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.

Explanation

Given total solution `= 200 +400 = 600mL`
`200mL` of `1M HCI` neutralise `400mL` of `0.5M NaOH mEq = N xx V`
`:. mM`or `mEq` of acid and base `= 200` (For monovalent neture)
i.e., `200mEq` of `HCI` react with `200mEq` of `NaOH` to produc heat `=DeltaH`
`1000mEq of HCI + 1000 mEq of NaOH` will given heat `=5 xx DeltaH` i.e., heat of neutralisation.
Now, `DeltaH`, i.e., heat produced during neutralisation of `200mEq` of acid and base
`=` Heat taken up by calorimeter `+` solution
`= m_(1)s_(1)DeltaT +m_(2)s_(2)DeltaT`
`=12 xx 4.4 xx 600 xx1xx4.4 = 2682.8 cal`
Heat of neutralisation `=- 5xx 2692.8 cal=-13.464 kcal.`