Question

Heat liberated when 100 ml of 1 N NaOH is neutralized by 300 ml of 1N HCl is

• A. 5.73 kJ
• B. 11.46 kJ
• C. 17.19 kJ
• D. 22.92 kJ

Answer 1

The correct option is A

5.73 kJ

NaOH is the limiting reactant (available in smaller proportion).
Heat of formation of water is $57.3kJ/mol$, this is the underlying reaction that happens when acids and bases neutralise too.
Also, normality is the same as normality here.

$\therefore$ Heat liberated $=\frac{57.3kJ}{1mol}\times \frac{1mol}{1L}\times \frac{1L}{1000mL}\times 100mL=5.73kJ$