Q20.34 P - Chemistry: Molecular Nature Of Matter And Change

Question

For each reaction, predict the sign and find the value of  ${\mathbf{\Delta S}}^{\mathbf{o}}$:

data-custom-editor="chemistry" $\begin{array}{c}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{3}\mathbf{N}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{NF}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{6}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{6}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\end{array}$

Step-by-Step Solution

Answer

1. The computed entropy of the reaction is  $-267.80\mathrm{J}/\mathrm{K}$, with  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$.
2.  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$, and the reaction's computed entropy is  $-278.40\mathrm{J}/\mathrm{K}$.
3.  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}>0$, and the reaction's computed entropy is  $972.42\mathrm{J}/\mathrm{K}$.

Step 1: Definition of thermodynamics .

The study of the relationships between heat, work, temperature, and energy is known as thermodynamics. The rules of thermodynamics define how energy evolves in a system and whether it can do beneficial work on its surroundings or not.

Step 2: Subpart (a).

The reaction produces  $2$ mol of liquid (solution)${\text{HNO}}_3$ and $1$ mol of NO gas from  $3$ mol of ${\text{NO}}_2$  gas and one mol of liquid  ${\text{H}}_{2}O$.

The entropy is determined by the physical state - since both gases and liquids are present, we make liquid and gas things from liquid and gas reactants, therefore the entropy cannot be exactly defined by physical state change. The number of mol of products and reactants, on the other hand, may be compared, and the entropy sign can be calculated.

$\begin{array}{c}{\mathrm{n}}_{\mathrm{products}}=\mathrm{n}\left({\mathrm{HNO}}_3\right)+\mathrm{n}\left(\mathrm{NO}\right)\\ =2\mathrm{mol}+1\mathrm{mol}\\ =3\mathrm{mol}\\ {\mathrm{n}}_{\mathrm{reactants}}=\mathrm{n}(\mathrm{N}{\mathrm{O}}_2)+\mathrm{n}({\mathrm{H}}_2\mathrm{O})\\ =3\mathrm{mol}+1\mathrm{mol}\\ =4\mathrm{mol}\\ {\mathrm{\Delta n}}_{\mathrm{rxn}}={\mathrm{\Delta n}}_{\mathrm{products}}-{\mathrm{\Delta n}}_{\mathrm{reactants}}\\ =-1\mathrm{mol}\end{array}$

The entropy drops as the number of moles decreases throughout the reaction:  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$.

Entropy is a state function, hence it's a state function.

$\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{products}}-\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{reactants}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=2{{\mathrm{S}}^o}_{{\mathrm{HNO}}_3}-1{{\mathrm{S}}^o}_{\mathrm{NO}}-1{{\mathrm{S}}^o}_{{\mathrm{NO}}_2}-1{{\mathrm{S}}^o}_{{\mathrm{H}}_2\mathrm{O}}\end{array}$ 

We can get the entropy values for each chemical in Appendix B and calculate the reaction molar entropy:

 $\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=2\mathrm{mol}\times 155.60\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}+1\mathrm{mol}\times 210.65\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-3\mathrm{mol}\times 239.90\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-1\mathrm{mol}\times 69.95\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=-267.80\frac{\mathrm{J}}{\mathrm{K}}\end{array}$

Therefore, the value is $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$ and the entropy is  $-267.80\mathrm{J}/\mathrm{K}$.

Step 3: Subpart (b).

 In the ${\text{NF}}_3$ creation reaction, we generate  $2$ mol of ${\text{NF}}_3$ gas from $1$mol of  ${\text{N}}_2$ and $3$ mol of ${\text{F}}_2$  gases.

The entropy of a system is determined by its physical state; in this example, all species are in the same gas phase. The number of mol of products and reactants, on the other hand, may be compared.

