211.3 \[\Delta H=16.73\]
You need to login to perform this action.
You will be redirected in
3 sec
