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Calculate the heat of neutralisation fro...

Calculate the heat of neutralisation from the following data:
`200mL` of `1M HCI` is mixed with `400mL` of `0.5M NaOH`. The temperature rise in calorimeter was found to be `4.4^(@)C`. Water equivalent of calorimeter is `12g` and specific heat is `1cal mL^(-1) degree^(-1)` for solution.

Text Solution

Verified by Experts

Meq. Of acid and base = 200
i.e. 200 Meq of HCl react with 200 Meq. Of NaOH to produce heat = `DeltaH`
`therefore 1000` Meq of HCl when react with 1000 Meq. Of NaOH will give heat = `5 xx DeltaH` = Heat of neutralization.
`=M_(1) xx S_(1)DeltaT + M_(2) xx S_(2)DeltaT`
`=12 xx 1 xx 4.4 + 600 xx 1 xx 4.4 = 2692.8` cal
`therefore` Heat of neutralization = `5 xx 2692.8` cal
`=-13.464` kcals
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