Whenever an acid is neutralised by a base, the net reaction is
`H^(o+)(aq)+overset(Theta)OH(aq)rarr H_(2)O(l),DeltaH =- 57.1 kJ`
Calculated the heat evolved for the following experiments?
a. `0.50 mol` of `HCI` solution is neutralised by `0.50mol` of `NaoH` solution.
b. `0.50mol` of `HNO_(3)` solution is mixed with `0.30mol` of `KOH` solution.
c. `100mL` of `0.2M HCI` is mixed with `100mL` fo `0.3M NaOH` solution.
d. `400mL` of `0.2M H_(2)SO_(4)` is mixed with `600mL` of `0.1M KOH` solution.

Explanation

According to the reaction
`H^(o+)(aq)+overset(Theta)OH(aq) rarr H_(2)O(l),DeltaH =- 57.1 kJ`
when `1mol` of `H^(o+)` ions and `1mol` of `overset(Theta)OH` ions are neutralised, `1mol` of water is formed and `57.1 kJ` of enegry is released.
a. `0.50 mol HCI -= 0.50 mol H^(o+)` ions
`0.50mol NaOH -= 0.50 mol overset(Theta)OH` ions
On mixing, `0.50 mol` of water is formed.
Heat evolved for the formation of `0.50mol` of water `=57.1 xx 0.5 = 28.55 kJ`
b. `0.50 mol HNO_(3)-= 0.50 mol H^(o+)` ions
`0.30 mol KOH -= 0.30 overset(Theta)OH` ions
i.e., `0.30 mol of H^(o+)` ions reacts with `0.30 mol` of `overset(Theta)OH` ions to form `0.30 mol` of water molecules
Heat evolved in the formation of `0.3 mol` of water `=57.1 xx 0.3 = 17.13kJ`
c. `100 mL` of `0.02 M HCI` will given
`((0.2)/(1000xx100)) = 0.02 mol of H^(o+)` ions and `100mL` of `0.3M NaOH` will given
`((0.3)/(1000)xx100) = 0.03 mol of overset(Theta)OH` ions i.e, `0.02mol` of `H^(o+)` ions reacts with `0.02` mole of `overset(Theta)OH` ions to form `0.02 mol` of water molecules.
Heat evolved in the formation of `0.02`mole of water `=0.02 xx 57.1 = 1.142 kJ`
d. `400mL of 0.2M H_(2)SO_(5)` will give
`((2x0.2)/(1000)xx400) = 0.16 mol of H^(o+)` ions and `600mL` of `0.1M KOH` will given
`((0.1)/(1000)xx600) = 0.60 mol of overset(Theta)OH` ions i.e, `0.06 mol of H^(o+)` ions reacts with `0.06 mol of overset(Theta)OH` ions to form `0.06 mol` of water molecules. Heat evolved in the formation of `0.06mol` of water `=0.06 xx 57.1 = 3.426 kJ`