Chapter 18, Problem 111AP

Carbon monoxide $\left(\text{CO}\right)$ and nitric oxide $\left(\text{NO}\right)$ are polluting gases contained in automobile exhaust. Under suitable conditions, these gases can be made to react to form nitrogen $\left({\text{N}}_{\text{2}}\right)$ and the less harmful carbon dioxide $\left({\text{CO}}_{\text{2}}\right)$. (a) Write an equation for this reaction. (b) Identify the oxidizing and reducing agents. (c) Calculate the for the reaction at $\text{25°C}$. (d) Under normal atmospheric conditions, the partial pressures are $P_{{\text{N}}_{\text{2}}}\text{ = 0}{\text{.80 atm, P}}_{\text{CO}}{}_{{}_{\text{2}}}\text{= 3}\text{.0 }\times {\text{ 10}}^{\text{-4}}\text{atm}\text{.}{P}_{\text{CO}}\text{= 5}\text{.0 }\times {\text{ 10}}^{\text{-5}}\text{atm}\text{. and }{P}_{\text{NO}}\text{= 5}\text{.0 }\times \text{ 10}$ atm. Calculate $Q_F$, and predict the direction toward which the reaction will proceed. (e) Will raising the temperature favor the formation of ${\text{N}}_{\text{2}}{\text{ and CO}}_{\text{2}}$ ?

Interpretation Introduction

Interpretation:

The values of ${\text{K}}_{\text{p}}$, $Q_{\text{p}}$, the direction of the reaction and the substance thatact as oxidizing and the reducing agent, are to be determined.

Concept introduction:

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

An atom with high value of standard reduction potential is a good oxidizing agent. An atom with low value of standard reduction potential is a good reducing agent.

The standard free energy of formation is the free energy change that occurs when one mole of the compound is combined from its constituent elements, in standard state.

A substance that gains electrons in the chemical reaction is called the oxidizing agent.

A substance that loses electrons in the chemical reaction is called the reducing agent

The standard free energy change is the difference of the sum of standard free energy change of products and the sum of the standard free energy change of reactants.

The equilibrium constant defines the ratio of concentration of product and reactants at equilibrium.

The relation between standard free energy change and equilibrium constant at 25^oC is given by relation

$\Delta G^{\text{o}}=-\text{R}T\ln {\text{K}}_{\text{p}}$.

The standard enthalpy changes of the reaction $\Delta H_{\text{rxn}}^{\text{o}}$ is calculated using expression as follows:

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

Answer to Problem 111AP

Solution:

a)

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

b)

$\text{NO}\left(g\right)$ is the oxidizing agent and $\text{CO}\left(g\right)$ is the reducing agent.

c)

The equilibrium pressure constant for the given reaction is $3\times {10}^{120}$.

d)

$Q_p$ is less than ${\text{K}}_{\text{p}}$. The reaction will proceed in the right direction.

e)

No, the formation of ${\text{N}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$ is not favored by increasing the temperature.

Explanation of Solution

a) Equation for the reaction of $\text{CO}\left(g\right)$ with $\text{NO}\left(g\right)$.

The equation for reaction of $\text{CO}\left(g\right)$ with $\text{NO}\left(g\right)$ is as follows:

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

In the above reaction, two moles of carbon monoxide react with two moles nitrogen oxide, resulting in the formation of dinitrogen and two moles of carbon dioxide.

b)Identification of oxidizing and reducing agents.

The equation for reaction of $\text{CO}\left(g\right)$ with $\text{NO}\left(g\right)$ is as follows:

$\text{2}\underset{2+}{\text{C}}\text{O}\left(g\right)\text{+ 2}\underset{2+}{\text{N}}\text{O}\left(g\right)\rightleftharpoons 2\underset{4+}{\text{C}}{\text{O}}_2\left(g\right)+\underset{0}{{\text{N}}_2}\left(g\right)$.

A substance that gains electrons in the chemical reaction is called the oxidizing agent.

A substance that loses electrons in the chemical reaction is called the reducing agent.

$\text{CO}\left(g\right)$ undergoes loss of electrons. It is the reducing agent. $\text{NO}\left(g\right)$ undergoes gain of electrons and is the oxidizing agent.

Therefore, $\text{NO}\left(g\right)$ is the oxidizing agent and $\text{CO}\left(g\right)$ is the reducing agent.

c) ${\text{K}}_{\text{p}}$ for the reaction at ${25}^{\text{o}}\text{C}$

The equation for reaction of $\text{CO}\left(g\right)$ with $\text{NO}\left(g\right)$ is as follows:

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

The standard free energy change for the system is calculated using the following expression:

$\Delta G_{\text{system}}^{\text{o}}=\Delta G_{\text{rxn}}^{\text{o}}.$

The standard free energy change for the reaction is calculated using the following expression:

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

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

The Gibbs energy change of the reaction is given by the following expression:

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

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

From appendix2, the values of the standard free energy of formation of compound are as follows:

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

$\Delta G_f^o\left[\text{NO}\left(g\right)\right]=86.7\text{kJ}/\text{mol}$;

$\Delta G_f^o\left[{\text{CO}}_2\left(g\right)\right]=-394.4\text{kJ}/\text{mol}$;

$\Delta G_f^o\left[\text{CO}\left(g\right)\right]=137.3\text{kJ}/\text{mol}$.

Substitute the value of the standard free energy of formation of compound and $\Delta G_{rxn}^o$ in the above expression.

