Question

When $50\, cm ^{3}$ of $0.2\, N$ $H_{2} SO _{4}$ is mixed with $50 cm ^{3}$ of $1 N\, KOH$, the heat liberated is :

• A. 573 kJ
• B. 573 J
• C. 11.46 kJ
• D. 57.3 kJ

Answer 1

Answer: (B)

Number of equivalents of KOH neutralized $=\frac{50 \times 0.2}{1000}=0.01$
The heat liberated for the neutralization of 1 equivalent of $KOH$ with 1 equivalent of sulphuric acid is $57.3 kJ$ or $57300 J$.
The heat liberated for the neutralization of 1 equivalent of $KOH$ with 1 equivalent of sulphuric acid is $57300 \times 0.01=573 J$