Saved Bookmarks
| 1. |
the heat of neutralisation of acetic acid and sodium hydroxide is -50.6 kJ eq^(-1). Find the heat of dissociation of CH_(3)COOH if the heat of neutralisation of a strong acid and a strong base is -55.9 Kj eq^(-1). |
|
Answer» Solution :We have, `DeltaH` (neutralisation) `=DeltaH ( "ionisation of" CH_(3)COOH)+DeltaH(H^(+)+OH^(-))` `:.DeltaH("ionisation of" CH_(3)COOH)=-50.6-(-55.9)=5.3 "kJ mol"^(-1)` [DeltaH (ionisation of NaOH)=0 as NaOH is a strong BASE] |
|