A) \[-13.7\,\,kcal\]
B) \[-\text{ }2.7\,\,kcal\]
C) \[+30.1\,\,kcal\]
D) \[+3.01\,\,kcal\]
Correct Answer: B
Solution :
(i)\[HF+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O+{{F}^{-}};\,\,\Delta H=-16.4\,\,kcal\] (ii)\[{{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O;\,\,\Delta H=-13.7\,\,kcal\] From Eqs. (i)-(ii) gives the required equation (ionisation of \[HF\] in water) \[HF\xrightarrow{{}}{{H}^{+}}+{{F}^{-}};\,\,\Delta H=-2.7\,\,kcal\]
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