Question

Calculate ${\mathbf{\text{ΔH}}}^{\mathbf{\text{o}}}$and ${\mathbf{\text{ΔS}}}^{\mathbf{\text{o}}}$at 25°C for the reaction
${\mathbf{\text{2SO}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}{\mathbf{\text{+O}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{2SO}}}_{\mathbf{\text{3}}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}$
at a constant pressure of 1.00 atm using thermodynamic data in Appendix 4. Also calculate ${\mathbf{\text{ΔH}}}^{\mathbf{\text{o}}}$and${\mathbf{\text{ΔS}}}^{\mathbf{\text{o}}}$ at 227°C and 1.00 atm, assuming that the constant pressure molar heat capacities for,${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}$,${\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}$and ${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}$are ${\mathbf{\text{39.9JK}}}^{\mathbf{\text{-1}}}{\mathbf{\text{mol}}}^{\mathbf{\text{-1}}}$,${\mathbf{\text{29.4JK}}}^{\mathbf{\text{-1}}}{\mathbf{\text{mol}}}^{\mathbf{\text{-1}}}$ ,and ${\mathbf{\text{50.7JK}}}^{\mathbf{\text{-1}}}{\mathbf{\text{mol}}}^{\mathbf{\text{-1}}}$respectively. (Hint: Construct a thermodynamic cycle, and consider how enthalpy and entropy depend on temperature.)

Short answer

The values of ${\mathbf{\text{ΔH}}}^{\mathbf{\text{o}}}$and ${\mathbf{\text{ΔS}}}^{\mathbf{\text{o}}}$at 25°Cand at227°Cfor the given reaction are as follows:

$\begin{array}{c}{\text{ΔH}}^{\text{o}}{\text{at25}}^{\text{o}}\text{C=}-198{\text{kJmol}}^{-1}\\ {\text{ΔS}}^{\text{o}}{\text{at25}}^{\text{o}}\text{C=}-187{\text{JK}}^{-1}{\text{mol}}^{-1}\\ {\text{ΔH}}^{\text{o}}{\text{at227}}^{\text{o}}\text{C}\text{=}-196.4{\text{kJmol}}^{-1}\\ {\text{ΔS}}^{\text{o}}{\text{at227}}^{\text{o}}\text{C}\text{=}-\text{191}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\end{array}$

Step-by-step solution

Step 1: 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.

Step 2: Determination of  ΔHoand ΔSoat 25°C 

Here, the given reaction is:

$2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right)$

And, from the given reaction,

$\begin{array}{c}\Delta H_f^{\circ }\left(S{O}_2\right)=-297{\text{kJmol}}^{-1}\\ \Delta H_f^{\circ }\left(O_2\right)=0{\text{kJmol}}^{-1}\\ \Delta H_f^{\circ }\left(S{O}_3\right)=-396{\text{kJmol}}^{-1}\end{array}$

Let’s compute the change in enthalpy,

$\begin{array}{l}\text{ΔH=}\sum {\text{ΔH}}_{\text{f}\left(\text{products}\right)}^{\text{o}}\text{-}\sum {\text{ΔH}}_{\text{f}\left(\text{reactants}\right)}^{\text{o}}\\ =2\times \Delta H_{f\left(S{O}_3\right)}^{\circ }-[2\times \Delta H_{f\left(S{O}_2\right)}^{\circ }+\Delta H_{f\left(O_2\right)}^{\circ }]\\ =2\times -396{\text{kJmol}}^{-1}-[2\times -297{\text{kJmol}}^{-1}+0]\\ =-198{\text{kJmol}}^{-1}\end{array}$

And, from the given reaction,

$\begin{array}{c}\Delta S_f^{\circ }\left(S{O}_2\right)=248{\text{JK}}^{-1}{\text{mol}}^{-1}\\ \Delta S_f^{\circ }\left(O_2\right)=205{\text{JK}}^{-1}{\text{mol}}^{-1}\\ \Delta S_f^{\circ }\left(S{O}_3\right)=257{\text{JK}}^{-1}{\text{mol}}^{-1}\end{array}$

