Chapter 15, Problem 15.89QP

Carbon monoxide (CO) and nitric oxide (NO) are polluting gases contained in automobile exhaust. Under suitable conditions, these gases can be made to react to form nitrogen (N₂) and the less harmful carbon dioxide (CO₂). (a) Write an equation for this reaction. (b) Identify the oxidizing and reducing agents. (c) Calculate the K_p for the reaction at 25°C. (d) Under normal atmospheric conditions, the partial pressures are $P_{{\text{N}}_{\text{2}}}=0.80\text{atm, }{P}_{{\text{CO}}_{\text{2}}}=3.0\times {10}^{-4}\text{atm, }$P_CO = 5.0 × 10⁻⁵ atm, and P_NO = 5.0 × 10⁻⁷ atm. Calculate Q_P and predict the direction toward which the reaction will proceed. (e) Will raising the temperature favor the formation of N₂ and CO₂?

Interpretation Introduction

Interpretation:

The different equilibrium terms should be calculated given the statements of atmospheric equilibrium reactions.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Oxidizing reagent: This type of process gains of electrons and is reduced in chemical reaction, also electron acceptor the oxidizing agent in normally in one of its higher oxidation states it will gain electrons (e^-) to be reduced.

Reducing reagent: This type of chemical process loses of electrons and oxidizing reactions, normally the reducing agent lower possible oxidation states and the electron donor. In other words loses of electrons undergoes for redox reaction. For example alkaline earth metals, carboxylic acids and sulfate compounds involved a reduction process.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Kp and Kc: This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

Exothermic reaction: This type of chemical reaction release energy by light or heat. Endothermic reaction: This type of chemical reactions that is accompanied by the absorption of heat.

Reaction quotient: This type of chemical equilibrium reaction proceeds likely to produced, given either the pressure (or) the concentration of the reactants and the products. The value can be compared to the equilibrium constant, to determine the direction of the reaction that is take place. Then reaction quotient (Qc) the indication of Q can be used to determine which direction will shift to reach of chemical equilibrium process.

Le Chatelier's Principle (Kp): The closed system is an increase in pressure, the equilibrium will shift towards the sides of the reaction with some moles of gas. The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

Answer to Problem 15.89QP

The atmospheric equilibrium reaction of given the different terms of process (a-e) are shown below.

$\begin{array}{l}\text{a)}\text{.}\text{2CO}\text{+}\text{2NO}\to {\text{2CO}}_{\text{2}}\text{+}{\text{N}}_{\text{2}}\text{-----[1]}\\ \text{b)}\text{.}\text{NO----Oxidizing,}\text{CO---Reducing}\\ \text{c)}\text{.}\text{Kp=}\text{3}{\text{.29×10}}^{\text{128}}\text{d)}\text{.}\text{Qp=}\text{1}{\text{.2×10}}^{\text{14}}\text{e)}\text{.}{\text{ΔH}}^{\text{0}}\text{=-746}\text{.8}\text{KJ/mol}\end{array}$

Explanation of Solution

To Identify: Write the equilibrium reactions of given the statements (a and b).

(a) and (b)

Draw and analyze the given equilibrium reactions (a) and oxidizing, reducing reactions (b).

$\begin{array}{l}\text{a)}\text{.}\text{2CO(g)}\text{+}\text{2NO(g)}\to {\text{2CO}}_{\text{2}}(g)\text{+}{\text{N}}_{\text{2}}(g)\text{-----[1]}\\ \text{b)}\text{.}{\text{NO---NO}}^{\text{-}}\text{Oxidizing}\text{agent}\text{(NO}\text{is}\text{gains}\text{of}\text{e}lectrons)\\ \text{CO---Gains of}\text{electrons}\text{(Reducing agent)}\text{. }\end{array}$

Analyzing Reaction (a): Above the reactions $\text{CO}$ and $\text{NO}$ reacted with gas phase conditions to produce a two moles of ${\text{CO}}_{\text{2}}$ and nitrogen ( ${\text{N}}_{\text{2}}$) gas, the balanced equilibrium reactions showed above. Further given statement (b) nitric oxide ( $\text{NO}$) is a very good oxidizing agent it will gains a one electrons to form a N₂ and carbon monoxide ( $\text{CO}$) is a reducing agent it will converted into ( ${\text{CO}}_{\text{2}}$) gas. The derived both equations (a and b) showed above.

To find: Calculate the partial pressure (Kp) values for given the equilibrium reaction at ${\text{25}}^{\text{0}}\text{C}$.

Calculate and analyze the (Kp) values for statement (c).

