Question

If $\Delta {H_c}^ \circ$ of the solid benzoic acid at $27{{\text{ }}^ \circ }C$ is $- x{\text{ kcal mo}}{{\text{l}}^{ - 1}}$ then $\Delta {E_c}^ \circ$ (in ${\text{kcal mo}}{{\text{l}}^{ - 1}}$) is:
$(i){\text{ - x + 0}}{\text{.9}}$
$(ii){\text{ - x + 0}}{\text{.3}}$
$(iii){\text{ - x - 0}}{\text{.9}}$
$(iv){\text{ - x - 0}}{\text{.3}}$

Answer 1

Here we will find the value of $\Delta {E_c}^ \circ$ by using the first law of thermodynamics. For that we will firstly write the combustion reaction for the combustion of solid benzoic acid then we will find the change in number of gaseous moles during the combustion reaction. With the help of the first law of thermodynamics we will find the value of change in internal energy.
Formula used:
$\Delta {H_c}^ \circ {\text{ = }}\Delta {E^ \circ }{\text{ + }}\Delta {{\text{n}}_g}RT$
Where, $\Delta {H_c}^ \circ$ is the change in enthalpy during the combustion, $\Delta {E^ \circ }$ is the change in internal energy, $\Delta {{\text{n}}_g}$ is the change in number of gaseous moles, R is universal gas constant and T is the temperature in kelvin.

Complete answer:
The combustion reaction of the solid benzoic acid $\left( {{C_6}{H_5}COOH} \right)$ can be written as:
${C_6}{H_5}COO{H_{(s)}}{\text{ + }}{{\text{O}}_2}_{(g)}{\text{ }} \to {\text{ C}}{{\text{O}}_{2(g)}}{\text{ + }}{{\text{H}}_2}{{\text{O}}_{(l)}}$
After balancing the above reaction we get the balanced combustion reaction of solid benzoic acid as:
${C_6}{H_5}COO{H_{(s)}}{\text{ + }}\dfrac{{15}}{2}{{\text{O}}_2}_{(g)}{\text{ }} \to {\text{ 7C}}{{\text{O}}_{2(g)}}{\text{ + 3}}{{\text{H}}_2}{{\text{O}}_{(l)}}$
We can find the change in number of gaseous moles during the reaction as:
$\Delta {n_g}{\text{ = }}{n_g}({\text{products}}){\text{ - }}{n_g}({\text{reactants}})$
From the above reaction the change in gaseous number of moles will be:
$\Rightarrow {\text{ }}\Delta {n_g}{\text{ = 7 - }}\dfrac{{15}}{2}$
$\Rightarrow {\text{ }}\Delta {n_g}{\text{ = - }}\dfrac{1}{2}$
Hence we get the value of $\Delta {{\text{n}}_g}$ equals to $- \dfrac{1}{2}$.
Now by using first law of thermodynamics we know that,
$\Delta {H_c}^ \circ {\text{ = }}\Delta {E^ \circ }{\text{ + }}\Delta {{\text{n}}_g}RT$
It can be reduced as:
$\Rightarrow {\text{ }}\Delta {E^ \circ }{\text{ = }}\Delta {H_c}^ \circ {\text{ - }}\Delta {{\text{n}}_g}RT$
Since it is given that $\Delta {H_c}^ \circ$ equals to $- x{\text{ kcal mo}}{{\text{l}}^{ - 1}}$, the value of R is $1.987{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ kcal mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{k}}^{ - 1}}$ and temperature is $\left( {273 + 27} \right){\text{ = 300 K}}$. Therefore on substituting the values we get,
$\Rightarrow {\text{ }}\Delta {E^ \circ }{\text{ = - x - }}\left( { - \dfrac{1}{2}} \right){\text{ }} \times {\text{ }}1.987{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ }} \times {\text{ 300}}$
$\Rightarrow {\text{ }}\Delta {E^ \circ }{\text{ = - x + 0}}{\text{.5 }} \times {\text{ 0}}{\text{.596 }}$
$\Rightarrow {\text{ }}\Delta {E^ \circ }{\text{ = - x + 0}}{\text{.298 }} \approx {\text{ - x + 0}}{\text{.3}}$
Hence the change in internal energy will be equal to $- x{\text{ + 0}}{\text{.3}}$. Therefore the correct option is $(ii){\text{ - x + 0}}{\text{.3}}$.

Note:
The value of universal gas constant, R must be noticed before use. Its unit must be in ${\text{ kcal mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{k}}^{ - 1}}$. The change in number of gaseous moles can only be found with the help of a balanced chemical reaction. While finding the change in number of gaseous moles of the reaction do not involve moles of compounds present in other states than gaseous states.