Question

A 150 mL of 0.4 $N$ HCI is neutralized with an excess of $N{H}_{4}OH$ in a Dewar vessel with a resulting rise in temperature of ${2.36}^{\circ }C$. If the heat capacity of Dewar and its content after the reaction is 315 cal/degree and if heat of neutralization in Kilocalorie $mo{l}^{-1}$ is $x$ then $-x/12$ is (nearest integer)__________.

Answer 1

A 150 mL of 0.4 N HCI corresponds to $150mL\times \frac{1L}{1000mL}\times 0.4mol/L=0.06mol$
The rise in the temperature is $2.36{}^{0}C$.
The heat capacity is 315 cal/degree.
The heat of neutralization for 0.06 mol is $315cal/degree\times 2.36{}^{0}C=743.4cal$.
The heat of neutralization for 1 mole is $\frac{-743.4cal}{0.06mol}=-12390cal=-12.39kcal=xkcal/mol$
Hence, $-x=12.39$

$-x/12\approx 1$