A) \[-13.7\text{ }kcal\]
B) \[-2.7\text{ }kcal\]
C) \[+30.1\text{ }kcal\]
D) \[+\text{ }3.01\text{ }kcal\]
Correct Answer: B
Solution :
(i) \[HF+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O+{{F}^{-}};\Delta \Eta =-16.4\,kcal\] (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) \[HF\xrightarrow{{}}{{H}^{+}}+{{F}^{-}};\Delta \Eta =-2.7kcal.\]
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