Chapter 18, Problem 85AP

Consider the reaction:

${\text{N}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\rightleftarrows 2\text{NO}\left(g\right)$

Given that $\Delta Gº$ for the reaction at $\text{25ºC is 173}\text{.4 kJ/mol,}$ (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 $\text{1100ºC}$. estimate $K_p$ for the given reaction. (d) As farmers know, lightning helps to produce a better crop. Why?

Interpretation Introduction

Interpretation:

The standard free energy change for the formation of $\text{NO}\left(g\right)$, the equilibrium pressure constant and the equilibrium pressure for the given reaction at a temperature of ${1100}^{\text{o}}\text{C}$, and the reason how lightening helps in the production of better crops are to be determined.

Concept introduction:

All the energy available in the system that is utilized in doing useful work is called Gibbs free energy.

The standard free energy change is the change in free energy that occurs under the standard state condition.

The standard free energy of formation is defined as the free energy change that occurs when one mole of a compound is combined from its constituent elements with each in the standard state.

Answer to Problem 85AP

Solution:

a)

$86.7\text{kJ}/\text{mol}$.

b)

$4\times {10}^{-31}$

c)

$3\times {10}^{-6}$.

d)

Lightening helps in the production of better crops because there is formation of nitrates and nitrite with soil mineral, which are good fertilizers.

Explanation of Solution

Given information: The reaction is as follows:

${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{NO}\left(g\right)$

Temperature ${\text{T}}_2={25}^o\text{C}$ and $\Delta G_{rxn}^o$ is the standard free energy change for the reaction, which is $173.4\text{kJ}/\text{mol}$.

a)The standard free energy of formation of NO

The equation for the reaction of ${\text{N}}_2$ with ${\text{O}}_2$ is as follows:

${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{NO}\left(g\right)$

The standard free energy change for the formation of $\text{NO}\left(g\right)$ is calculated by using the expression as follows:

$\Delta G_{rxn}^o={\displaystyle \sum n}\Delta G_f^o\left(\text{product}\right)-{\displaystyle \sum m\Delta G_f^o}\left(\text{reactant}\right)$

Here, $\Delta G_{rxn}^o$ is the standard free energy change for the reaction, $\Delta G_f^o$ is the standard free energy of the formation of the compound, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactants.

The Gibbs energy for the reaction is as follows:

$\Delta G_{rxn}^o=\left(2\times \Delta G_f^o\left[\text{NO}\left(g\right)\right]\right)-\left(\Delta G_f^o\left[{\text{N}}_2\left(g\right)\right]+\Delta G_f^o\left[{\text{O}}_2\left(g\right)\right]\right)$

The standard free energy of formation of any element in its stable allotropic form at $1\text{atm}$ and ${25}^o\text{C}$ is zero.

From appendix 2, the values of the standard free energy of formation of the compounds are as follows:

$\Delta G_f^o\left[{\text{O}}_2\left(g\right)\right]=0\text{kJ}/\text{mol}$

$\Delta G_f^o\left[{\text{N}}_2\left(g\right)\right]=0\text{kJ}/\text{mol}$

Substitute the values of the standard free energy of formation of the compoundsand $\Delta G_{rxn}^o$ in the above expression as follows:

$\begin{array}{c}173.4\text{kJ/mol}=\left(2\times \Delta G_f^o\left[\text{NO}\left(g\right)\right]\right)-\left(\left[0\right]+\left[0\right]\right)\\ \Delta G_f^o\left[\text{NO}\left(g\right)\right]=\left(\frac{173.4\text{kJ/mol}}{2}\right)\\ =86.7\text{kJ}/\text{mol}\end{array}$

Therefore, thestandard free energy change for the formation of $\text{NO}\left(g\right)$ is $86.7\text{kJ}/\text{mol}$.

b) $K_p$ of the reaction

The equation for the reaction of ${\text{N}}_2$ with ${\text{O}}_2$ is as follows:

${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{NO}\left(g\right)$

The equilibrium pressure constant is calculated by using the expression as follows:

$\Delta G_{}^o=-RT\ln K_p$

Here, $\text{R}$ is the ideal gas constant, $T$ is the temperature, $K_p$ is the equilibrium pressure constant, and $\Delta G^o$ is the Gibbs free energy change.

Substitute the values of $\text{R}$, $T$ and $\Delta G^o$ in the above expression as follows:

$\begin{array}{c}\ln {\text{K}}_{\text{p}}=\left(\frac{-173.4\times {10}^3\text{J}/\text{mol}}{\left(8.314\text{J}/\text{K}\cdot \text{mol}\right)\left(298\text{K}\right)}\right)\\ =-69.98\\ {\text{K}}_{\text{p}}=e^{-69.98}\\ =4\times {10}^{-31}\end{array}$

Therefore, the equilibrium pressure constant is $4\times {10}^{-31}$.

