Question

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temerature increase of `0.7^(@)`C was measured for the beaker and its contents (Expt. 1 ) . Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ `"mol"^(-1)`), the experiment could be used to measure the calorimeter constant.
In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid `(K_(a)=2.0xx10^(-5))` was mixed with 100 mL of 1.0 M NaOH . (under identical conditions of Expt.1) where hte temperature rise of `5.6^(@)C` was measured.
(Consider heat capacity of all solutions as `4.2 J g^(-1) K^(-1)` and density of all solutions as 1.0 m `mL^(-1)`)
Enthalpy of dissociation (in kJ `"mol"^(-1)`) of acetic acid obtained from Expt. 2 is
A. `1.0`
B. `10.0`
C. `24.5`
D. `51.4`

Answer 1

Correct Answer - A
Moles of HCl or NaOH neutralized (n)
`=100xx1` mmol = 0.1 mole
Heat evolved = 0.1 mole `xx` 57.0 kJ `"mol"^(-1)`
= 5.7 kJ = 5700 J
Heat used to increase temperature of the solution (200 mL ) by `5.7^(@)`
`=200 xx 4.2 xx 5.7`
(sp. heat capacity of solution `= 4.2 JK^(-1)g^(-1)`)
= 4788 J
`:.` Heat used to increase the temperature of the calorimeter
`=5700-4788 J = 912 J`
ms `Delta t = 912`
or ms `(5.7) = 912`
or ms (calorimeter constant ) `= (912)/(5.7) = 160 J//^(@)C`
In Expt. 2, heat evolved by neutralization of `CH_(3)CO OH` with NaOH
`{:(=,200xx4.2xx5.6,+,160xx5.6),(,("Absorbed by solution"),,("Absorbed by calorimeter")):}`
`= 4704+896 J = 5600 J`
`CH_(3)CO OH` present in 100 mL of 2.0 M solution
`=100xx2 ` mmol = 0.2 mol
NaOH present in 100 mL of 1.0 M NaOH
`=100xx1` mmol = 0.1 mol
`:. CH_(3)CO OH` neutralized by NaOH = 0.1 mol
Thus, heat evolved by neutralisation of 0.1 mol of `CH_(3)CO OH = 5600 J`
`:.` Heat used in the dissociation of 0.1 mol of
`CH_(3)CO OH =5700-5600J = 100 J `
`:.` Enthalpy of dissociation of acetic acid per mole = 1000 J = 1 kJ