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The heats fo neutralization of `HCl` with `NH_(4) OH` and `NaOH` with `CH_(3) COOH` are `- 51.4 kJ eq^(-1)` and `- 50.6 kJ eq^(-1)`, respectively. The heat of neutralization of acetic acid with `NH_(4) OH` will beA. `-44.6 kJ eq^(-1)`B. `-50.6 kJ eq^(-1)`C. `-51.4 kJ eq^(-1)`D. `-57.4 kJ eq^(-1)` |
Answer» Correct Answer - A `(i) HCl+NH_(4)OHtoNH_(4)Cl+H_(2)O`, `DeltaH=-51.4 kJ` `(ii) CH_(3)COOH+NaOHtoCH_(3)COONa+H_(2)O`, `DeltaH=-50.6 kJ` `(iii) CH_(3)COOH+NH_(4)OHtoCH_(3)COONH_(4)+H_(2)O`, `DeltaH=?` We know `(iv) HCl+NaOH to NaCl+H_(2)O`, `DeltaH=-57.4 kJ` To get `(iii)` add `(i)` and `(ii)` and subtract `(iv)`. (Please note that salts formed in case `(i)` and `(ii)` are almost completely ionised. Also suppose that the salt formed in case `(iii)` is also ionised). `DeltaH=(-51.4)+(-50.6)-(-57.4)kJ` `=-51.4-50.6+57.4=-44.6 kJ` |
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