Question

Calculate $∆\mathrm{H}°$and $∆\mathrm{S}°$ at 25°C for the reaction
$2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{SO}}_3\left(\mathrm{g}\right)$
at a constant pressure of 1.00 atm using thermodynamic data in Appendix 4. Also calculate $∆\mathrm{H}°$and $∆\mathrm{S}°$at 227°C and 1.00 atm, assuming that the constant pressure molar heat capacities for${\mathrm{SO}}_2$, ${\mathrm{O}}_2$and ${\mathrm{SO}}_3$are 39.9${\mathrm{JK}}^{-1}$ ,29.4${\mathrm{JK}}^{-1}$ ,and 50.7${\mathrm{JK}}^{-1}$respectively. (Hint: Construct a thermodynamic cycle, and consider how enthalpy and entropy depend on temperature.)

Short answer

The values of$∆\mathrm{H}°$ and$∆\mathrm{S}°$ at 25°Cand at 227°Cfor the given reaction are as follows:

$$\begin{gathered} ∆\mathrm{H}°\mathrm{at}25°\mathrm{C}\mathrm{is}=-198{\mathrm{kJmol}}^{-1} \\ ∆\mathrm{S}°\mathrm{at}25°\mathrm{C}\mathrm{is}=-187\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \\ ∆\mathrm{H}°\mathrm{at}227°\mathrm{C}\mathrm{is}=-196{\mathrm{kJmol}}^{-1} \\ ∆\mathrm{S}°\mathrm{at}227°\mathrm{C}\mathrm{is}=-191\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \end{gathered}$$

Step-by-step solution

Introduction to the concept

Change in enthalpy of the reaction is defined as the difference between the enthalpies of the products and reactants.

The entropy of the reaction is the difference between the entropy of the products and the entropy of the reactants.

Determination of ∆H°and ∆S°at 25°C 

Here is the given reaction.

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

And from the given,

$$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{SO}}_2\right)=-297{\mathrm{kJmol}}^{-1} \\ ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{SO}}_2\right)=0{\mathrm{kJmol}}^{-1} \\ ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{SO}}_3\right)=-396{\mathrm{kJmol}}^{-1} \end{gathered}$$

Let’s compute the change of enthalpy$$\begin{gathered} ∆\mathrm{H}=\underset{}{\sum ∆{\mathrm{H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{°}-\sum ∆{\mathrm{H}}_{\mathrm{f}\left(\mathrm{reactants}\right)}^{°}} \\ =2\times ∆{\mathrm{H}}_{\mathrm{f}\left({\mathrm{SO}}_3\right)}^{°}-[2\times ∆{\mathrm{H}}_{\mathrm{f}\left({\mathrm{SO}}_2\right)}^{°}+∆{\mathrm{H}}_{\mathrm{f}\left({\mathrm{O}}_2\right)}^{°}] \\ =-198{\mathrm{kJmol}}^{-1} \end{gathered}$$

and from the given,

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{f}}^{°}\left({\mathrm{SO}}_2\right)=248{\mathrm{kJmol}}^{-1} \\ ∆{\mathrm{S}}_{\mathrm{f}}^{°}\left({\mathrm{SO}}_2\right)=205{\mathrm{kJmol}}^{-1} \\ ∆{\mathrm{S}}_{\mathrm{f}}^{°}\left({\mathrm{SO}}_3\right)=257{\mathrm{kJmol}}^{-1} \end{gathered}$$

Let’s compute the change of entropy.

$$\begin{gathered} ∆\mathrm{S}=\underset{}{\sum ∆{\mathrm{S}}_{\mathrm{f}\left(\mathrm{products}\right)}^{°}-\sum ∆{\mathrm{S}}_{\mathrm{f}\left(\mathrm{reactants}\right)}^{°}} \\ =2\times ∆{\mathrm{S}}_{\mathrm{f}\left({\mathrm{SO}}_3\right)}^{°}-[2\times ∆{\mathrm{S}}_{\mathrm{f}\left({\mathrm{SO}}_2\right)}^{°}+∆{\mathrm{S}}_{\mathrm{f}\left({\mathrm{O}}_2\right)}^{°}] \\ =-187{\mathrm{kJmol}}^{-1} \end{gathered}$$

