Chapter 15, Problem 15.95QP

Consider the reaction

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

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 given reaction. (d) As farmers know, lightning helps to produce a better crop. Why?

Interpretation Introduction

Interpretation:

Calculate the standard free energy $({\text{ΔG}}_{\text{f}}^{\text{0}})$ and partial pressure (Kp) values for given the $\text{NO}$ nitric oxide formation reactions with respective temperature at ${\text{25}}^{\text{0}}\text{C}$.

Concept Introduction:

Standard free energy$({\text{ΔG}}_{\text{f}}^{\text{0}})$: This thermodynamic function associated with a chemical reaction that can be used to do work. The standard free energy of formation of a compound change of Gibbs free energy the formation of 1 mole of substance from its component elements, at their standard states (For example the obtained product is more stable at 25⁰C). 

Entropy$({\text{ΔG}}^0)$: 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.

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.

Le Chatelier's Principle (Kp): In a closed system if there is an increase in pressure, then the equilibrium will shift towards the side of the reaction having smaller number of moles.  The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

To find: Calculate the $({\text{ΔG}}_{\text{f}}^{\text{0}})$ values and partial pressure (Kp) for given the statement (a and b) of equilibrium reaction.

Answer to Problem 15.95QP

The equilibrium constant (Kp), standard free energy values $({\text{ΔG}}_{\text{f}}^{\text{0}})$ and enthalpy changes $({\text{ΔG}}^0\&\text{ΔG})$ values are given the statement of nitric oxide chemical reaction is showed below.

$\begin{array}{l}{\text{N}}_{\text{2}}\text{(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons 2NO(g)\\ \text{Standard}\text{free}\text{energy}{\text{ΔG}}_{\text{f}}^{\text{0}}=4.0\times {10}^{-30}\\ \text{The}\text{equilibrium}\text{constant (Kc)}\\ {\text{K}}_{\text{P}}\text{=}\frac{{(P_{NO})}^2}{(P_{N_2})(P_{O_2})}=4.0\times {10}^{-30}\\ {\text{ΔH}}^{\text{0}}=180.8\text{KJ/mol}\\ {\text{K}}_{\text{2}}=2.62\times {10}^{-6}\end{array}$

Explanation of Solution

(a)

Calculate and analyze the standard free energy $({\text{ΔG}}_{\text{f}}^{\text{0}})$ and (Kp) values.

Consider the following equation (1)

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

The equilibrium constant is related to the to the standard free energy change by the followed above equation (1). Than the given statement of values (K_P, R and T) are substituted in same equation.

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

Let us consider the statement (b)

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

The given equilibrium concentration reaction the different reactants and products all exists in the single phase and this equilibrium reaction expression contains single conditions like gases phase, so this equilibrium reaction homogenies equilibrium reaction. The thermodynamic terms are constant can also be represented by Kp, Kc, R, T and ${\text{ΔG}}^0$ were the “P” partial pressure, Kc represents molar concentration R gas constant T, temperature and ${\text{ΔG}}^0$ entropy. The standard free energy values $({\text{ΔG}}^0)$ were derived and showed above equation.

To find: Calculate the standard enthalpy changes (${\text{ΔH}}^{\text{0}}$), industrial and crop application of given the statement (c and d).

(b)

Calculate the standard enthalpy changes ${\text{ΔH}}^{\text{0}}$.

$\begin{array}{l}{\text{N}}_{\text{2}}\text{(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons 2NO(g)\\ \text{Here}\\ \text{Standard enthalpy changes equation }\\ \text{In}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{=}\frac{{\text{ΔH}}^{\text{0}}}{\text{R}}\left(\frac{{\text{T}}_{\text{2}}{\text{-T}}_1}{{\text{T}}_{\text{1}}\times {\text{T}}_{\text{2}}}\right)-----[2]\\ {\text{ΔH}}^{\text{0}}\text{total}\text{no}\text{.}\text{of}\text{product}\text{formation}\text{2×}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{(NO)}\\ \text{The equilibrium values of NO= 90}\text{.4}\text{KJ/mol}\to (thisvaluesformequilibriumtables)\\ \text{2×}{\text{ΔH}}_{\text{f}}^{\text{o}}\text{(NO)}=2(90.4KJ/mol)=180.8KJ/mol\\ Therespactivevaluesaresubstitutedequation(1)\\ \text{In}\frac{{\text{K}}_{\text{2}}}{\text{4}\text{.0}{\text{×10}}^{\text{-31}}}\text{=}\frac{\text{180}{\text{.8×10}}^{\text{3}}\text{J/mol}}{\text{8}\text{.314}\text{J/K}\text{.mol}}\left(\frac{\text{1373K-298K}}{\text{(1373)(298K)}}\right)\\ =\frac{\text{180}{\text{.8×10}}^{\text{3}}\text{J/mol}}{\text{8}\text{.314}\text{J/K}\text{.mol}}\times (0.00262)(or)(2.62\times {10}^2)\\ {\text{K}}_{\text{2}}\text{=2}{\text{.62×10}}^{\text{-6}}\end{array}$

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 N₂ and two products are generated in ${\text{K}}_{\text{2}}\text{=2}{\text{.62×10}}^{\text{-6}}$, the derived pressure equilibrium reaction are placed above.

Statement (d): The equilibrium reactions lightening supply the energy necessary to derive this reaction, concerting the two most abundant gases in the atmosphere into $\text{NO}\left(\text{g}\right)$. The $\text{NO}$ gas dissolves in the rain, which carries it into the soil where it is converting into nitrate bacterial action. The particular or fixed nitrogen is a necessary nutrient for plants.