Heat of neuralisation is amount of heat evolved or absorbed when `1g-` equivalent of an acid reacts with `1g-` equivalent of a base in dilute solution . If weak acid or weak bae are neutralised, the heat released during neutralisation is somewhat lesser than `-13.7kcal` or `-57.27 kJ` . Het of neutralisation is also referred as heat of formation of water from `H^(+)` and `OH^(-)` ions `i.e., H^(+)+OH^(-) rarr H_(2)O, DeltaH=-13.7kcal. ` `200mL` of `0.1M NaOH` is mixed with `100mL` of `0.1M H_(2)SO_(4)` in 1 experiment. In `II` experiment `100mL` of `0.1M NaOH` is mixed with `50mL` of `0.1M H_(2)SO_(4)`. Select the correct statements: `(1)` heat liberated in each of the two reactions is `274cal`. `(2)` heat liberated in `I` is `274 cal` and in `II` is `137 cal`. ,brgt `(3)` temperature rise is `I` reaction is equal to the temperature rise in `II`. `(4)` temperature rise in `I` reaction is equal to the temperaure rise in `II`A. `1,3`B. `2,4`C. `2,3`D. `1,4`

Heat of neuralisation is amount of heat evolved or absorbed when `1g-`

1.	Heat of neuralisation is amount of heat evolved or absorbed when `1g-` equivalent of an acid reacts with `1g-` equivalent of a base in dilute solution . If weak acid or weak bae are neutralised, the heat released during neutralisation is somewhat lesser than `-13.7kcal` or `-57.27 kJ` . Het of neutralisation is also referred as heat of formation of water from `H^(+)` and `OH^(-)` ions `i.e., H^(+)+OH^(-) rarr H_(2)O, DeltaH=-13.7kcal. ` `200mL` of `0.1M NaOH` is mixed with `100mL` of `0.1M H_(2)SO_(4)` in 1 experiment. In `II` experiment `100mL` of `0.1M NaOH` is mixed with `50mL` of `0.1M H_(2)SO_(4)`. Select the correct statements: `(1)` heat liberated in each of the two reactions is `274cal`. `(2)` heat liberated in `I` is `274 cal` and in `II` is `137 cal`. ,brgt `(3)` temperature rise is `I` reaction is equal to the temperature rise in `II`. `(4)` temperature rise in `I` reaction is equal to the temperaure rise in `II`A. `1,3`B. `2,4`C. `2,3`D. `1,4`
Answer» Correct Answer - b `1 Meq.` of `NaOH=20, Meq . Of H_(2)SO_(4)=20` ltbr. `II Meq. Of NaOH=10, Meq. ofH_(2)SO_(4)=10` In `I,DeltaH=-(13.7xx10^(3)xx20)/(1000)` `=274 cal. `taken by `300mL, ` In `II: DeltaH=-(13.7xx10^(3)xx10)/(1000)` `=137cal. `taken by `150 mL`

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