Question

When $50c{m}^3$ of $0.2N$ $H_{2}S{O}_4$ is mixed with $50c{m}^3$ of $1NKOH$, 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 $57.3kJ$ or $57300J$.
The heat liberated for the neutralization of 1 equivalent of $KOH$ with 1 equivalent of sulphuric acid is $57300\times 0.01=573J$