Chapter 17, Problem 17.71QP

Consider the reaction

${\text{N}}_2(g)+O_2(g)\rightleftharpoons 2\text{NO}(g)$

Given that ΔG° for the reaction at 25°C is 173.4 kJ/mol, (a) calculate the standard free energy of formation of NO, and (b) calculate K_P of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is 1100°C, estimate K_P for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why?

Interpretation Introduction

Interpretation:

The standard free energy $\Delta G°$ value should be derived given nitric oxide formation reaction at ${\text{25}}^{°}\text{C}$.

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four 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.  In non-spontaneous reaction there is an increase of free energy in the system.

Free energy$({\text{ΔG}}^0)$: In thermodynamics free energy or Gibbs free energy is the energy that is used to express the total energy content of a system.  According to second law of thermodynamics, in all spontaneous process is associated with the decrease in free energy of the system.  That is the change in free energy will be negative.

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 $\text{G}$. All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Explanation of Solution

The equilibrium constant is related to the to the standard free energy change by the given equation (1).

$\begin{array}{l}{\text{N}}_{\text{2}}\text{(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons 2NO(g)\\ \text{ΔG}\text{=-RTln}\text{K-------------[1]}\\ \text{We rewrite the above equation}\\ \text{ΔG}=\frac{\text{Given}\text{value}s}{\text{Product}}=\frac{173.4\text{KJ/mol}}{2}(\because 2NOProduct)\\ \text{ΔG}=86.7\text{kJ/mol}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium constant $Kp$ value should be calculated for given reaction.

Concept Introduction:

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 $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, $\text{"n"}$ is the number of moles

The calculate the standard free $({\text{ΔG}}_{\text{f}}^{\text{0}})$ value for given reaction is shown below,

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.08314 (or) 8}\text{.314}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

First calculate the equilibrium constant ($Kp$) value for given reaction

$\begin{array}{l}{\text{N}}_{\text{2}}\text{(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons 2NO(g)\\ {\text{ΔG}}_{\text{f}}^{\text{0}}\text{=-RTln}\text{K-------------[1]}\\ Here\text{R}\text{=}\text{0}\text{.08314}\text{J/K}\text{.}\text{mol (or) -8}\text{.314}\text{J/K}\text{.}\text{mol}\\ \text{Substituted}\text{for}\text{(Kp,}\text{R}\text{and}\text{T)}\text{values}\text{are}\text{equation}\text{(1)}\\ \text{Given the values 173}\text{.4KJ/mol}\\ 173.4\times {10}^3\text{J/mol}=-(8.\text{314J/K}\text{.}\text{mol})(298K)lnKp\\ \text{Rewrite the above equation (1)}\\ {\text{lnK}}_{\text{p}}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{173.4\times {10}^3\cancel{\text{J}}\text{/mol})}{-(8.314\cancel{\text{J}}\text{/}\cancel{\text{K}}\cdot \text{mol)}(298\cancel{K})}\\ {\text{lnK}}_{\text{p}}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{173.4\times {10}^3)}{-(2477.572)}\\ {\text{lnK}}_{\text{p}}\text{ = -69}\text{.99}\\ {\text{K}}_{\text{p}}=e^{-\text{69}\text{.99}}\\ {\text{K}}_{\text{p}}\text{=}4.{\text{0×10}}^{-31}\end{array}$

Given $NO$ formation reaction equilibrium constant value is $4.{\text{0×10}}^{-31}$

Interpretation Introduction

Interpretation:

The nitric oxide starting formation has to determine.

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four 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

Explanation of Solution

 We calculate the enthalpy changes for given reaction

$\begin{array}{l}{\text{N}}_{\text{2}}\text{(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons 2NO(g)\\ \text{Here}\\ \text{Standard enthalpy changes equation }\\ \text{In}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{=}\frac{{\text{ΔH}}^{\text{0}}}{\text{R}}\left(\frac{{\text{T}}_{\text{2}}{\text{-T}}_1}{{\text{T}}_{\text{1}}\times {\text{T}}_{\text{2}}}\right)----[1]\\ {\text{ΔH}}^{°}\text{total}\text{no}\text{.}\text{of}\text{product}\text{formation}\text{2×}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{(NO)}\\ \text{The equilibrium values of NO= 86}\text{.7}\text{kJ/mol}\\ \text{2×}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{(NO)}=2(\text{86}\text{.7}\text{kJ/mol})=173.4\text{kJ/mol}\\ Therespactivevaluesaresubstitutedequation(1)\\ \text{In}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.0}{\text{×10}}^{\text{-31}}}\text{=}\frac{173.4\times {10}^3\text{J/mol}}{\text{8}\text{.314}\text{J/mol}\cdot \text{K}}\left(\frac{\text{1373K-298K}}{\text{(1373)(298K)}}\right)\\ \text{In}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.0}{\text{×10}}^{\text{-31}}}\text{=}\frac{173.4\text{J/mol}}{\text{8}\text{.314}\text{J/mol}\cdot \text{K}}\left(\frac{\text{1075K}}{\text{409154K}}\right)\\ \text{In}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.0}{\text{×10}}^{\text{-31}}}\text{=(208563}\text{.86)}\left(\text{2}\text{.672}\times {\text{10}}^{\text{-3}}\text{K}\right)\\ \text{In}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.0}{\text{×10}}^{\text{-31}}}\text{=}5572.82\text{K}\\ {\text{K}}_{\text{2}}{\text{=3×10}}^{\text{-7}}\end{array}$

Interpretation Introduction

Interpretation:

The lighting helps to produce a better crop, the reason behind this has to explained. 

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four 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.  In non-spontaneous reaction there is an increase of free energy in the system.

Non-spontaneous reaction: This type of reaction explain as, endergonic reaction (mean by heat absorption non-spontaneous process) or unfavourable reaction in a chemical reaction in which the standard change in free energy is positive and energy is absorbed.

Explanation of Solution

The lighting promotes the formation of $NO$ (from $N_2$ and $O_2$ in the air), which eventually leads to the formation of nitrate ions $(N{O}_3^{-})$ as essential nutrition for plants.