Question

Question: (a) Use data from Appendix D to calculate the standard entropy change at$\mathbf{25}\mathbf{°}\mathbf{}\mathbf{C}$for the reaction

.${\mathbf{\text{CH}}}_{\mathbf{3}}\mathbf{\text{COOH}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{NH}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{CH}}}_{\mathbf{3}}{\mathbf{\text{NH}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

(b)$\mathbf{1}\mathbf{}\mathbf{mol}$each of solid acetamide, ${\mathbf{\text{CH}}}_{\mathbf{3}}{\mathbf{\text{CONH}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$,and water, ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}$, react to give the same products. Will the standard entropy change be larger or smaller than that calculated for the reaction in part (a)?

Short answer

The final value of standard entropy is given below:

(a) .$∆\mathrm{S}°=112.76\mathrm{J}/\mathrm{Kmol}$

(b) Larger.

Step-by-step solution

Given data

The value of standard entropies is given below:

$$\begin{gathered} \mathrm{S}°({\mathrm{CH}}_3\mathrm{COOH},\mathrm{g})=282.4{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{NH}}_3,\mathrm{g})=192.34{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{CH}}_3{\mathrm{NH}}_2,\mathrm{g})=243.30{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{CO}}_2,\mathrm{g})=213.63{\mathrm{JK}}^1{\mathrm{mol}}^{-1} \end{gathered}$$

And other standard entropy is,$\mathrm{S}°({\mathrm{H}}_2,\mathrm{g})=130.57{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

Concept of the third law of thermodynamics

Regarding the attributes of closed systems in thermodynamic equilibrium, the third law of thermodynamics states: When a system's temperature approaches absolute zero, its entropy approaches a constant value.

Calculation of change of entropy

(a)

The change in entropy is calculated with the help of the formula:

$∆\mathrm{S}°=\underset{}{\sum \mathrm{S}{°}_{\mathrm{products}}}-\underset{}{\sum \mathrm{S}}{°}_{\mathrm{reactants}}$

Where, $\underset{}{\sum \mathrm{S}{°}_{\mathrm{products}}}$entropy of the product and $\underset{}{\sum \mathrm{S}}{°}_{\mathrm{reactants}}$is the entropy of the recant.

Put the value of given data in equation (i).

$$\begin{gathered} ∆\mathrm{S}°=243.30+213.63+130.57-282.4-192.34 \\ ∆\mathrm{S}°=112.76\mathrm{J}/\mathrm{Kmol} \end{gathered}$$

(b) Explanation of standard entropy

The entropy is explained as the "level" of randomness.

For gaseous compounds, it is high because the molecules are free to move around, bump into each other, etc. In other words, they have higher randomness (entropy).
Liquid compounds have less of a level of randomness since the particles can still move around, but the bond between them is much stronger than in gasse.

When talking about solid compounds, the level of randomness is the lowest. The particles are fixated in a strict structure, not allowing the random allocation or movement. This means that they have lower randomness (entropy).

Conclusion of standard entropy

The sum of the entropies of the products stays the same, while the sum of the entropies of the reactants will be much lower.

Therefore, this change will result in a larger entropy change.