Question

Heat of dissociation for ${H_2}O$ in ${H^ + }$ and $O{H^ - }$ ions is:
A. -57.27 KJ/mol
B. +57.27 KJ/mol
C. -50 KJ/mol
D. +50 KJ/mol

Answer 1

Hint:
According to laws of thermochemistry there are various types of enthalpies of reactions. A chemical reaction can be either exothermic or endothermic.

Complete step by step solution:
The heat of formation of any compound is equal in magnitude and of opposite sign to the heat of dissociation of that compound at the given temperature and pressure. Dissociation of ${H_2}O$ will evolve enthalpy of neutralization; it is the enthalpy change when one equivalent of an acid is neutralized by a base in dilute solution. This is constant and its value is -13.7 kcal for neutralization of any strong acid and strong base since in dilute solutions they completely dissociate into ions for example dissociation of hydrogen.
     ${H^ + }(aq) + O{H^ - }(aq) \to {H_2}O(l)$
     $\Delta {\rm H} = - 13.7 \times 4.18$ KJ=-57.27 KJ/mol
The value of enthalpy of neutralization is converted into kilo joule from kilo calorie and we know that
     1kcal=4.18 KJ
The heat of dissociation is always positive as breaking a bond requires the input of energy (positive change in enthalpy) and energy is released (negative change in enthalpy) while forming a bond. If the heat is absorbed by the system (q>0) then the reaction is said to be endothermic and $\Delta $E and$\Delta $H value is given a positive sign. If the heat is evolved (q<0) the reaction is said to be exothermic, and $\Delta $E and $\Delta $H are given a negative sign. Therefore the correct option is option B (+57.27 Kj/mol).

Additional information: For weak acid and bases, heat of neutralization is different because they are not completely dissociated and during dissociation some heat is absorbed. So total heat evolved during neutralization will be less.

Note: Take care of the sign convention.