Given information: The reaction is as follows:

${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{NO}\left(g\right)$

Temperature ${\text{T}}_2={1100}^o\text{C}$, temperature ${\text{T}}_1={25}^o\text{C}$ and $\Delta G_{rxn}^o$ is the standard free energy change for the reaction, which is $173.4\text{kJ}/\text{mol}$

c)The equilibrium pressure for the given reaction

The equation for the reaction of ${\text{N}}_2$ with ${\text{O}}_2$ is as follows:

${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{NO}\left(g\right)$

The standard enthalpy change of the reaction ($\Delta H_{rxn}^o$) is calculated by using the expression as follows:

$\Delta H_{rxn}^o={\displaystyle \sum n}\Delta H_f^o\left(\text{product}\right)-{\displaystyle \sum m\Delta H_f^o}\left(\text{reactant}\right)$

Here, $\Delta H_{rxn}^o$ is the standard enthalpy change for the reaction, $\Delta H_f^o$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactants.

The enthalpy for the reaction is as follows:

$\Delta H_{rxn}^o=\left(2\times \Delta H_f^o\left[\text{NO}\right]\right)-\left(\Delta H_f^o\left[{\text{N}}_2\right]+\Delta H_f^o\left[{\text{O}}_2\right]\right)$

From appendix 2, the values of the standard enthalpy change for the formation of the substances are as follows:

$\Delta H_f^o\left[\text{NO}\left(g\right)\right]=\text{}90.4\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{N}}_2\left(g\right)\right]=0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{O}}_2\left(g\right)\right]=0\text{kJ}/\text{mol}$

Substitute the values of the standard enthalpy change for the formation of the substances in the above expression as follows:

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(2\right)\times \left(90.4\text{kJ}/\text{mol}\right)\right]-\left[\left(0\right)+\left(0\right)\right]\\ =180.8\text{kJ}/\text{mol}\end{array}$

The temperatures are converted from degree Celsius to Kelvin as follows:

$\begin{array}{c}\text{Temperatute}\left(T_1\left(\text{K}\right)\right)=\left({\text{25}}^o\text{C}+273\right)\\ =298\text{K}\end{array}$

$\begin{array}{c}\text{Temperatute}\left(T_2\left(\text{K}\right)\right)\text{=}\left({\text{1100}}^o\text{C}+273\right)\\ =1373\text{K}\end{array}$

The equilibrium pressure for this reaction at a temperature of ${1100}^o\text{C}$ is calculated by using the expression as follows:

$\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times \left(\frac{1}{T_1}-\frac{1}{T_2}\right)=\frac{\Delta H^o}{\text{R}}\times \left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Here, $K_1$ is the equilibrium pressure at temperature $T_1$, $K_2$ is the equilibrium pressure at temperature $T_2$, $\text{R}$ is the gas constant, and $\Delta H^o$ is the standard enthalpy change for the reaction.

Substitute the values of $K_1$, $K_2$, $T_1$, $T_2$, and $\text{R}$ in the above expression and solve as follows:

$\begin{array}{l}\ln \frac{{\text{K}}_2}{4\times {10}^{-31}}=\left(\frac{180.8\times {10}^3\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}}\right)\times \left(\frac{1373\text{K}-298\text{K}}{\left(\text{298K}\right)\left(1373\text{K}\right)}\right)\\ \text{}\text{}\text{}\ln \frac{{\text{K}}_2}{4\times {10}^{-31}}=13086.36\times \left(\frac{1075\text{K}}{\left(\text{298K}\right)\left(1373\text{K}\right)}\right)\\ \text{}\text{}\text{}\ln \frac{{\text{K}}_2}{4\times {10}^{-31}}=\left(13086.36\times 2.6\times {10}^{-3}\right)\\ \text{}=34.38\end{array}$

The value of ${\text{K}}_2$ is as follows:

$\begin{array}{l}\text{}\text{}\ln \frac{{\text{K}}_2}{4\times {10}^{-31}}=34.38\\ {\text{K}}_2=3\times {10}^{-6}\end{array}$

Hence, the equilibrium pressure at a temperature of ${1100}^o\text{C}$ is $3\times {10}^{-6}$.

d)Lightning helps to produce better crops

In the atmosphere, nitrogen molecules are present. Two nitrogen atoms are tightly boundedto each other. In lightening, electrical energy is generated, which has enough energy to break the bonds of the nitrogen atoms in the air. It combines with oxygen to form $\text{NO}\left(g\right)$. $\text{NO}\left(g\right)$ easily dissolves in the rain and fallsto the Earth. It combines with the minerals in the soil to form nitrates and nitrite. It is a type offertilizer that helps in the production of better crops.

The equation for the reaction of ${\text{N}}_2$ with ${\text{O}}_2$ is as follows:

${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons \text{NO}\left(g\right)$

Hence, lightening helps in the production of better crops because there is formation of nitrates and nitrite with soil mineral, which are fertilizers.