Question

What would be the heat released when: (i) 0.25 mol of hydrochloric acid solution is neutralised by 0.25 mol of sodium hydroxide solution? (ii) 0.02 mol of sulphuric acid solution is mixed with 0.2 mol of potassium hydroxide solution?

Answer 1

Here acid is Present in excess. So 0.2 mole of KOH will be completely neutralized by acid.

Means, 0.2 mole of base completely neutralized by acid.

Enthalpy of neutralization ( for strong acid base reaction at 25℃ ) is 57.62 kJ per mole.

So, heat evolved = 57.62x0.2 = 11.524 kJ.

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Answer 2

(i) Given, 0.25 mol of hydrochloric acid solution is neutralized by 0.25 mol of sodium hydroxide solution.

At $25^{ \circ  }$C for strong acid base reaction, the enthalphy of neutralization is 57.62KJ per mole .

To find: How much heat released.  

                                  0.25 mole HCl + 0.25 mole NaOH,

Then the net reaction is,

${ H }^{ + } (0.25 mole) + { OH }^{ - }(0.25mole) \rightarrow { H }_{ 2 }O (0.25 mole)$

Therefore the heat released is $57.62  \times  0.25 KJ =14.4 KJ$

(ii) Given, 0.02 mol of sulphuric acid solution is mixed with 0.2 mol of potassium hydroxide solution.

To find: How much heat released.  

0.02 mole ${ H }_{ 2 }{ SO }_{ 4 }$  + 0.2 mole KOH,

1 mole of ${ H }_{ 2 }{ SO }_{ 4 }$ contains 2 mole of ${ H }^{ + }$

Therefore 0.02 mole of ${ H }_{ 2 }{ SO }_{ 4 }$ contains 0.04 mole of ${ H }^{ + }$

Then, the net reaction is ,

${ H }^{ + } (0.04 mole) + { OH }^{ - }(0.2mole) \rightarrow { H }_{ 2 }O$ (0.04 mole)

Therefore, the heat released is $57.62 \times 0.04 KJ = 2.304 KJ.$