A) \[-44\,\,kcal\]
B) \[44\,\,kcal\]
C) \[-22\,\,kcal\]
D) \[22\,\,kcal\]
Correct Answer: C
Solution :
\[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\xrightarrow{{}}HCl\] \[\Delta H=\Sigma B{{E}_{\text{reactants}}}-\Sigma B{{E}_{\text{products}}}\] \[=\left[ \frac{1}{2}BE({{H}_{2}})+\frac{1}{2}BE{{(Cl)}_{2}} \right]-BE(HCl)\] \[=\left[ \left( \frac{1}{2}\times 104 \right)+\left( \frac{1}{2}\times 58 \right) \right]-103\] \[=(52+29)-103\] \[=-22\,\,kcal\]You need to login to perform this action.
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