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100 cm^(3) of 0.5N HCI solution at 299.9...

`100 cm^(3)` of `0.5N HCI` solution at `299.95 K` was mixed with `100 cm^(3)` of ` 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final temperature was found to be `302.65K`. Calculate the enthalpy of neutralisation of `HCI`. Water equivalent of thermos flask is `44 g`.

Text Solution

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The initial average temperature of the acid and the
`=(299.95+299.75)/(2)=299.85K`
rise in temperature `=(302.65-299.65)=2.80K`
Heat evolved during neutralisation
`=(100+1000+44)xx4.184xx2.8=2858.5J`
`therefore` Enthalpy of neutralisation `=-(2858.5)/(100)xx1000xx(1)/(0.50)`
`=-57.17kJ`
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