Question

The${{K}_{w}}$ of water at two different temperature is:
T = ${{25}^{0}}C$
$1.08\text{ x 1}{{\text{0}}^{-14}}$
T = ${{50}^{0}}C$
$\text{5}\text{.474 x 1}{{\text{0}}^{-14}}$
Assuming that $\Delta H$ of any reaction is independent of temperature, calculate the enthalpy of neutralization of a strong acid and a strong base.(in kJ/mol)

Answer 1

The values given above denote the dissociation constant for water at given temperatures. This value depends only on the temperature of the surroundings and not on the concentration of external atoms or molecules added like acid or base. The energy released during neutralization reaction is the enthalpy in terms of magnitude.

Complete step-by-step answer:
Enthalpy is a property of a thermodynamic system, that is a convenient state function used as a means of measurement in various chemical and biological systems at constant pressure.
A state function is a property such that its value does not depend on the path taken by the physical quantity to reach a specific value. Enthalpy is one such state function.
We will now calculate the enthalpy of neutralisation of strong acid and strong base.
It is known to us that the natural log of dissociation constant of water is directly related to temperature and enthalpy. The formula is thus,
$\ln \dfrac{{{K}_{{{w}_{2}}}}}{{{K}_{{{w}_{1}}}}}\text{ = }\dfrac{\Delta H}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$
Substituting the values given in the equation above, we get:
$\ln \dfrac{5.474\text{ x 1}{{\text{0}}^{-14}}}{\,1.08\text{ x 1}{{\text{0}}^{-14}}}\text{ = }\dfrac{\Delta H}{8.314}\left( \dfrac{1}{298}-\dfrac{1}{323} \right)$
$\text{ }\\!\\!\Delta\\!\\!\text{ H = 51952}\text{.6 J = 51}\text{.95 kJ mo}{{\text{l}}^{\text{-1}}}$
Therefore, the enthalpy of neutralization of a strong acid and a strong base is $\text{51}\text{.95 kJ mo}{{\text{l}}^{\text{-1}}}$.

Additional information: Thermodynamics is a branch of physics that deals with the physical quantities heat, work and temperature and their relation to radiation and energy released.
The behaviour of these quantities is in accordance with the laws of thermodynamics. The 3 laws of thermodynamics give a quantitative description using the above physical quantities at a microscopic level. In contrast to state function, functions that depend on the path taken by the quantity to go between two values are called path functions. Both these functions are often used in thermodynamics.

Note: We can never define the absolute enthalpy of a substance or a reaction. This is the reason why we consider change in enthalpy for a reaction like enthalpy of formation, enthalpy of atomisation etc.