Chapter 15, Problem 15.76QP

Interpretation Introduction

Interpretation:

Calculate the partial pressure values all species of reactant and product given the equilibrium reaction at ${372}^{\text{ο}}\text{C}$.

Concept Introduction:

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 (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.

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.

Answer to Problem 15.76QP

The reactant and product partial equilibrium pressure (Kp) values for the given ${\text{NH}}_{\text{3}}$ formation reaction is shown below.

  $\begin{array}{l}{\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Kp=}\frac{{(P_{N{H}_3})}^2}{{(P_{H_2})}^3(P_{N_2})}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=0}\text{.860}\text{atm}{\text{; P}}_{H_{\text{2}}}\text{=0}\text{.366atm}{\text{;P}}_{{\text{NH}}_{\text{3}}}\text{=4}\text{.40}\times {\text{10}}^{\text{-3}}\text{atm}\end{array}$

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 Kp, were the Kp 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 next method.

To find: Calculate the each partial pressure (Kp) values for given the statement of equilibrium reaction.

Calculate and analyze the (Kp) values at $\text{375}{}^{\text{0}}\text{C}$.

Consider the given equilibrium reactions

  $\begin{array}{l}{\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\rightleftharpoons {\text{2NH}}_{\text{3}}\text{(g)}\\ \text{Initial (atm): }0.8620.3730\\ \text{Change (atm): }\text{-x}-3x+2x\\ ----------------------------\\ \text{Eqilibrium (atm):}(0.862-x)(0.373-3x)2x\\ Thepartialpressureequationhas\\ \text{Kp=}\frac{{(P_{N{H}_3})}^2}{{(P_{H_2})}^3(P_{N_2})}-----[1]\\ \text{Given equilibrium constan values}\text{(Kp) 4}\text{.31}\times {\text{10}}^{\text{-4}},\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=0}\text{.862}\text{atm}{\text{; P}}_{{\text{H}}_{\text{2}}}\text{=0}\text{.373atm}\\ \text{The respactive values are substituted equation (1)}\\ \text{4}\text{.31}\times {\text{10}}^{\text{-4}}=\frac{{(2x)}^2}{{(0.373-3x)}^3(0.862-x)}\\ \text{Here (3x)}\text{is very small when compared to the (0}\text{.373}{\text{H}}_{\text{2}}\text{) values }\\ \text{and that x is very small when comapred to 0}{\text{.862 of (N}}_{\text{2}}\text{)}\text{values}\\ \text{0}\text{.373-3x}\approx \text{0}\text{.373}\text{and}\text{0}\text{.862-x}\approx \text{0}\text{.862}\\ \text{This values are substituted equation (1)}\\ \text{4}\text{.31}\times {\text{10}}^{\text{-4}}\approx \frac{{(2x)}^2}{{(0.373)}^3(0.862)}\\ \text{We solved here (x)}\\ \text{4}\text{.31}\times {\text{10}}^{\text{-4}}\approx \frac{4x^2}{(0.0518)(0.862)}=\frac{4x^2}{0.04465}\\ \text{Changeing above equation}\\ {\text{4x}}^{\text{2}}=\text{4}\text{.31}\times {\text{10}}^{\text{-4}}\times (0.04465)\\ {\text{4x}}^{\text{2}}=1.928\times {10}^{-5}\\ x^2=\frac{1.928\times {10}^{-5}}{4}=4.82\times {10}^{-6}\\ \text{Taking for square root both sides}\\ \text{x=}\text{2}\text{.195}\times {\text{10}}^{\text{-3}}(Roundingtwosiginificantvalues)\\ x=2.2\times {\text{10}}^{\text{-3}}\\ \text{The equilibrium pressure are}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=[0}\text{.862-}(2.2\times {\text{10}}^{\text{-3}})]\text{atm=}\text{0}\text{.860}\text{atm}\\ {\text{P}}_{{\text{H}}_{\text{2}}}\text{=[0}\text{.373-}(3)(2.2\times {\text{10}}^{\text{-3}})]\text{atm=}\text{0}\text{.366}\text{atm }\\ {\text{P}}_{{\text{NH}}_3}\text{=(2)}(2.2\times {\text{10}}^{\text{-3}})\text{atm=}4.40\times {\text{10}}^{\text{-3}}\text{atm }\end{array}$

The given ammonia formation reaction the respective reactant to give products all exists in the same phase  and this equilibrium reaction expression contains single conditions like gases phase, the equilibrium constant can also be represented by Kp, were the “P” partial pressure. The each partial pressure values are Kp derived given equation at $3{\text{75}}^{\text{0}}\text{C}$ as showed above.

Conclusion

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