A) \[-\,22\text{ kcal mo}{{\text{l}}^{-1}}\]
B) \[-\,44\text{ kcal mo}{{\text{l}}^{-1}}\]
C) \[+\,44\text{ kcal mo}{{\text{l}}^{-1}}\]
D) \[+\,22\text{ kcal mo}{{\text{l}}^{-1}}\]
E) \[-\,11\text{ kcal mo}{{\text{l}}^{-1}}\]
Correct Answer: A
Solution :
\[{{H}_{2}}+C{{l}_{2}}\xrightarrow{{}}2HCl\] Enthalpy of formation of\[HCl\] \[=\frac{(104+58)-(2\times 103)}{2}\] \[=\frac{162-206}{2}=-\frac{44}{2}\] \[=-\,22\,\text{kcal}\,\text{mo}{{\text{l}}^{-1}}\]You need to login to perform this action.
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