Question

A 150 c.c. portion of 0-4 N HCl is neutralised with an excess of NH4OH on a
Dewar vessel with a resulting rise in temperature of 2.36°C. If the heat capacity
of the Dewar and its contents after the reaction is 315 Cal/degree, calculate heat
of neutralisation in calories/mole.
​

Answer 1

Answer:

A 150 mL of 0.4 N HCI corresponds to 150mL×  

1000mL

1L

​  

×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×2.36  

0

C=743.4cal.

The heat of neutralization for 1 mole is  

0.06mol

−743.4cal

​  

=−12390cal=−12.39kcal=xkcal/mol

Hence, −x=12.39

−x/12≈1