Question

Calculate the heat of neutralization from the following data.
200 ml of 1 M $HCl$ is mixed with 400 ml of 0.5 M $NaOH$ and the temperature rise in calorimeter was found to be ${4.4}^{\circ }\text{C}$ water equivalent of calorimeter is 12 g and specific heat is $2{\text{cal ml}}^{-1}{\text{degree}}^{-1}$ for solution. Report the absolute value of the answer you get in kcal

Answer 1

Milliequivalents of acid and base = 200 meq
200 meq of $HCl$ react with 200 meq of NaOH to produce heat $=\Delta H$.
$\therefore$ 1000 meq of $HCl$ when react with 1000 meq of $NaOH$ will give heat $=5\times \Delta H$
= Heat of neutralization
Now, heat produced during neutralisation of 200 meq of acid and base.
= Heat taken up by calorimeter + solution
$=M_1\times S_1\Delta T+M_2\times S_2\Delta T$
$=12\times 2\times 4.4+600\times 2\times 4.4=5385.6\text{cal}$
This is the heat sent out by the system (the neutralisation reaction) to the surrounding (the calorimeter which gets heated). Hence, to calculate the heat of neutralisation, $q=-5385.6\text{cal}$
$\therefore$ Heat of neutralization $=5\times q=5\times (-5385.6)=-26.928\text{kcal}$