| \[C{{l}_{2}}(g)\to 2Cl(g);\] | \[242.3kJ\text{ }mo{{l}^{-1}}\] |
| \[{{I}_{2}}(g)\to 2I(g);\] | \[151.0kJ\text{ }mo{{l}^{-1}}\] |
| \[ICl(g)\to I(g)+Cl(g);\] | \[211.3kJ\text{ }mo{{l}^{-1}}\] |
| \[{{I}_{2}}(s)\to {{I}_{2}}(g);\] | \[62.76kJ\text{ }mo{{l}^{-1}}\] |
A) +244.8 kJ \[mo{{l}^{-1}}\]
B) -14.6 kJ \[mo{{l}^{-1}}\]
C) -16.8kJ \[mo{{l}^{-1}}\]
D) + 16.8kJ \[mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[{{I}_{2}}+C{{l}_{2}}\to 2ICl\] \[\Delta H={{e}_{{{I}_{2}}S\to g}}+{{e}_{I-\,I}}+{{e}_{Cl-\,Cl}}-2\times {{e}_{1-\,Cl}}\] \[=62.76+151.0+242.3-2\times 211.3\] \[=33.46kJ\] for \[2\text{ }mol\text{ }ICl\] \[\therefore \Delta H/mol=\frac{33.46}{2}=16.73kJ\,mo{{l}^{-1}}\]
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