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.

Options

A. `-13.464 kcal`
B. `-12.02 kcal`
C. `-15.262 kcal`
D. `-20 . 92 kcal`

Answer

A

Explanation

Meq. of acid and base = 200
i.e. 200 Meq. of HCl react with 200 Meq. of NaOH to produce heat = `DeltaH`
`:.` 1000 Meq. of HCl when react with 1000 Meq. of NaOH will give heat
`= 5 xx DeltaH` = Heat of neutralization
Now, heat produced during neutralisation of 200 Meq. of acid and base
= heat taken up by calorimeter + solution = `M_1 xx S_1DeltaT + M_2 xx S_2DeltaT`
`= 12 xx 1 xx 4.4 + 600 xx 1 xx 4.4 = 2692.8 cal `
`:. "Heat of neutralization" = 5 xx 2692.8 cal = – 13.464 kcals`.