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Describe the measurement of Delta U by b...

Describe the measurement of `Delta U` by bomb calorimeter.

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When a reaction occurs in an aqueous solution, the reaction is siad to occur at constant pressure (atmospheric pressure). Then the heat liberated or absorbed is the heat of reaction at constant pressure `(Delta H)` . It can be measures by using a calorimeter.
Let us take the example of enthalpy of neutralisation of an acid by a base. The apparatus consists of a polythene bottle fitted with a coke carrying a thermometer and a stirrer.
Let us consider the determination of enthalpy of neutalisation of hydrochloric acid by sodium hydroxide solution.
PROCEDURE :
(a) `50 cm^(3)` of 0.1 N HCl is taken in a breaker and kept in a water bath taken in an insulated vessel.
(b) `50 cm^(3)` of 0.1 N NaoH is taken in another beaker and kep in the same water bath.
(c ) When the two solutions attains the same initial temperature as that of the water bath, the temperature is noted. Let it be `t_(1).^(@)C`
HCl in the first beaker is poured into the ploythene bottle. Immediately NaOH solution kep in the second beaker is added into the polythene bottle. THe neutralisation reaction occurs with the liberation of heat. The temperature rises. When it reaches a constant temperature, it is noted. Let it be `t_(2).^(@)C`. It is the final temperature.
CALCULATION :
Heat capacity of the polythene bottle is negligible and not considered. The amount of NaCl formed in the reaction is negligible. Its heat capacity is also negligible. The heat liberated in the reaction is taken as completely absorbed by `100 cm^(3)` of water.
Rise in temperature `= (t_(2) - t_(2))^(@)C`
Mass of water = 50 + 50 = 100 g
Specific heat capacity of water `= 4.184 Jg^(-1)`
Heat liberated in the reaction = heat gained by water
`= "mass" xx "specific heat" xx "rise in temperature"`
`= 100 xx 4.184 xx (t_(2) - t_(1))`
= xJ
Heat liberated by the neutralisation of `50 cm^(3)` of 0.1 N HCl = x J
Heat liberated by the neutralisation of `1000 cm^(3)` of 1N HCl will be `(x xx 1000 xx 1)/(50 xx 0.1) = QJ`
Enthalpy of neutralisation of HCl by NaOH = -QJ.
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