Chapter 15, Problem 15.60QP

The dissociation of molecular iodine into iodine atoms is represented as

I₂(g) ⇄ 2I(g)

At 1000 K, the equilibrium constant K_c for the reaction is 3.80 × 10⁻⁵. Suppose you start with 0.0456 mole of I₂ in a 2.30-L flask at 1000 K. What are the concentrations of the gases at equilibrium?

Interpretation Introduction

Interpretation:

The equilibrium concentration $\left(\text{Kc}\right)$ should be calculated given the gas phase (I₂) reaction.

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.

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.

Heterogeneous equilibrium: This equilibrium reaction does not depend on the amounts of pure solid and liquid present, in other words heterogeneous equilibrium, substances are in different phases.

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.

Vaporized equilibrium: This conversion of liquid in gaseous phase is known as vaporization process. At starting the rate of condensation is less than the rate of evaporation but as evaporation continues the concentration of gaseous molecule in the vapour phase increase.

Answer to Problem 15.60QP

The equilibrium concentration $\left(\text{Kc}\right)$ values for given the gases phase I₂(g) into 2I(g) equilibrium reactions is shown below.

$\begin{array}{l}{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2I(g)}\\ \text{Kc=}\frac{{[I]}^2}{[I_2]}=3.80\times {10}^{-5}\text{and}\text{Kp=}\frac{{(P_I)}^2}{(P_{I_2})}\\ \text{The equilibrium}\text{concentrations}\text{[I]=}\text{8}\text{.58}\times {\text{10}}^{\text{-4}}\text{M}\text{and}{\text{[I}}_{\text{2}}\text{]=0}\text{.0194}\text{M}\end{array}$

Explanation of Solution

To find: The each reactant product equilibrium concentration should be identified given the gas phase reaction.

Write and Analyze the given gas phase chemical equilibrium reaction.

$\text{a)}\text{.}{\text{I}}_{\text{2}}(g)\rightleftharpoons \text{2I(g)}----[DessociationReaction]$

The given equilibrium reaction has a homogenous process, then the equilibrium constant can also be represented by Kp, were the Kp represents partial pressure. Then the product molecule partial pressure ${\text{(Reactant I}}_{\text{2}}\text{,Product}\text{2I)}$ is derived in step-2.

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

Calculate and analyze the (Kp) values at $1000{}^{\text{0}}\text{C}$.

We derived here (Kp) values of (I₂) dissociation reaction

First we derived the initial concentration of (I₂) is

$\begin{array}{l}\text{The }\text{initial}\text{concentration}\text{of}{\text{(I}}_{\text{2}}\text{)=}\frac{\text{0}\text{.0456}mol}{2.30L}=0.0198M\\ \text{here the 1}\text{mole}\text{of}{\text{(I}}_{\text{2}}\text{) dissocoating}\text{to}\text{2 moles}\text{of (I)}\text{atoms}\\ \text{Let}\text{(x)}\text{amount }\text{in}\text{mol/L}\text{od}{\text{(I}}_{\text{2}}\text{)}\text{dissociated}\text{. }\\ \text{The equilibrium }\text{concentration}\text{of}\text{(I)}\text{atoms must}\text{be}\text{=2x}\end{array}$

$\begin{array}{l}\text{Here set up the (ICE) table }\\ {\text{Let (x) be the decrease in concentration of (I}}_{\text{2}}\text{and}2\text{I)}\\ {\text{I}}_{\text{2}}(g)\rightleftharpoons 2\text{I(g)}\\ \text{Initial (M): }0.01980.000\\ \text{Change (M): }\text{-x}+2x\\ ----------------------------\\ \text{Eqilibrium (M):}(0.0198-x)2x\end{array}$

We consider the equilibrium expression in terms of the equilibrium concentration.

$\begin{array}{l}\text{The equilibrium constant solve}\text{for}\text{(x)}\\ \text{Kc=}\frac{{[I]}^2}{[I_2]}-----[1]\\ The(ICE)tablevaluesaresubstitutedequation(1)\\ =\frac{{(2x)}^2}{(0.0198-x)}=\frac{4x^2}{(0.0198-x)}\\ \text{Given}(Kc)valuesare3.80\times {10}^{-5}\\ \text{Hence,}\\ 3.80\times {10}^{-5}=\frac{4x^2}{(0.0198-x)}------[2]\\ Rewritetheaboveequation(2)\\ 4x^2=(3.80\times {10}^{-5})x-(7.52\times {10}^{-7})=0\\ \text{We solved a quadratic equation from}{\text{ax}}^{\text{2}}\text{+bx+c=0}\\ \frac{-b\pm \sqrt{b^2-4ac}}{2a}-----[3]\\ a=4,b=3.80\times {10}^{-5}c=-7.52\times {10}^{-7}\\ \text{This values are substituted equation (3)}\\ x=\frac{(-3.80\times {10}^{-5})\pm \sqrt{{(3.80\times {10}^{-5})}^2-4(4)(-7.52\times {10}^{-7})}}{2(4)}\\ x=\frac{(-3.80\times {10}^{-5})\pm (3.47\times {10}^{-3})}{8}\\ x=4.29\times {10}^{-4}Mandx=-4.29\times {10}^{-4}M\end{array}$

The obtained second (x) values are negative concentration, this physically impossible so we omitted this values. First (x) value is correct one.

$\begin{array}{l}\text{The (x) values}\text{are}\text{substituted (ICE) equilibrium values}\\ \text{[I]= 2X=}\text{(2)(4}\text{.29}\times {\text{10}}^{\text{-4}}\text{M)}\\ \text{=}8.58\times {\text{10}}^{\text{-4}}\text{M}\\ {\text{[I}}_{\text{2}}\text{]=}\text{(0}\text{.0198-x)M}\\ \text{=(0}\text{.0198-4}\text{.29}\times {\text{10}}^{\text{-4}}\text{)}\\ \text{=}0.0194M\end{array}$

The given iodine dissociation equilibrium reaction the respective reactant to give the two moles of products in the gas 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 molar concentration values are Kp derived given the gas phase reaction at ${1000}^{\text{0}}\text{C}$ as showed above.

Conclusion

The molar concentration (M) values are derived given the iodine $\left({\text{I}}_{\text{2}}\right)$ dissociation equilibrium reactions.