 $\begin{array}{c}{\mathrm{n}}_{\mathrm{products}}=\mathrm{n}\left({\mathrm{NF}}_3\right)\\ =3\mathrm{mol}\\ {\mathrm{n}}_{\mathrm{reactants}}=\mathrm{n}\left({\mathrm{N}}_2\right)+\mathrm{n}({\mathrm{F}}_2)\\ =1\mathrm{mol}+3\mathrm{mol}\\ =4\mathrm{mol}\\ {\mathrm{\Delta n}}_{\mathrm{rxn}}={\mathrm{\Delta n}}_{\mathrm{products}}-{\mathrm{\Delta n}}_{\mathrm{reactants}}\\ =-1\mathrm{mol}\end{array}$

Because the number of moles in the system decreases as the reaction progresses, the system becomes less disordered and the entropy decreases:  ${\mathrm{\Delta S}}_{\mathrm{rxn}}<0$.

Entropy is a state function, hence it's a state function.

 $\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{products}}-\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{reactants}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=2{{\mathrm{S}}^o}_{{\mathrm{NF}}_3}-1{{\mathrm{S}}^o}_{{\mathrm{N}}_2}-3{{\mathrm{S}}^o}_{{\mathrm{F}}_2}\end{array}$

We can determine the entropy values for each chemical in Appendix B and estimate the reaction molar entropy:

 $\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=2\mathrm{mol}\times 260.60\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-1\mathrm{mol}\times 191.50\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-3\mathrm{mol}\times 202.70\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=-278.40\frac{\mathrm{J}}{\mathrm{K}}\end{array}$

Therefore, the value is  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$ and the entropy is  $-278.40\mathrm{J}/\mathrm{K}$.

Step 4: Subpart (c).

In the carbohydrate (glucose) combustion process, we get 6 mol of ${\text{CO}}_2$ and 6 mol of ${\text{H}}_{2}O$ gases from 1 mol of solid  ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ and 6 mol of  ${\text{O}}_2$ gas.

Because both gases and solids are present, we can only make gas substances from solid and gas reactants in this situation, hence the entropy increases dramatically when gas substances are formed from the more ordered solid. The number of mol of products and reactants, on the other hand, may be compared, and the entropy sign can be calculated.

$\begin{array}{c}{\mathrm{n}}_{\mathrm{products}}=\mathrm{n}\left({\mathrm{CO}}_2\right)+\mathrm{n}({\mathrm{H}}_2\mathrm{O})\\ =6\mathrm{mol}+6\mathrm{mol}\\ =12\mathrm{mol}\\ {\mathrm{n}}_{\mathrm{reactants}}=\mathrm{n}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)+\mathrm{n}({\mathrm{O}}_2)\\ =1\mathrm{mol}+6\mathrm{mol}\\ =7\mathrm{mol}\\ {\mathrm{\Delta n}}_{\mathrm{rxn}}={\mathrm{\Delta n}}_{\mathrm{products}}-{\mathrm{\Delta n}}_{\mathrm{reactants}}\\ =5\mathrm{mol}\end{array}$

The entropy grows as the number of moles in the reaction increases:  ${\mathrm{\Delta S}}_{\mathrm{rxn}}>0$.

Entropy is a state function, hence it's a state function.

 $\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{products}}-\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{reactants}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=6{{\mathrm{S}}^o}_{{\mathrm{CO}}_2}+6{{\mathrm{S}}^o}_{{\mathrm{H}}_2\mathrm{O}}-1{{\mathrm{S}}^o}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}-6{{\mathrm{S}}^o}_{{\mathrm{O}}_2}\end{array}$

We can get the entropy values for each chemical in Appendix B and calculate the reaction molar entropy:

$\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=6\mathrm{mol}\times 213.70\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}+6\mathrm{mol}\times 188.72\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-1\mathrm{mol}\times 212.10\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-6\mathrm{mol}\times 205.00\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=972.42\frac{\mathrm{J}}{\mathrm{K}}\end{array}$

Therefore, the value is ${\mathrm{\Delta S}}_{\mathrm{rxn}}>0$ and the entropy is  $972.42\mathrm{J}/\mathrm{K}$.