Chapter 10, Problem 150CP

Interpretation Introduction

Interpretation:

The value of ΔH⁰ and ΔS⁰at 25⁰C and 227⁰C should be predicted.

Concept Introduction:

The difference between the enthalpy of products and reactants is known as change in enthalpy of the reaction.

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

Answer to Problem 150CP

The value of ΔH⁰at 25⁰C = -198 kJmol-1

The value of ΔS⁰at 25⁰C = -187 JK-1

The value of ΔH⁰at 227⁰C = - 196.4 kJ

The value of ΔS⁰at 227⁰C = - 191 JK-1

Explanation of Solution

The given reaction is shown below:

  ${\text{2SO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2SO}}_{\text{3}}\left(\text{g}\right)$

Given information:

ΔH⁰_f(SO₂) = -297 kJmol-1

ΔH⁰_f(O₂) = 0 kJmol-1

ΔH⁰_f(SO₃) = -396 kJmol-1

The calculation of change of enthalpyat 25⁰C for the reaction is shown below:

  $\begin{array}{c}\text{ΔH}\text{=}{\displaystyle \sum {\text{ΔH}}^{\text{0}}{}_{\text{f}\left(\text{products}\right)}\text{-}{\displaystyle \sum {\text{ΔH}}^{\text{0}}{}_{\text{f}\left(\text{reactants}\right)}}}\\ \text{=}\text{2×}{\text{ΔH}}^{\text{0}}{}_{\text{f}\left({\text{SO}}_{\text{3}}\right)}\text{-}\left[\text{2}\text{×}{\text{ΔH}}^{\text{0}}{}_{\text{f}\left({\text{SO}}_{\text{2}}\right)}\text{+}{\text{ΔH}}^{\text{0}}{}_{\text{f}\left({\text{O}}_{\text{2}}\right)}\right]\\ \text{=}\text{2}\text{×}\text{-396}{\text{kJmol}}^{\text{-1}}\text{-}\left[\text{2}\text{×}\text{-297}{\text{kJmol}}^{\text{-1}}\text{+}\text{0}{\text{kJmol}}^{\text{-1}}\right]\\ \text{=}\text{-198}{\text{kJmol}}^{\text{-1}}\end{array}$

Given information

ΔS⁰_f(SO₂) = 248 JK-¹mol-1

ΔS⁰_f(O₂) = 205 JK-¹mol-1

ΔS⁰_f(SO₃) = 257 JK-¹mol-1

The calculation of change in entropy at 25⁰C for the reaction is shown below:

  $\begin{array}{c}\text{ΔS}\text{=}{\displaystyle \sum {\text{ΔS}}^{\text{0}}{}_{\text{f}\left(\text{products}\right)}\text{-}{\displaystyle \sum {\text{ΔS}}^{\text{0}}{}_{\text{f}\left(\text{reactants}\right)}}}\\ \text{=}\text{2×}{\text{ΔS}}^{\text{0}}{}_{\text{f}\left({\text{SO}}_{\text{3}}\right)}\text{-}\left[\text{2}\text{×}{\text{ΔS}}^{\text{0}}{}_{\text{f}\left({\text{SO}}_{\text{2}}\right)}\text{+}{\text{ΔS}}^{\text{0}}{}_{\text{f}\left({\text{O}}_{\text{2}}\right)}\right]\\ \text{=}\text{2}\text{×}\text{257}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{-}\left[\text{2}\text{×}\text{248}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+}\text{205}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\\ \text{=}\text{- 187}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Given information:

C_p(SO₂) = 39.9 JK-¹mol-1

C_p(O₂) = 29.4 JK-¹mol-1

C_p(SO₃) = 50.7 JK-¹mol-1

The calculation of change in molar capacities is shown below:

  $\begin{array}{c}{\text{ΔC}}_{\text{p}}\text{=}{\displaystyle \sum {\text{C}}_{\text{p}\left(\text{products}\right)}\text{-}{\displaystyle \sum {\text{C}}_{\text{p}\left(\text{reactants}\right)}}}\\ \text{=}\text{2}\text{×}{\text{C}}_{\text{p}\left({\text{SO}}_{\text{3}}\right)}\text{-}\left[\text{2}\text{×}{\text{C}}_{\text{p}\left({\text{SO}}_{\text{2}}\right)}\text{+}{\text{C}}_{\text{p}\left({\text{O}}_{\text{2}}\right)}\right]\\ \text{=}\text{2}\text{×}\text{50}\text{.7}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{-}\left[\text{2}\text{×}\text{39}\text{.9}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+}\text{29}\text{.4}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\\ \text{=}\text{7}\text{.8}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

The Kirchhoff’s equation is shown below:

  $\begin{array}{c}{\text{ΔC}}_{\text{p}}\text{=}\frac{{\text{ΔH}}_{\text{2}}\text{-}{\text{ΔH}}_{\text{1}}}{{\text{T}}_{\text{2}}\text{-}{\text{T}}_{\text{1}}}\\ \text{7}\text{.8}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\frac{{\text{ΔH}}_{\text{2}}\text{-}\left(\text{-198}{\text{kJmol}}^{\text{-1}}\right)}{\text{500}\text{K}\text{-}\text{298}\text{K}}\\ \text{7}\text{.8}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\frac{{\text{ΔH}}_{\text{2}}\text{-}\left(\text{-198}{\text{kJmol}}^{\text{-1}}\right)}{\text{202K}}\\ \text{1575}\text{.6}{\text{Jmol}}^{\text{-1}}\text{=}{\text{ΔH}}_{\text{2}}\text{-}\left(\text{-198}{\text{kJmol}}^{\text{-1}}\right)\\ \text{1}\text{.5756}{\text{kJmol}}^{\text{-1}}\text{=}{\text{ΔH}}_{\text{2}}\text{+}\text{198}{\text{kJmol}}^{\text{-1}}\\ {\text{ΔH}}_{\text{2}}\text{=}\text{-}\text{196}\text{.4}{\text{kJmol}}^{\text{-1}}\end{array}$

First calculate the entropy to cool the reactants to 298 K as follows:

  $\Delta S_{500\to 298}=n{C}_{p,S{O}_2}\ln \frac{T_2}{T_1}+n{C}_{p,O_2}\ln \frac{T_2}{T_1}$

Putting the values,

  $\begin{array}{l}\Delta S_{500\to 298}=\left(\left(2\right)\left(39.9\right)\ln \frac{298}{500}+\left(1\right)\left(29.4\right)\ln \frac{298}{500}\right)\text{ J/K}\\ \text{=}-\text{56}\text{.5 J/K}\end{array}$

Now, entropy change to heat the product to 500 K can be calculated as follows:

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

The entropy change for the reaction at 500 K will be:

  $\Delta S=\left(-56.5-187+52.5\right)\text{ J/K}=-191\text{ J/K}$

Conclusion

The value of ΔH⁰at 25⁰C = -198 kJmol-1

The value of ΔS⁰at 25⁰C = -187 JK-1

The value of ΔH⁰at 227⁰C = - 196.4 kJ

The value of ΔS⁰at 227⁰C = - 191 JK-1