Let’s ,compute the change in entropy

$\begin{array}{l}\text{ΔS=}\sum {\text{ΔS}}_{\text{f}\left(\text{products}\right)}^{\text{o}}\text{-}\sum {\text{ΔS}}_{\text{f}\left(\text{reactants}\right)}^{\text{o}}\\ =2\times \Delta S_{f\left(S{O}_3\right)}^{\circ }-[2\times \Delta S_{f\left(S{O}_2\right)}^{\circ }+\Delta S_{f\left(O_2\right)}^{\circ }]\\ =2\times 257{\text{JK}}^{-1}{\text{mol}}^{-1}-[2\times 248{\text{JK}}^{-1}{\text{mol}}^{-1}+205{\text{JK}}^{-1}{\text{mol}}^{-1}]\\ =-187{\text{JK}}^{-1}{\text{mol}}^{-1}\end{array}$

Step 3: Determination of change in molar capabilities

From the given reaction,

$\begin{array}{c}{C}_P\left(S{O}_2\right)=39.9{\text{JK}}^{-1}{\text{mol}}^{-1}\\ C_P\left(O_2\right)=29.4{\text{JK}}^{-1}{\text{mol}}^{-1}\\ C_P\left(S{O}_3\right)=50.7{\text{JK}}^{-1}{\text{mol}}^{-1}\end{array}$

Let’s compute the change in molar capabilities,

$\begin{array}{c}{\text{ΔC}}_{\text{P}}\text{=}\sum {\text{C}}_{\text{P}\left(\text{products}\right)}\text{-}\sum {\text{C}}_{\text{P}\left(\text{reactants}\right)}\\ =2\times C_{P\left(S{O}_3\right)}-[2\times C_{P\left(S{O}_2\right)}+C_{P\left(O_2\right)}]\\ =2\times 50.7{\text{JK}}^{-1}{\text{mol}}^{-1}-[2\times 39.9{\text{JK}}^{-1}{\text{mol}}^{-1}+29.4{\text{JK}}^{-1}{\text{mol}}^{-1}]\\ =7.8{\text{JK}}^{-1}{\text{mol}}^{-1}\end{array}$

Step 4: Determination of Kirchhoff’s equation

Let us compute the Kirchhoff’s equation below,

$\begin{array}{c}{\text{ΔC}}_{\text{P}}\text{=}\frac{{\text{ΔH}}_{\text{2}}{\text{-ΔH}}_{\text{1}}}{{\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}}\\ 7.8{\text{JK}}^{-1}{\text{mol}}^{-1}=\frac{\Delta H_2-(-198{\text{kJmol}}^{-1})}{500\text{K}-\text{298K}}\\ 1575.6{\text{Jmol}}^{-1}=\Delta H_2-(-198{\text{kJmol}}^{-1})\\ 1.575{\text{kJmol}}^{-1}=\Delta H_2+198{\text{kJmol}}^{-1}\\ \Delta H_2=-196.4{\text{kJmol}}^{-1}\end{array}$

Let’s compute for entropy to 298 K,

$\begin{array}{c}\Delta S_{500\to 298}=n{C}_{P,S{O}_2}\ln \times \left(\frac{T_2}{T_1}\right)+n{C}_{P,O_2}\ln \times \left(\frac{T_2}{T_1}\right)\\ \Delta S_{500\to 298}=(2\times \left(39.9\right)\times \ln \times \left(\frac{298}{500}\right)+1\times \left(29.4\right)\times \ln \times \left(\frac{298}{500}\right))\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\\ =-56.5\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\end{array}$

Then,let’s compute for entropy to 500 K,

$\begin{array}{c}\Delta S_{\text{298}\to 500}=n{C}_{P,S{O}_2}\ln \times \left(\frac{T_2}{T_1}\right)\\ \Delta S_{\text{298}\to 500}=(2\times \left(50.7\right)\times \ln \times \left(\frac{500}{298}\right))\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\\ =52.5\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\end{array}$

Thus,the entropy changeis computed as follows:

$\begin{array}{c}\Delta S=(-56.6-187+52.5)\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\\ \text{=}-\text{191}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\end{array}$

Therefore,the valuesare:

$\begin{array}{c}{\text{ΔH}}^{\text{o}}{\text{at25}}^{\text{o}}\text{C=}-198{\text{kJmol}}^{-1}\\ {\text{ΔS}}^{\text{o}}{\text{at25}}^{\text{o}}\text{C=}-187{\text{JK}}^{-1}{\text{mol}}^{-1}\\ {\text{ΔH}}^{\text{o}}{\text{at227}}^{\text{o}}\text{C}\text{=}-196.4{\text{kJmol}}^{-1}\\ {\text{ΔS}}^{\text{o}}{\text{at227}}^{\text{o}}\text{C=}-\text{191}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\end{array}$