What would be the heat released when 0.5 mole of HCl in solution is neutralised by 0.25 mole of KOH?

Explanation

In this case, both HCl and KOH are strong electrolytes and dissociate completely in solution. Therefore,
0.5 mole of HCl = 0.5 mole of `H^(+)-=` 0.5 gram equivalents of `H^(+)`
0.5 moel of KOH `-=0.5` mole of `OH^(-)-= 0.5` gram equivalents of `OH^(-)`.
When one gram equivalent of `H^(+)` ions combine with one gram equivalent of `OH^(-)` ions to form one mole of `H_2O`, the heat evolved is 57.1 kJ, i.e.,
`underset("1 gm eq.")(H^(+)(aq)) + underset("1 gm eq.")(OH^(-)(aq)) to underset("1 mole")(H_2O(l)) , DeltaH = - 57.1 kJ`
The combination of 0.5 gram equivalent each of `H^(+)` and `OH^(-)` will produce 0.5 mole of `H_2O (l)`.
`:.` Heat evolved in the formation of 0.5 mole of `H_2O`
`57.1 xx 0.5 = 28.5 kJ`