Determination of change in molar capabilities

From the given,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{p}}\left({\mathrm{SO}}_2\right)=248{\mathrm{kJmol}}^{-1} \\ {\mathrm{C}}_{\mathrm{p}}\left({\mathrm{SO}}_2\right)=205{\mathrm{kJmol}}^{-1} \\ {\mathrm{C}}_{\mathrm{p}}\left({\mathrm{SO}}_3\right)=257{\mathrm{kJmol}}^{-1} \end{gathered}$$

Let’s compute the change in molar capabilities.

$$\begin{gathered} {\mathrm{C}}_{\mathrm{p}}={\mathrm{C}}_{\mathrm{p}}\underset{}{{}_{\left(\mathrm{products}\right)}^{}-{{\mathrm{C}}_{\mathrm{p}}}_{\mathrm{f}\left(\mathrm{reactants}\right)}^{}} \\ =2\times {{\mathrm{C}}_{\mathrm{p}}}_{\mathrm{f}\left({\mathrm{SO}}_3\right)}^{}-[2\times {{\mathrm{C}}_{\mathrm{p}}}_{\left({\mathrm{SO}}_2\right)}^{}+{{\mathrm{C}}_{\mathrm{p}}}_{\left({\mathrm{O}}_2\right)}^{}] \\ =7.8{\mathrm{kJmol}}^{-1} \end{gathered}$$

Determination of Kirchhoff’s equation

Let us compute Kirchhoff’s equation below.

$∆{\mathrm{C}}_{{\mathrm{p}}_{}}=\frac{∆{\mathrm{H}}_2-∆{\mathrm{H}}_1}{{\mathrm{T}}_2-{\mathrm{T}}_1}$

$$\begin{gathered} 1575.6{\mathrm{Jmol}}^{-1}=∆{\mathrm{H}}_2-(-198{\mathrm{kJmol}}^{-1}) \\ ∆{\mathrm{H}}_2=-196.4{\mathrm{kJmol}}^{-1} \end{gathered}$$

Let’s compute for entropy to 298 K.

$$\begin{gathered} ∆{\mathrm{S}}_{500\to 298}={\mathrm{nC}}_{\mathrm{P},{\mathrm{SO}}_2}\ln \times \left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)+{\mathrm{nC}}_{\mathrm{P},{\mathrm{O}}_2}\ln \times \left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right) \\ ∆{\mathrm{S}}_{500\to 298}=-56.5\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \end{gathered}$$

Then let’s compute for entropy to 500 K.

$$\begin{gathered} ∆{\mathrm{S}}_{298\to 500}={\mathrm{nC}}_{\mathrm{P},{\mathrm{SO}}_2}\ln \times \left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right) \\ ∆{\mathrm{S}}_{298\to 500}=2\times 50.7\times \ln \times \left(\frac{500}{298}\right)\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \\ =52.5\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \end{gathered}$$

Thusthe entropy changeis computed as follows:

$$\begin{gathered} ∆\mathrm{S}=(-56.6-187+52.5)\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \\ =-191\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \end{gathered}$$

Therefore,the valuesare

$$\begin{gathered} ∆\mathrm{H}°\mathrm{at}25°\mathrm{C}\mathrm{is}=-198{\mathrm{kJmol}}^{-1} \\ ∆\mathrm{S}°\mathrm{at}25°\mathrm{C}\mathrm{is}=-187\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \\ ∆\mathrm{H}°\mathrm{at}227°\mathrm{C}\mathrm{is}=-196{\mathrm{kJmol}}^{-1} \\ ∆\mathrm{S}°\mathrm{at}227°\mathrm{C}\mathrm{is}=-191\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \end{gathered}$$