Question

Enthalpy of neutralisation of acetic acid by NaOH is -50.6kJ\[mo{l^{ - 1}}\] . Calculate $\Delta H$ for ionisation of $C{H_3}COOH$. Given, the heat of neutralisation of a strong acid with a strong base is -55.9kJ \[mo{l^{ - 1}}\] .
A. $\Delta H$ = \[ + 5.3kJmo{l^{ - 1}}\]
B. $\Delta H$ = \[ - 5.3kJmo{l^{ - 1}}\]
C. $\Delta H$ = \[ - 10.6kJmo{l^{ - 1}}\]
D. None of these

Answer 1

Hint: The ionisation enthalpy is the amount of energy which is required to remove the electron from the outermost orbital. It depends on the penetration effect, shielding effect and electronic configuration. The penetration effect is the proximity of the electron in the orbital of the nucleus.

Complete step by step answer:
Let us first discuss the definition of each term.
The enthalpy (heat) of neutralization $\Delta {H_n}$is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.
Enthalpy of Ionization of elements is the amount of energy that an isolated gaseous atom requires to lose an electron in its ground state.
We know that,
Enthalpy (heat) of neutralization + Enthalpy of Ionization = heat of neutralisation of a strong acid with a strong base
$\Delta H$+ $\Delta {H_i}$ = $\Delta {H_n}$
Here,
The enthalpy (heat) of neutralization of acetic acid by NaOH
$C{H_3}COOH + N{a^ + } + O{H^ - } \to C{H_3}CO{O^ - } + N{a^ + } + {H_2}O$ $\Delta {H_n}$ = -50.6kJ \[mo{l^{ - 1}}\]
Heat of neutralisation of a strong acid with a strong base
${H^ + } + N{a^ + } + O{H^ - } \to N{a^ + } + {H_2}O$${H^ + }$ $\Delta H$ = -55.9kJ \[mo{l^{ - 1}}\]
We need to calculate Enthalpy of Ionization of acetic acid
$C{H_3}COOH \Leftrightarrow C{H_3}CO{O^ - } + {H^ + }$ $\Delta {H_i}$ =?
By using above mentioned formula we get,
-55.9kJ \[mo{l^{ - 1}}\] + $\Delta {H_i}$ = -50.6kJ \[mo{l^{ - 1}}\]
$\Delta {H_i}$ = -50.6kJ \[mo{l^{ - 1}}\] + 55.9kJ \[mo{l^{ - 1}}\]
$\Delta {H_i}$ = 5.3kJ \[mo{l^{ - 1}}\]
During ionization of acetic acid, a small amount of heat (5.3kJ \[mo{l^{ - 1}}\] ) is absorbed. Therefore, it’s a positive value
Therefore, the Enthalpy of Ionization of acetic acid $\Delta {H_i}$ is 5.3kJ \[mo{l^{ - 1}}\] .

So, the correct answer is Option A.

Note: The ionisation enthalpy increases across the period. As the size of the atom decreases because the attraction between the nucleus and electron increases. While down the group the ionisation enthalpy decreases. It so because the size of the atom increases as the shell increases down the group.