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Types of Redox Reactions

When 50 cm3...

Question

When $50 c m^{3}$ of $0.2 N H_{2} S O_{4}$ is mixed with $50 c m^{3}$ of $1 N K O H$ , the heat liberated is:

11.46 k J

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57.3 k J

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573 k J

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573 J

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Solution

The correct option is D $573 J$
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 $57300 \times 0.01 = 573 J$

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