Question

The amount of the heat released when $20\,mL$ $0.5\, M\, NaOH$ is mixed with $100\, mL\, 0.1\, M\, HCl$ is $x\, kJ$. The heat of neutralization is

• A. $- 100\, x\, kJ/mol$
• B. $- 50\, x \,kJ/mol$
• C. $+ 100\, x \,kJ/mol$
• D. $50\, x \,kJ/mol$

Answer 1

Answer: (A)

$- 100\, x\, kJ/mol$

Answer 2

Millimoles (or milliequivalents) of
$NaOH =20 \times 0.5=10$
Millimoles (or milliequivalents) of
$HCl =100 \times 0.1=10$
$\therefore \underset{\overset{10}{\text{milliequivalents}}}{NaOH}+ \underset{\overset{10}{\text{milliequivalents}}}{HCl} \rightarrow NaCl + H _{2} O$
Thus, heat released when 10 milliequivalents of $HCl$ are neutralised by 10 millimoles of $NaOH =\times kJ$. But heat of neutralisation is heat released when 1 equivalent of $HCl$ is neutralised by $1$ equivalent of $NaOH$.
$\therefore \Delta_{\text {neu }} H=-\frac{x}{10 \times 10^{-3}}$ or $-100 \times kJ / mol$