Question

The heat of neutralization of oxalic acid is $- 25.4$ ${\text{Kcalmo}}{{\text{l}}^{{\text{ - 1}}}}$ using strong base, $NaOH$. Hence the enthalpy change of the process is:
${H_2}Cr{O_4} \to 2{H^ + } + {C_2}{O_4}^{2 - }$ is?
A.$2.0$ ${\text{KCal}}$
B.$- 11.8$ ${\text{KCal}}$
C.$1.0$ ${\text{KCal}}$
D.$- 1.0$ ${\text{KCal}}$

Answer 1

To solve this question, knowledge on Enthalpy change of reactions is essential. The change in enthalpy of neutralization $\left( {\Delta {{\text{H}}_{\text{n}}}} \right)$is the change in enthalpy that occurs one equivalent of an acid reacts with one equivalent of base to undergo neutralization and form salt and water. We shall find the heat of neutralisation for one mole of the acid and then use that to calculate the enthalpy change from the reaction.

Complete step by step solution:
Whenever an acid and base react together, the salts formed remain in the solution in the dissolved state. So the neutralization can be represented as:
${H^ + } + {A^ - } + {B^ + } + O{H^ - } \to {H_{\text{2}}}O + {A^ - } + {B^ + }$
So the change in the heat of neutralization of the reaction between an acid and a base is also equal to the heat of formation of one mole of water which is equal to $- 13.7$ ${\text{Kcal/equivalent}}$
Oxalic acid being a dibasic acid can release two protons on reacting with a base but it being a weak acid, not all the molecules are dissociated. Some remain dissociated in water. A part of the heat of neutralization is utilized to dissociate these undissociated molecules and hence the amount of energy release in the process of neutralization is less than$- 25.4$ ${\text{Kcalmo}}{{\text{l}}^{{\text{ - 1}}}}$.
Since two molecules of water are released, so the heat of formation of water or the neutralization of oxalic acid = $- 13.7 \times 2$=$- 27.4$ ${\text{Kcalmo}}{{\text{l}}^{{\text{ - 1}}}}$

The enthalpy change of the process = $- 25.4 - ( - 27.4) = 2$ ${\text{Kcalmo}}{{\text{l}}^{{\text{ - 1}}}}$

Note:
1.When one gram equivalent of $HCl$is neutralized by one gram equivalent of $NaOH$, both the solutions being dilute, $57.1$ kJ of heat is released.
2.As the bonds of an acid and a base are dissociated to form salt and water, so the energy stored in the bonds of both of them is utilized for the bond formation of salt and water. Hence the heat of neutralization is equal to the heat of formation of water, as the salt remains in the dissociated state mostly.