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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 KOH,...

When 50 cm3c{{m}^{3}} of 0.2 N H2SO4{{H}_{2}}S{{O}_{4}}​ is mixed with 50 cm3c{{m}^{3}} of 1 N KOH, the heat liberated is:
A. 11.46 kJ
B. 57.3 kJ
C. 573 kJ
D. 573 J

Explanation

Solution

There is a relationship between heat liberated during the neutralization reaction with the number of equivalents of the base and it is as follows.
heat liberated during the neutralization reaction = (number of equivalents of the base) (Heat liberated by the acid during the neutralization)

Complete answer:
- In the question it is asked to calculate the heat liberated When 50 cm3c{{m}^{3}} of 0.2 N H2SO4{{H}_{2}}S{{O}_{4}} ​ is mixed with 50 cm3c{{m}^{3}} of 1 N KOH.
- With the given data in the question we can calculate the number of equivalents of the KOH is neutralized by 0.2 N H2SO4{{H}_{2}}S{{O}_{4}}.
- Number of equivalents of the KOH neutralized =50×0.21000=0.01=\dfrac{50\times 0.2}{1000}=0.01 .
- Means 0.01 equivalents of KOH is going to be neutralized with 0.2 N H2SO4{{H}_{2}}S{{O}_{4}} with 50 cm3c{{m}^{3}} volume.
- Now we have to calculate the heat liberated during the mixing of the 50 cm3c{{m}^{3}} of 0.2 N H2SO4{{H}_{2}}S{{O}_{4}} ​ with 50 cm3c{{m}^{3}} 1 N KOH.
- We know that the heat which is going to be released when 1 equivalent of the KOH is going to react with 1 equivalent of sulphuric acid is 57.3 kJ or 57300 J.
- Now the heat liberated during the mixing of the 50 cm3c{{m}^{3}} of 0.2 N H2SO4{{H}_{2}}S{{O}_{4}} with 50 cm3c{{m}^{3}} 1 N KOH.
= (57300) (0.01)
= 573 J.

So, the correct option is D.

Note:
We should know the heat which is going to be released during the reaction of the I equivalent of the KOH with one equivalent of the sulphuric acid. Without this we cannot calculate the heat liberated when different concentrations of the acid are going to react with different concentrations of the base.