Question

Predict the sign of $\mathbf{∆}\mathbf{S}\mathbf{°}$and then calculate$\mathbf{∆}\mathbf{S}\mathbf{°}$for each of
the following reactions.

1. $\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\left(g\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{3}{\mathbf{S}}_{\mathbf{rhombic}}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$
2. $\mathbf{2}{\mathbf{SO}}_{\mathbf{3}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)$
   
3. ${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{Fe}\left(s\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$
   

Short answer

Answer

1. For the given reaction $2{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\to 3{\mathrm{S}}_{\mathrm{rhombic}}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$, the signof ${\mathbf{\Delta S}}^o$is Negative and the valueis $\mathbf{-}\mathbf{186}\mathbf{J}\mathbf{/}\mathbf{Kmol}$
2. For the given reactionlocalid="1664269660870" $2{\mathrm{SO}}_3\left(\mathrm{g}\right)\to 2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$, the signof$\mathbf{∆}\mathbf{S}\mathbf{°}$ is Positive and the valueis $\mathbf{+}\mathbf{187}\mathbf{J}\mathbf{/}\mathbf{Kmol}$.
3. For the given reaction,${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)\to 2\mathrm{Fe}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$the signof $\mathbf{∆}\mathbf{S}\mathbf{°}$is hard to predict and the value is $\mathbf{138}\mathbf{J}\mathbf{/}\mathbf{Kmol}$.

Step-by-step solution

Step 1: Introduction to the Concept

The ratio of thermal energy to the temperature at which work cannot be done is known as entropy. It's also known as the measure of a system's molecular disarray. It has a lot of properties and state functions.

The number of microstates for a system is connected to entropy, and microstate is defined as the number of possible arrangements for the system. Entropy is affected by the phase of a substance in distinct physical states; the order of entropyis indicated by: ${\mathbf{S}}_{\mathbf{solid}}\mathbf{<}{\mathbf{S}}_{\mathbf{liquid}}\mathbf{<}{\mathbf{S}}_{\mathbf{gas}}$

With the use of the aforementioned data, the sign of the entropy should be predicted.

Prediction of the sign of ∆S°

a)

Given reaction is:$2{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\to 3{\mathrm{S}}_{\mathrm{rhombic}}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Because the number of moles of reactant in the gaseous state is greater than the number of moles of product in the gaseous state, the sign of entropy is negative in this situation. As a result, the quantity of random substances in the solution is reduced by the comparatively ordered precipitate S_rhombic(s). The sign of entropy is negative due to decreases in randomness.

Calculation of ∆S°

The mathematical equation for the standard entropy value at room temperature is given as follows,

$\mathbf{∆}\mathbf{S}\mathbf{°}\mathbf{298}\mathbf{=}\mathbf{\sum }_{\mathbf{n}}\mathbf{s}\mathbf{°}\mathbf{298}\left(\mathrm{Products}\right)\mathbf{-}\mathbf{\sum }_{\mathbf{p}}\mathbf{s}\mathbf{°}\mathbf{298}\left(\mathrm{Reactants}\right)$

In the balanced chemical equation,n and pindicate the coefficients of reactants and products.

Thus the value of standard entropy is given as,

For data-custom-editor="chemistry" ${\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)=206\mathrm{J}/\mathrm{Kmol}$

For data-custom-editor="chemistry" ${\mathrm{SO}}_2\left(\mathrm{g}\right)=248\mathrm{J}/\mathrm{Kmol}$

For S_rhombic(s)=32J/K mol

For data-custom-editor="chemistry" ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=189\mathrm{J}/\mathrm{Kmol}$

