Chapter 18, Problem 18.51QP

(a)

Interpretation Introduction

Interpretation:

To derive the entropy, enthalpy $({\text{ΔH}}^{\text{ο}}and{\text{ΔS}}^{\text{ο}})$ values for given gas phase equilibrium reaction (K₁ and K₂) at ${\text{25}}^{°}\text{C}$. 

Concept Information:

Gibbs free energy (G): The thermodynamic quantity to the $({\text{ΔG}}^0)$ enthalpy of a system process and minutes the product of entropy and the absolute temperature.  The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Free energy$({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Explanation of Solution

Derive the equilibrium constants and different temperature for give standard state conditions $({\text{ΔH}}^{\text{ο}}and{\text{ΔS}}^{\text{ο}})$ equation.

Let us consider the following thermodynamic equations,

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2NO}}_{\text{2}}{}_{\text{(g)}}{\text{ΔH}}^{\text{ο}}\text{=58}\text{.0}\text{KJ/mol}\\ {\text{Given the two differnt temprature (T}}_{\text{1}}\text{and}{\text{T}}_{\text{2}}\text{)}\\ {\text{ΔG}}_1^{\text{ο}}={\text{ΔH}}^{\text{ο}}-{\text{T}}_{\text{1}}{\text{ΔS}}^{\text{ο}}=-{\text{RTInK}}_{\text{1}}\text{----[1]}\\ {\text{ΔG}}_{\text{2}}^{\text{ο}}{\text{=ΔH}}^{\text{ο}}{\text{-T}}_{\text{2}}{\text{ΔS}}^{\text{ο}}{\text{=-RTInK}}_{\text{2}}\text{-}---[2]\\ RearrangingEquations(1)and(2)weget\\ {\text{ln K}}_{\text{1}}=\frac{{\text{-ΔH}}^{\text{ο}}}{{\text{RT}}_{\text{1}}}+\frac{{\text{ΔS}}^{\text{ο}}}{R}------[3]\\ {\text{ln K}}_{\text{2}}=\frac{{\text{-ΔH}}^{\text{ο}}}{{\text{RT}}_2}+\frac{{\text{ΔS}}^{\text{ο}}}{R}------[4]\\ Subtractingequation(3)and(4)gives\\ {\text{ln K}}_2-{\text{ln K}}_1=\left(\frac{{\text{-ΔH}}^{\text{ο}}}{{\text{RT}}_2}+\frac{{\text{ΔS}}^{\text{ο}}}{R}\right)-\left(\frac{{\text{-ΔH}}^{\text{ο}}}{{\text{RT}}_{\text{1}}}+\frac{{\text{ΔS}}^{\text{ο}}}{R}\right)\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}=\frac{{\text{ΔH}}^{\text{ο}}}{\text{R}}\left(\frac{1}{{\text{T}}_{\text{1}}}-\frac{1}{{\text{T}}_{\text{2}}}\right)\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}=\frac{{\text{ΔH}}^{\text{ο}}}{\text{R}}\left(\frac{{\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right)\end{array}$

(b)

Interpretation Introduction

Interpretation:

To derive the equilibrium constant $(\text{K})$ values for given gas phase equilibrium reaction (K₁ and K₂) at ${\text{25}}^{°}{\text{C, 65}}^{°}\text{C}$. 

Concept Information:

Thermodynamics is the branch of science that relates heat and energy in a system.  The laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

$\text{ΔG = }\Delta {\rm H}\text{- T}\Delta \text{S}$

Where,

  $\text{ΔG }$ is the change in free energy of the system

  $\Delta {\rm H}$ is the change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta \text{S}$ is the change in entropy in the system.

The equilibrium constant can be calculated by using following formula,

  $\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}=\frac{{\text{ΔH}}^{\text{ο}}}{\text{R}}\left(\frac{{\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right)$

Explanation of Solution

Using the above equation that we can derived, we can calculate the equilibrium constant at $65°C$

  $\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}=\frac{{\text{ΔH}}^{\text{ο}}}{\text{R}}\left(\frac{{\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right)$

  $\begin{array}{l}\text{Hence calculated the equilibrium constant at}{\text{65}}^{\text{0}}\text{C}\\ {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2NO}}_{\text{2}}{}_{\text{(g)}}{\text{ΔH}}^{\text{ο}}\text{=58}\text{.0}\text{KJ/mol}\\ \text{Given the statement of values are }\\ {\text{K}}_{\text{1}}\text{=4}\text{.63}\times {\text{10}}^{\text{-3}},{\text{T}}_{\text{1}}\text{=}\text{273+25=298K,}\\ {\text{K}}_{\text{2}}\text{=}\text{?}{\text{T}}_{\text{2}}\text{=273+65=338}\text{K}\\ \text{Hence we derived}\text{the equilibrum constant}\text{values of}{\text{(K}}_{\text{2}}\text{)}\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.63}\times {\text{10}}^{\text{-3}}}=\frac{\text{58}{\text{.0×10}}^{\text{3}}}{\text{8}\text{.314}\text{J/K×mol}}\left(\frac{\text{338K-298K}}{(338K)(298K)}\right)\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.63}\times {\text{10}}^{\text{-3}}}=\frac{\text{58}{\text{.0×10}}^{\text{3}}}{\text{8}\text{.314}\text{J/K×mol}}\left(\frac{40}{100724}\right)\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.63}\times {\text{10}}^{\text{-3}}}=\frac{\text{58}{\text{.0×10}}^{\text{3}}}{\text{8}\text{.314}\text{J/K×mol}}(3.9717\times {10}^{-4})\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.63}\times {\text{10}}^{\text{-3}}}=(6.9761\times {10}^{-5})(3.9717\times {10}^{-4})\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.63}\times {\text{10}}^{\text{-3}}}=2.77\\ {\text{K}}_{\text{2}}=0.074\end{array}$

$K_2>K_1$

As we would predict for a positive $\Delta H°$, if  increase in temperature will shift the equilibrium toward the endothermic reaction, that is decomposition of $N_2{O}_4$