$\begin{array}{l}\Delta G_{\text{rxn}}^{\text{o}}=\left(2\times \left[-394.4\text{kJ}/\text{mol}\right]+\left[0\right]\right)-\left(2\times \left[-137.3\text{kJ}/\text{mol}\right]+2\times \left[86.7\text{kJ}/\text{mol}\right]\right)\\ =-687.6\text{kJ}/\text{mol}\text{.}\end{array}$

Therefore, standard free energy change for the reaction is $-687.6\text{kJ}/\text{mol}$.

The equilibrium pressure constant is calculated using the following expression:

$\Delta G^{\text{o}}=-RT\ln {\text{K}}_{\text{p}}$.

Here, $\text{R}$ is the constant, $T$ is the temperature, ${\text{K}}_{\text{p}}$ is the overall equilibrium constant, and $\Delta G^o$ is the Gibbs free energy change.

Substitute the value of $\text{R}$, $T,$ and $\Delta G^o$ in the above expression.

$\begin{array}{c}\ln {\text{K}}_{\text{p}}=-\left(\frac{-6.876\times {10}^5\text{J}/\text{mol}}{\left(8.314\text{J}/\text{K}\cdot \text{mol}\right)\left(\text{298K}\right)}\right)\\ \ln {\text{K}}_{\text{p}}=277.5\\ {\text{K}}_{\text{p}}={\text{e}}^{277.5}\\ {\text{K}}_{\text{p}}=3\times {10}^{120}.\end{array}$

Therefore, equilibrium pressure constant for the given reaction is $3\times {10}^{120}$.

Given information:

The partial pressure of ${\text{N}}_{\text{2}}$ is $P_{N_2}=0.80\text{atm}$, the partial pressure of ${\text{CO}}_{\text{2}}$ is $P_{C{O}_2}=3\times {10}^{-4}\text{atm}$, the partial pressure of $\text{CO}$ is $P_{CO}=5\times {10}^{-5}\text{atm}$, the partial pressure of $\text{NO}$ is $P_{NO}=5\times {10}^{-5}\text{atm,}$    and temperature $\text{T}={25}^o\text{C}$.

d) $Q_p$ for reaction and direction of reaction toward which it will proceed

The reaction quotient expression for given the reaction is as follows:

$Q_p=\frac{\left(P_{N_2}\right){\left(P_{C{O}_2}\right)}^2}{{\left(P_{CO}\right)}^2{\left(P_{NO}\right)}^2}$.

Here, $P_{N_2}$ is a partial pressure of ${\text{N}}_{\text{2}}$, $P_{C{O}_2}$ is the partial pressure of ${\text{CO}}_{\text{2}}$, $P_{CO}$ is the partial pressure, $P_{NO}$ is the partial pressure of $\text{NO,}$ and $Q_p$ is the reaction quotient.

Substitute the value of $P_{N_2}$, $P_{C{O}_2}$, $P_{CO}$ and $P_{NO}$ in the above expression.

$\begin{array}{l}{Q}_p=\left(\frac{\left(0.80\right){\left(3\times {10}^{-4}\right)}^2}{{\left(5\times {10}^{-5}\right)}^2{\left(5\times {10}^{-7}\right)}^2}\right)\\ =1.2\times {10}^{14}.\end{array}$

The direction of the reaction is identified by comparing the value of reaction quotient and equilibrium constant.

If $Q_p$ is greater than ${\text{K}}_{\text{p}}$, the reaction will proceed in the left direction.

If $Q_p$ is less, then the reaction will proceed in the right direction.

For the given reaction, $Q_p$ is less than ${\text{K}}_{\text{p}}$. Thus, the reaction will proceed in the right direction.

Therefore, $Q_p$ is less than ${\text{K}}_{\text{p}}$. The reaction will proceed in the right direction.

e)Raising temperature favor the formation of ${\text{N}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$.

The equation for reaction of $\text{CO}\left(g\right)$ with $\text{NO}\left(g\right)$ is as follows:

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

The standard enthalpy changes of the system $\Delta H^{\text{o}}$ is calculated using expression as follows:

$\Delta H^{\text{o}}=\Delta H_{\text{rxn}}^{\text{o}}.$

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

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

Here, $\Delta H_{\text{rxn}}^{\text{o}}$ 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 products, and $m$ is the stoichiometric coefficient of reactants.

The enthalpy change for the reaction is as follows:

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

From appendix2, the standard enthalpy changes of formation for the substance are as follows:

$\Delta H_f^o\left[\text{CO}\left(g\right)\right]=\text{}-110.5\text{kJ}/\text{mol;}$

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

$\Delta H_f^o\left[{\text{CO}}_2\left(g\right)\right]=-393.5\text{kJ}/\text{mol;}$

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

Substitute the standard enthalpy change of the formation value of the substance in the above expression.

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{o}}=\left[2\times \left(-393.5\text{kJ}/\text{mol}\right)+\left(0\right)\right]-\left[2\times \left(90.4\text{kJ}/\text{mol}\right)+2\times \left(-110.5\text{kJ}/\text{mol}\right)\right]\\ =-746.8\text{kJ}/\text{mol}\\ \Delta H^{\text{o}}=-746.8\text{kJ}/\text{mol}\text{.}\end{array}$

Therefore, the standard enthalpy change for the given reaction is $-746.8\text{kJ}/\text{mol}$.

The standard enthalpy change for the given reaction is negative. As the temperature increases, the value of ${\text{K}}_{\text{p}}$ will decreaseson increasing the amount of reactants and decreasing the amount of products. No, the formation of ${\text{N}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$ is not favored by increasing the temperature.