Chapter 15.3, Problem 6PPA

For the reaction

   ${\text{N}}_{\text{2}}\left(g\right)+{\text{3H}}_{\text{2}}\left(g\right)\rightleftarrows {\text{2NH}}_{\text{2}}\left(g\right)$

  K_C is 2.3 × 10⁻² at 375°C. Calculate K_P for the reaction at this temperature.

Interpretation Introduction

Interpretation:

Calculate the partial pressure $\left({\text{K}}_{\text{p}}\right)$ values of given the equilibrium constant of given the ammonia $\left({\text{NH}}_{\text{3}}\right)$ formation reaction at ${375}^{\text{ο}}\text{C}$.

Concept Introduction:

$\left({\text{K}}_{\text{p}}\right)$and ${\text{K}}_{\text{c}}$ : This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas $\left({\text{K}}_{\text{p}}\right)$ is defined by the partial pressures of the gasses inside a closed system.

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Le Chatelier's Principle $\left({\text{K}}_{\text{p}}\right)$: 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.

$\left({\text{K}}_{\text{p}}\right)$:  The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

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.

Solution: The reactant and product partial equilibrium pressure $\left({\text{K}}_{\text{p}}\right)$ and equilibrium constant values for the given ${\text{NH}}_{\text{3}}$ formation reaction is shown below.

$\begin{array}{l}{\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Kc=}\frac{{[N{H}_3]}^2}{[N_2]{[H_2]}^3}=1.2\\ {\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Kp=}\frac{{(P_{N{H}_3})}^2}{{(P_{H_2})}^3(P_{N_2})}\\ \text{Kp=}8.1\times {10}^{-6}\\ \end{array}$

Explanation of Solution

To find: The equilibrium reaction should be identified given the statement.

Analyze the chemical equilibrium reaction.

$\begin{array}{l}{\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}----[FormationReaction]\\ {\text{2NH}}_{\text{3}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\text{---------[Decomposition Reaction]}\end{array}$

The given equilibrium concentration reaction is the combined reaction is the product of the constants for this component reaction. This equilibrium reaction expression contains different conditions like solid phase into gases phase, so this process homogenous equilibrium the equilibrium constant can also be represented by $\left({\text{K}}_{\text{p}}\right)$, were the $\left({\text{K}}_{\text{p}}\right)$ represents partial pressure. Then the each (reactant and product) molecule partial pressure $\left({\text{N}}_{\text{2}}{\text{, H}}_{\text{2}}{\text{and NH}}_{\text{3}}\right)$ is derived in further method.

To find: Calculate the each partial pressure $\left({\text{K}}_{\text{p}}\right)$ and equilibrium constant ${\text{K}}_{\text{c}}$ values for given the statement of equilibrium reaction.

Calculate and analyze the $\left({\text{K}}_{\text{p}}\right)$ and ${\text{K}}_{\text{c}}$ values at $\text{373}{}^{\text{0}}\text{C}$.

Consider the given equilibrium reactions

$\begin{array}{l}\text{a)}.{\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\text{----------}[Equilibriumassociation]\\ \text{b)}{\text{.2NH}}_{\text{3}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)----\text{[Equilbrium dissociation]}\\ Weconsiderthereaction(b),andwritetheKprelatioshipforgivenbelow\\ \text{Kp=}\frac{{(P_{N{H}_3})}^2}{{(P_{H_2})}^3(P_{N_2})}\text{=}\frac{\text{Product}}{\text{Reactant}}\\ \text{Consider}\text{the following}\text{equation}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{Kc(RT)}}^{\text{Δn}}-----[1]\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.08206}}\\ \text{Here}\text{Δn}\text{=}2\text{-4}\text{=}-2\text{(Product - Reactant) consider reaction (a)}\\ =\Delta n=-2\\ \text{Here}\text{Δn}\text{=}-2\text{(Product - Reactant)}\\ \text{and}\text{given}\text{Temprature}{375}^{\text{0}}\text{C}\\ \text{T}\text{=}\text{(375+ 273)}\text{K}\\ =648\text{K}\\ \text{Given Kc value 2}\text{.3}\times {\text{10}}^{-2}\\ \text{The respactive values are substituted for equation (2)}\\ \text{Kp=}\text{Kc[0}\text{.08206}\text{L×atm/K}{\text{.mol)×T]}}^{\text{Δn}}-------[2]\\ \text{Kp=}(2.3\times {10}^{-2}){(}^{0.08206}\\ \text{Kp=}8.1{\text{×10}}^{\text{-6}}\end{array}$

The equilibrium constant values are derived given the equilibrium statement (a) and its temperature.

Conclusion

The each of reactant and product partial pressure $\left({\text{K}}_{\text{p}}\right)$ and equilibrium constant values are derived given the ammonia $\left({\text{NH}}_{\text{3}}\right)$ equilibrium reactions.