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InterviewSolution
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What would be the heat released when: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` |
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Answer» 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` |
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