Let us consider the following partial equilibrium equation

$\begin{array}{l}\text{a)}\text{.}\text{2CO}\text{+}\text{2NO}\to {\text{2CO}}_{\text{2}}\text{+}{\text{N}}_{\text{2}}\\ {\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{.0826}\text{l}\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{)}\\ \ln =(-ve(\log )\text{State Function})\\ \text{a)}\text{.}\text{2CO}\text{+}\text{2NO}\to {\text{2CO}}_{\text{2}}\text{+}{\text{N}}_{\text{2}}\\ {\text{ΔG}}_{\text{f}}^{\text{0}}={\text{CO}}_{\text{2}}\text{=}394.4\text{KJ/mol}\\ {\text{ΔG}}_{\text{f}}^{\text{0}}\text{=}{\text{N}}_{\text{2}}\text{=}137.3\text{KJ/mol}\text{(This}\text{va}luesfromequilibriumtable)\\ {\text{ΔG}}_{\text{f}}^{\text{0}}\text{=}\text{CO=}86.1\text{KJ/mol,}\text{NO=}\text{86}\text{.7}\text{KJ/mol}\\ {\text{ΔG}}_{\text{f}}^{\text{0}}=(Product-Reac\tan t)\\ {\text{ΔG}}^{\text{0}}={\text{ΔGf}}^{\text{0}}\text{(Products)-}{\text{Δf}}^0(Reac\tan ts)\\ (or){\text{ΔG}}_{\text{rxn}}^0=\sum n{\text{ΔGf}}^{\text{0}}\text{(Products)-}\sum m{\text{Δf}}^0(Reac\tan ts)\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=2{\text{ΔG}}_{\text{f}}^0(C{O}_2)+{\text{ΔG}}_{\text{f}}^0(N_2)-2{\text{ΔG}}_{\text{f}}^0(CO)-2{\text{ΔG}}_{\text{f}}^0(NO)\\ \text{The respactive values are substituted for above equlibrium reaction (1)}\\ {\text{ΔG}}^{\text{0}}=(2)(-394.4\text{KJ/mol})+(0)-(2)(-137.3\text{KJ/mol})-(2)(86.7\text{KJ/mol})\\ {\text{ΔG}}^{\text{0}}=-687.6\text{KJ/mol}\\ \text{Let}\text{us consider following equation (1)}\\ {\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{Rewrite the above equation }\\ \text{InK}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{6.876\times {10}^5\text{KJ/mol})}{(8.314\text{J/K}\text{mol)}(298K)}\because -(-687.6\text{KJ/mol)}\text{Change to}\text{6}\text{.876}\times {\text{10}}^{\text{5}}\text{KJ/mol}\\ \text{InK}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{6.876\times {10}^5\text{KJ/mol})}{(2477.572)}=(HereR=8.314,273+52=298K)\\ \text{InK}\text{=}\text{0}\text{.00277}\text{(or)}\text{277}\text{.5}\\ \text{Here K=}\text{In277}\text{.5}\text{;}\text{Kp=}\text{3}{\text{.29×10}}^{\text{120}}(\because e^{277.5}=3.286)\\ \text{Kp=}\text{3}{\text{.29×10}}^{\text{120}}\end{array}$

The given reaction is homogenous equilibrium reaction proceeds in same conditions (for example both reactant and product are proceeds in gas phase). Further the total equilibrium that is given CO₂ and N₂ two products are generated in $\text{Kp=3}\text{.29}\times {\text{10}}^{\text{120}}$, the derived pressure equilibrium reaction are placed above.

To Identify: Given the statements (d and e) reaction quintet (Qr) temperature must be analyzed.

Explanation:

(d) and (e)

Write and analyze the given equilibrium reactions (d and e).

$\begin{array}{l}\text{a)}\text{.}\text{2CO}\text{+}\text{2NO}\to {\text{2CO}}_{\text{2}}\text{+}{\text{N}}_{\text{2}}\text{-----[1]}\\ \text{Apply for partial pressure values for reaction (1)}\\ \text{2CO(g)}\text{+}\text{2NO(g)}\to {\text{2CO}}_{\text{2}}(g)\text{+}{\text{N}}_{\text{2}}(g)\\ \text{Kp=}\frac{(P_{N_2}){(P_{C{O}_2})}^2}{{(P_{CO})}^2{(P_{NO})}^2}=\frac{\text{Product}}{\text{Reactant}}-----[2]\\ \text{Given the statemnt of vslues are substituted equation (2)}\\ \text{Kp=}\frac{(0.80){(3.0\times {10}^{-4})}^2}{{(5.0\times {10}^{-5})}^2{(5.0\times {10}^{-7})}^2}\\ \text{Kp=}1.2\times {10}^{14}\end{array}$

Let us consider a statement (e).

$\begin{array}{l}\Delta H^0={\text{2ΔH}}_{\text{f}}^{\text{o}}{\text{(CO}}_{\text{2}}\text{)+}{\text{2ΔH}}_{\text{f}}^{\text{o}}{\text{(N}}_{\text{2}}\text{)-}{\text{2ΔH}}_{\text{f}}^{\text{o}}{\text{(CO)-2ΔH}}_{\text{f}}^{\text{o}}\text{(NO)}\\ \text{The respactive values are substituted for above equlibrium reaction (1)}\\ {\text{ΔH}}^{\text{0}}=(2)(-393.5\text{KJ/mol})+(0)-(2)(-110.5\text{KJ/mol})-(2)(90.4\text{KJ/mol})\\ {\text{ΔH}}^{\text{0}}=-746.8\text{KJ/mol}\end{array}$

Statement (d): The equal amount of CO and NO to produce CO₂ and N₂ this equilibrium reaction placed above if we increased pressure equilibrium will favor the reaction (product) decrease the total number of moles of gas. Here equilibrium should be shifted left to the right side, and produced more $\left(\text{CO}\right)$ at the expense of ${\text{(CO}}_{\text{2}}\text{)}$.

Statement (e): Given the equilibrium process is endothermic (heat absorption reaction) and this equilibrium shifted into left side, when temperature decreased moreover this reaction get positive $\text{(-ΔH)}$ values and its equilibrium constant (Kc) and equilibrium pressure (Kp) should become lesser values.

Conclusion

The different statements of equilibrium process are derived given the atmospheric equilibrium reactions.