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

Answer 1

The correct option is D $573J$

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 $57300\times 0.01=573J$