Question

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.

Answer 1

According to the reaction,
`H^(+)(aq.)+OH^(-)(aq)toH_(2)O(l),DeltaH=-57.1kJ`
when 1 mole of `H^(+)` ions and 1 mole of `OH^(-)` ions are neutralised 1, mole of water is formed and 57.1 kJ of energy is released.
(i). 0.50 mole `HCl-=0.50" mole "H^(+)` ions
0.50 mole `NaOH-=0.50" mole "OH^(-)` ions
On mixing, 0.50 mole of water is formed
heat evolved for the formation of 0.50 mole of water
`=57.1xx0.5=28.55kJ`
(ii) 0.50 mole `HNO_(3)-=0.50" mole "H^(+)` ions
i.e., 0.30 mole of `H^(+)` ions react with 0.30 mole of `OH^(-)` ions to form 0.30 mole of water of water molecules.
Heat evolved in the formation of 0.3 mole of water
`=57.1xx0.3=17.13kJ`
(iii) 100 mL of 0.2 M HCl will give
`((0.2)/(1000)xx100)=0.02" mole of "H^(+)` ions ltBrgt and 100 mL of 0.3 M NaOH will give
`((0.3)/(1000)xx100)=0.03" mole "of" "OH^(-)` ions
i.e., 0.02 mole of `H^(+)` ions react with 0.02 mole of `OH^(-)` ions to form 0.02 mole of water molecules.
Heat evovled int he foramtion of 0.02 mole of water
`=0.02xx57.1=1.142kJ`
(iv). 400 mL of 0.2 M `H_(2)SO_(4)` will give
`((2xx0.2)/(1000)xx400)=0.16` mole of `H^(+)` ions
and 600 mL of 0.1 M KOH will give
`((0.1)/(1000)xx600)=0.06` mole of `OH^(-)` ion
i.e., 0.06 mole of `H^(+)` ions react with 0.06 mole of `OH^(-)` ions to form 0.06 mole of water molecules.
Heat evolved in the formation of 0.06 mole of water `=0.06xx57.1=3.426kJ`