A) ? 44 kcal
B) 44 kcal
C) ? 22 kcal
D) 22 kcal
Correct Answer: C
Solution :
Aim: \[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\to HCl\] \[\Delta H=\sum B.E{{.}_{\text{(Products)}}}-\sum B.E{{.}_{\text{(Reactants)}}}\] \[=B.E.(HCl\,)-\left[ \frac{1}{2}B.E.({{H}_{2}})+\frac{1}{2}B.E.(C{{l}_{2}}) \right]\] \[=-103-\left[ \frac{1}{2}(\,-104\,)+\frac{1}{2}(\,-58) \right]\] \[=-103-(\,-52-29\,)=-22\,\,kcal\].You need to login to perform this action.
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