A) \[-\text{ }44\text{ }kcal\]
B) \[-\text{ }88\text{ }kcal\]
C) \[-22\text{ }kcal\]
D) \[-\text{ }11\text{ }kcal\]
Correct Answer: C
Solution :
\[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\xrightarrow{{}}HCl\] \[\Delta H=\Sigma BE\](reactants)\[-\Sigma BE\](products) \[=BE(HCl)-\left[ \frac{1}{2}BE({{H}_{2}})+\frac{1}{2}BE(C{{l}_{2}}) \right]\] \[=-103-\left[ \frac{1}{2}\times (-104)+\frac{1}{2}\times (-58) \right]\] \[=-103-[52-29]\] \[=-103+81\] \[=-22kcal\]
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