Question

$150$ ml of $0.5N$ nitric acid solution at ${25.35}^{o}C$ was mixed with $150$ ml of $0.5N$ sodium hydroxide solution at the same temperature. The final temperature was recorded to be ${28.77}^{o}C$. The heat of neutralisation of nitric acid with sodium hydroxide is:

• A. -12.64 kcal
• B. -11.98 kcal
• C. -13.68 kcal
• D. -12.68 kcal

Answer 1

Answer: (C)

-13.68 kcal

The temperature change $\Delta T=28.77{}^{o}C-25.35{}^{o}C=3.42{}^{o}C$
The specific heat of water is 1 cal/g.K.
The density of the solution is approximately equal to the density of water which is 1 g/ml
The volume of solution will be $150+150=300ml$
The mass of solution will be $300ml\times 1g/ml=300g$
The heat released during neutralisation
$q=mS\Delta T=-300\times 1\times 3.42=1026cal=1.026kcal$
The number of moles (gram equivalents) of nitric acid will be
$Molarity\times Volume=0.150\times 0.5=0.075$ moles

[Note: n-factor is 1. So, molarity is numerically equal to normality]

Heat of neutralisation of nitric acid with sodium hydroxide will be
$=\frac{-1.026}{0.075}=-13.68kcal/mol$