Q20.33 P - Chemistry: Molecular Nature Of Matter And Change

Question

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

$\begin{array}{c}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{3}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{N}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{F}{\mathbf{e}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{Fe}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}{\mathbf{P}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{P}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{10}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\end{array}$

Step-by-Step Solution

Answer

1. The computed entropy of the reaction is $-172.35\mathrm{J}/\mathrm{K}$ , 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  $141.56\mathrm{J}/\mathrm{K}$.
3.  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}>0$, and the reaction's computed entropy is  $-837.10\mathrm{J}/\mathrm{K}$.

Step 1: Standard entropy of a reaction.

The standard entropy of a reaction is the difference in the total entropy of gaseous products and the total entropy of reactants.

Step 2: Subpart (a)

In the NO decomposition process, one mol of each of the gases ${\text{NO}}_2$ and  ${\text{N}}_{2}O$ is produced from three mol of NO gas.

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.

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

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

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

$\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{products}}-\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{reactants}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}={{\mathrm{S}}^o}_{{\mathrm{N}}_2\mathrm{O}}+{{\mathrm{S}}^o}_{{\mathrm{NO}}_2}-3{{\mathrm{S}}^o}_{\mathrm{NO}}\end{array}$

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

$\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=1\mathrm{mol}\times 219.70\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}+1\mathrm{mol}\times 239.90\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-3\mathrm{mol}\times 210.65\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=-172.35\frac{\mathrm{J}}{\mathrm{K}}\end{array}$

Therefore, the sign is $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$  having the value of  $-172.35\mathrm{J}/\mathrm{K}$.

Step 3: Subpart (b).

The reaction produces two mol of solid  ${\mathrm{Fe}}_2{\mathrm{O}}_3$ and three mol of vapor  ${\mathrm{H}}_2\mathrm{O}$ from three mol of ${\mathrm{H}}_2$ gas and one mol of solid  ${\mathrm{Fe}}_2{\mathrm{O}}_3$.

The entropy is affected by the physical state; because both gases and solids are present, it is more difficult to anticipate the sign of the entropy in this scenario.

The number of mol of products and reactants, on the other hand, may be compared, and the entropy sign determined.

It should be noted that each molecule of iron (III) oxide contains two Fe atoms, and since there is an additional source of Fe in the process, we get:

$\begin{array}{c}\mathrm{n}\left(\mathrm{Fe}\right)=\mathrm{n}(\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3)\\ =2\mathrm{mol}\end{array}$

Then, we get:

$\begin{array}{c}{\mathrm{n}}_{\mathrm{products}}=\mathrm{n}\left(\mathrm{Fe}\right)+\mathrm{n}\left({\mathrm{H}}_2\mathrm{O}\right)\\ =3\mathrm{mol}+2\mathrm{mol}\\ =5\mathrm{mol}\\ {\mathrm{n}}_{\mathrm{reactants}}=\mathrm{n}({\mathrm{H}}_2)+\mathrm{n}(\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3)\\ =3\mathrm{mol}+2\mathrm{mol}\\ =5\mathrm{mol}\\ {\mathrm{\Delta n}}_{\mathrm{rxn}}={\mathrm{\Delta n}}_{\mathrm{products}}-{\mathrm{\Delta n}}_{\mathrm{reactants}}\\ =0\mathrm{mol}\end{array}$

The number of moles does not vary as the reaction progresses. Then, because ${\mathrm{H}}_2\mathrm{O}$ is more complicated than ${\mathrm{H}}_2$ gas in terms of molecular complexity (because solids have a relatively low entropy), the complexity grows with the reaction, corresponding to the increase in the reaction entropy:  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}>0$.

The entropy is a state function, then we may say:

$\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{products}}-\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{reactants}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=2{{\mathrm{S}}^o}_{\mathrm{Fe}}+3{{\mathrm{S}}^o}_{{\mathrm{H}}_2\mathrm{O}}-3{{\mathrm{S}}^o}_{{\mathrm{H}}_2}-1{{\mathrm{S}}^o}_{{\mathrm{Fe}}_2{\mathrm{O}}_3}\end{array}$

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

$\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=2\mathrm{mol}\times 27.30\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}+3\mathrm{mol}\times 188.72\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-3\mathrm{mol}\times 130.60\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-1\mathrm{mol}\times 87.40\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=141.56\frac{\mathrm{J}}{\mathrm{K}}\end{array}$

Therefore, the sign is  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}>0$ having the value of  $141.56\mathrm{J}/\mathrm{K}$.

 

Step 4: Subpart (c).

We acquire $1$ mol of solid ${\mathrm{P}}_4{\mathrm{H}}_{10}$ from $1$mol of solid ${\mathrm{P}}_4$ and  $5$ mol of  $O_2$ gas in the process.

The entropy of a system is determined by its physical state; as both gases and solids are present, we build a solid substance from solid and gas reactants. As a result, the system's disorder is minimised, and the entropy is reduced. Furthermore, the number of mol of products and reactants may be compared, as well as the entropy sign.

$\begin{array}{c}{\mathrm{n}}_{\mathrm{products}}=\mathrm{n}\left({\mathrm{P}}_4{\mathrm{H}}_{10}\right)\\ =1\mathrm{mol}\\ {\mathrm{n}}_{\mathrm{reactants}}=\mathrm{n}({\mathrm{P}}_4)+\mathrm{n}({\mathrm{O}}_2)\\ =1\mathrm{mol}+5\mathrm{mol}\\ =6\mathrm{mol}\\ {\mathrm{\Delta n}}_{\mathrm{rxn}}={\mathrm{\Delta n}}_{\mathrm{products}}-{\mathrm{\Delta n}}_{\mathrm{reactants}}\\ =-5\mathrm{mol}\end{array}$

The entropy falls as the number of moles decreases during the process:  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$.

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

 $\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{products}}-\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{reactants}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=1{{\mathrm{S}}^o}_{{\mathrm{P}}_4{\mathrm{H}}_{10}}-1{{\mathrm{S}}^o}_{{\mathrm{P}}_4}-5{{\mathrm{S}}^o}_{{\mathrm{O}}_2}\end{array}$

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

 $\begin{array}{c}\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=1\mathrm{mol}\times 229.0\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-1\mathrm{mol}\times 41.1\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}-5\mathrm{mol}\times 205.0\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\\ \mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}=-837.10\frac{\mathrm{J}}{\mathrm{K}}\end{array}$

Therefore, the sign is  $\mathrm{\Delta }{{\mathrm{S}}^o}_{\mathrm{rxn}}<0$ having the value of  $-837.10\mathrm{J}/\mathrm{K}$.