Chapter 15, Problem 15.145QP

The equilibrium constant K_c for the reaction

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

is 0.83 at 375°C. A 14.6-g sample of ammonia is placed in a 4.00-L flask and heated to 375°C. Calculate the concentrations of all the gases when equilibrium is reached.

Interpretation Introduction

Interpretation:

To calculate the equilibrium concentration values are given homogenies equilibrium of ammonia (NH₃) dissociation reaction with respective pressure and temperature at ${375}^{\text{ο}}\text{C}$.

Concept Introduction:

Equilibrium concentration: If Kc and the initial concentration for a reaction and calculate for both equilibrium concentration, and using the (ICE) chart and equilibrium constant and derived changes in respective reactants and products.

Pressure effect in equilibrium: 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.

Temperature affect in equilibrium: This process chemical shifts changes (or) towards the product or reactant, which can be determined by studying the reaction and deciding whether it is exothermic or endothermic.

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 15.145QP

The reactant and product each equilibrium concentration (ICE) values for the given ${\text{NH}}_{\text{3}}$ ammonia dissociation reaction is shown below.

$\begin{array}{l}{\text{2NH}}_{\text{3}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\\ \text{Kc=}\frac{[N_2]{[H_2]}^3}{{[N{H}_3]}^2}\\ [N{H}_3]=0.042\text{M,}{\text{[N}}_{\text{2}}\text{]=}\text{0}\text{.086}\text{M}\text{and}{\text{[H}}_{\text{2}}\text{]=}\text{0}\text{.26}\text{M}\end{array}$

Explanation of Solution

To find: The equilibrium concentration should be identified given the gases phase reaction.

Analyze the chemical equilibrium reaction.

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

Given the gas phase equilibrium concentration reaction is the combined reaction; it is the product of the constants for this component reaction. This equilibrium reaction expression contains same conditions like gas phase. Hence this process homogenous equilibrium further the equilibrium constant can also be represented by Kc, were the Kp represents partial pressure. Then the each (reactant and product) molecule equilibrium concentration $\left({\text{N}}_{\text{2}}{\text{, H}}_{\text{2}}{\text{and NH}}_{\text{3}}\right)$ are discussed and explained the fallowing below method.

To find: Calculate the each concentration values for given the equilibrium constant (Kc) of ammonia dissociation reaction.

Calculate and analyze the respective concentration values at $\text{375}{}^{\text{0}}\text{C}$.

First let’s calculate the initial concentration of ammonia.

$\begin{array}{l}[N{H}_3]=\frac{14.6g\times \frac{\text{1}\text{}\text{mol}\text{of}{\text{NH}}_{\text{3}}}{\text{17}\text{.03}\text{g}\text{of}{\text{NH}}_{\text{3}}}}{4.00L}[\therefore MolarMassofN{H}_3=17.03g]\\ =\frac{14.6g\times 0.05871}{4.00L}=\frac{0.8571}{4.00L}=0.214\\ \text{The}\text{concentration}\text{of}\text{[}N{H}_3]=0.214M\end{array}$

Further we set up a ICE table to represent the equilibrium concentrations. With respect the amount of (NH₃) that reacts as (2x) fallowed by,

$\begin{array}{l}{\text{2NH}}_{\text{3}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}(g)+{\text{3H}}_{\text{2}}(g)\\ \text{Initial (M): }0.21400\\ \text{Change (M): }\text{-2x}+x+3x\\ ----------------------------\\ \text{Eqilibrium (M):}(0.214-2x)x3x\\ \text{Substitute}\text{into}\text{the }\text{equilibrium}\text{expression}\text{to}\text{solve}\text{for}\text{(x)}\text{.}\\ \text{Kc=}\frac{[N_2]{[H_2]}^3}{{[N{H}_3]}^2}\\ \text{The}\text{respactive equilbrium}\text{concentration}\text{values}\text{are}\\ N_2=+x,H_2=3xandN{H}_3=0.214-2x\\ \text{The}\text{partial}\text{presure}\text{values}\text{of}{\text{(NH}}_{\text{3}}\text{)=}\text{0}\text{.83}\\ \text{The respactive values are substituted equation (1)}\\ 0.83=\frac{(x){(3x)}^2}{{(0.214-2x)}^2}=\frac{27x^4}{{(0.214-2x)}^2}-------[2]\\ \text{Taking}\text{the square root}\text{of both sides}\text{of the}\text{above}\text{equation}\text{(2)}\\ \text{0}\text{.91=}\frac{5.20x^2}{0.214-2x}-----[3]\\ Rearrangingequation(3)\\ 5.20x^2+1.82x-0.195=0\\ Solvingthequadraticequationgivestheaboveequation\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=0.086Mandx=-0.44M\\ \text{Here}\text{the possitive}\text{root}and\text{equilibrium}\text{concetrations}\text{are}\\ N{H}_3=0.214-2x(x=0.086M)\\ [N{H}_3]=0.214-2(0.086)=0.0442M\\ [N_2]=0.086M\\ H_2=3x;3(0.086)=0.26M\end{array}$

The given ammonia dissociation reaction the respective reactant to give a two moles of products, and this reaction proceeds in 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 reactant and product concentration values are derived given above equation at $3{\text{75}}^{\text{0}}\text{C}$ finally the equilibrium constant of following methods as showed above.

Conclusion

The each of reactant and product equilibrium concentration values are derived given the gas phase ammonia $\left({\text{NH}}_{\text{3}}\right)$ dissociation equilibrium reactions.