Question

The heat of formation of water from ${H^ + }$ and $O{H^ - }\;$ is :
A.$13.06{\text{kcal}}$
B.$- 13.06{\text{kcal}}$
C.$16.32{\text{kcal}}$
D.$- 16.32{\text{kcal}}$

Answer 1

To answer this question, recall the concept of the heat of formation. It is also known as reaction Enthalpy. It is defined as the difference in the enthalpy of a specific chemical reaction that is obtained at a constant pressure. The Arrhenius equation can be used to calculate the heat of formation.
The formula used: Arrhenius equation: $2.303{\text{log }}\dfrac{{{{\text{k}}_{\text{2}}}}}{{{{\text{k}}_{\text{1}}}}}{\text{ }} = \dfrac{{{{\Delta H}}}}{{\text{R}}}\left[ {\dfrac{1}{{{{\text{T}}_{\text{2}}}}} - \dfrac{{\text{1}}}{{{{\text{T}}_{\text{1}}}}}} \right]$, where ${{\text{k}}_{\text{2}}}{\text{, }}{{\text{k}}_{\text{1}}}$ are rate constants are different condition, $\Delta {\text{H}}$is the heat of formation, ${\text{R}}$ is the universal gas constant, ${{\text{T}}_{\text{2}}}{\text{, }}{{\text{T}}_{\text{1}}}$are temperatures at different conditions.

Complete step by step answer:
According to the definition of the heat of formation we can say that it is the thermodynamic unit of measurement applied in measuring the total amount of energy per mole either produced or released in a reaction.
At ${25^o}C$, pH = 7 $\Rightarrow {K_w} = {10^{ - 14}}$
At {60^o}C,\;{\text{ pH}} = 6.5$$$$ \Rightarrow {K_w} = {10^{ - 13}}.
The reaction for which $\Delta H$needs to be calculated is
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$ .
Substituting the given values into the Arrhenius equation with appropriate units we get,
$2.303{\text{log}}\dfrac{{{{10}^{ - 13}}}}{{{{10}^{ - 14}}}} = \dfrac{{{{\Delta H}}}}{{\text{R}}}\left[ {\dfrac{{35}}{{333 \times 298}}} \right]$.
Rearranging the values and solving we get
$\Delta {\text{H}} = 13.06{\text{kcal}}$.
${H^ + } + O{H^ - } \to H_2^{}O{\text{ }}$results in a ${\text{ }}\Delta {\text{H}} = - 13.06{\text{kcal}}$.

The correct option is B.

Note:
The Arrhenius equation is not only simple but a remarkably accurate formula too. Not only it is used to study reaction rates but also to model the temperature variance of permeation, diffusion and solubility coefficients, and other chemical processes over moderate temperature ranges. You should know the importance of the Arrhenius equation. ${\text{RT}}$ is the average kinetic energy, and the exponent is just the ratio of the activation energy ${{\text{E}}_{\text{a}}}$ to the average kinetic energy. Larger this ratio, the smaller the rate. It can be concluded that high temperature and low activation energy favour larger rate constants, and thus speed up the reaction.