Let’s substitute the values,

data-custom-editor="chemistry" $$\begin{gathered} \Delta S^{o}298=\left(\left(3\times S^{o}298\right)\times \left(S_{\text{rhombic}}\left(s\right)\right)+2\times S^{o}298\left(H_{2}O\left(g\right)\right)\right)\backslash \mathrm{hfill} \\ -\left(\left(2\times S^{o}298\right)\times \left(H_{2}S\left(g\right)\right)+1\times S^{o}298\left(S{O}_2\left(g\right)\right)\right) \end{gathered}$$

data-custom-editor="chemistry" $∆\mathrm{S}°298=186\mathrm{J}/\mathrm{kmol}$

Therefore,data-custom-editor="chemistry" $\mathbf{∆}\mathbf{S}\mathbf{°}\mathbf{is}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{Kmol}$

Prediction of the sign of ∆S°

b)

Given reaction is:$2{\mathrm{SO}}_3\left(\mathrm{g}\right)\to 2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

The sign of entropy is positive in this situation because the number of moles of products in the gaseous state is greater than the number of moles of reactants.

Calculation of ∆S°

The standard entropy value at room temperature is calculated as follows:

$\Delta S^{o}298=\sum n{S}^{o}298\left(Products\right)-\sum p{S}^{o}298\left(Reactants\right)$

In the balanced chemical equation, n and p indicate the coefficients of reactants and products.

Thus the value of standard entropy is given as,

For ${\mathrm{SO}}_2\left(\mathrm{g}\right)=248\mathrm{J}/\mathrm{Kmol}$

For${\mathrm{O}}_2\left(\mathrm{s}\right)=205\mathrm{J}/\mathrm{Kmol}$

For ${\mathrm{SO}}_3\left(\mathrm{g}\right)=257\mathrm{J}/\mathrm{Kmol}$

Let’s substitute the values,

$$\begin{gathered} \Delta S^{o}298=\left(\left(2\times S^{o}298\right)\times \left(S{O}_2\left(g\right)\right)+1\times S^{o}298\left(O_2\left(g\right)\right)\right) \\ -\left(\left(2\times S^{o}298\right)\times \left(S{O}_3\left(g\right)\right)\right) \end{gathered}$$

$∆\mathrm{S}°298=+\mathrm{J}/\mathrm{Kmol}$

Therefore$\mathbf{∆}\mathbf{S}\mathbf{°}\mathbf{}\mathbf{is}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{Kmol}$

Prediction of the sign of ∆S°

c)

Given reaction is:${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)\to 2\mathrm{Fe}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The number of moles of products in the gaseous state equals the number of moles of reactants in the gaseous state in this example. As a result, predicting the sign of entropy is difficult.

Calculation of ∆S°

The standard entropy value at room temperature is calculated as follows:

$\Delta S^{o}298=\sum n{S}^{o}298\left(Products\right)-\sum p{S}^{o}298\left(Reactants\right)$

In the balanced chemical equation, n and pindicate the coefficients of reactants and products.

Thus the value of standard entropy is given as,

For ${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)=90\mathrm{J}/\mathrm{Kmol}$

For ${\mathrm{H}}_2\left(\mathrm{s}\right)=121\mathrm{J}/\mathrm{Kmol}$

For ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=\mathrm{J}/\mathrm{Kmol}$

For $\mathrm{Fe}\left(\mathrm{s}\right)=27\mathrm{J}/\mathrm{Kmol}$

Let’s substitute the values,

$$\begin{gathered} \Delta S^{o}298=\left(\left(2\times S^{o}298\right)\times \left(Fe\left(s\right)\right)+3\times S^{o}298\left(H_{2}O\left(g\right)\right)\right) \\ -\left(\left(1\times S^{o}298\right)\times \left(F{e}_2{O}_3\left(s\right)\right)+3\times S^{o}298\left(H_2\left(g\right)\right)\right) \end{gathered}$$

$∆\mathrm{S}°298=138\mathrm{J}/\mathrm{Kmol}$

Therefore,$\mathbf{∆}\mathbf{S}\mathbf{°}\mathrm{is}\mathbf{138}\mathbf{J}\mathbf{/}\mathbf{Kmol}$