What would be the heat released when:

Options

A. `0.5` mol of `HCI` is neutralised with `0.5` mol of `NaOH`
B. `0.5` mol of `HNO_(3)` is neutralised with `0.3` mol of `NaOH`
C. `100 ml` of `0.2M HCI` + `200 ml` of `0.2 M KOH`
D. `200 ml` of `0.1 M H_(2)SO_(4)` + `150 ml` of `0.2 M KOH`

Explanation

a. `0.5 mol` of `HCI` is neutralised by `0.5mol` of `NaOH`
`:. DeltaH =- 57.1 xx 0.5 =- 28.55 kJ`
b. `0.3 mol` of `HNO_(3)` will neutralise `0.3` mol of `NaOH`.
`:. DeltaH =- 57.1 xx 0.3 =- 17.1 kJ`
c. `HCI = 100 xx 0.2 = 20 mmol or 20 mEq`
`KOH = 200 xx 0.2 = 40 mmol or 40 mEq`
`:. 20 mEq` will neutralise
or `20 xx 10^(-3) Eq` will neutralise
`DeltaH = (-57.1 xx 20 xx10^(-3)) =- 1.14 kJ`
d. `H_(2)SO_(4) = 200 xx 0.1 xx2` (since `H_(2)SO_(4)` is diabasic)
`= 40 mEq`
`KOH = 150 xx 0.2 = 30 mEq`
`:. 30 mEq or 30 xx 10^(-3) Eq` will neutralise
`:. DeltaH =- 57.1 xx 30 xx 10^(-3) =- 1.713 kJ`