Chapter 15, Problem 15.137QP

Consider the equilibrium reaction described in Problem 15.30. A quantity of 2.50 g of PCl₅ is placed in an evacuated 0.500-L flask and heated to 250°C. (a) Calculate the pressure of PCl₅, assuming it does not dissociate. (b) Calculate the partial pressure of PCl₅ at equilibrium. (c) What is the total pressure at equilibrium? (d) What is the degree of dissociation of PCl₅? (The degree of dissociation is given by the fraction of PCl₅ that has undergone dissociation.)

Interpretation Introduction

Interpretation:

The different equilibrium (Kp, Kc) constants should be calculated given the ${\text{PCl}}_{\text{5}}$ dissociation reaction.

Concept Introduction:

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.

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.

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.

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 15.137QP

The equilibrium constant (Kp and Kc) values are given the statement of (PCl₅) homogenous reaction is shown below.

$\begin{array}{l}{\text{PCl}}_{\text{5}}\text{(g)}\stackrel{}{\to }{\text{PCl}}_{\text{3}}\text{+}{\text{Cl}}_{\text{2}}\\ {\text{K}}_c=\frac{{\text{(PCl}}_{\text{3}})(C{l}_2)}{({\text{PCl}}_5)}\frac{[Product]}{[Reac\tan t]}\\ {\text{K}}_p=\frac{{\text{(P}}_{{\text{PCl}}_{\text{3}}})(P_{C{l}_2})}{(P_{{\text{PCl}}_5})}\\ \text{a)}\text{.}\text{Molar}\text{mass=}\text{1}\text{.03, b)}\text{.}{\text{P}}_{{\text{PCl}}_5}=0.39\text{atm,}\\ \text{c)}\text{.}{\text{P}}_{\text{total}}=1.67a\text{tm,}\text{d)}\text{.}\text{Degreee}\text{of}\text{dessociation}\text{=62}\text{.0\%}\end{array}$

Explanation of Solution

To find: Calculate the molar mass values and partial pressure $(Kc)$ values for given the statement of PCl5 dissociation reaction.

Calculate and analyze the molar mass and $(Kp)$ values with respective statement (a and b).

Let us considers the (ICE) equilibrium method

$\begin{array}{l}\text{Let}\text{us consider the given statement (a)}\\ \text{We derived the statement}\text{(a)}\\ \text{The molar mass of }{\text{PCl}}_{\text{5}}=208.2g/mol\\ {\text{PCl}}_{\text{5}}\text{(g)}\rightleftharpoons {\text{PCl}}_{\text{3}}\text{+}{\text{Cl}}_{\text{2}}\\ \text{Acoording to the ideal gas equation }\\ \text{Pv=nRT---------------[1]}\\ \text{Change the above equation (2) and derived partial pressure values}\\ \text{P=}\frac{nRT}{V}------[2][Here\text{P=Partial pressure of reactant}\text{and}\text{product}]\\ \text{n=}2.50gm,R=0.08206\text{L×atm}\text{/K×mol,}\text{T=(275+250=523K}),\text{V=0}\text{.500}\text{L}\\ \text{Given}\text{the}\text{statement (b) values are substituted above equation (2)}\\ \text{P=}\frac{nRT}{V}=\left(\frac{2.50g\times \frac{1mol}{208.2g}}{0.500L}\right)\left(0.08206\frac{L.atm}{K.mol}\right)(523K)\\ =\left(2.50\times \frac{4.8030}{0.500L}\right)\left(0.08206\times 523\right)\\ =\left(2.50\times 9.606\right)(42.9173)\\ =(24.015)(42.9173)\\ =1.03\text{atm}\end{array}$

We derive here equilibrium constant values are ${\text{PCl}}_{\text{5}}$.

$\begin{array}{l}\text{Hare, }\\ {\text{PCl}}_{\text{5}}\text{(g)}\rightleftharpoons {\text{PCl}}_{\text{3}}(g)\text{+}{\text{Cl}}_{\text{2}}\text{(g) }\\ \text{Initial (atm): }1.0300\\ \text{Change (atm): }\text{-x}+x+x\\ ------------------------\\ \text{Eqilibrium (atm):}1.03-xxx\\ \text{Solving for the equilibrium constant: }\\ {\text{K}}_p=\frac{{\text{(P}}_{{\text{PCl}}_{\text{3}}})(P_{C{l}_2})}{(P_{{\text{PCl}}_5})}\text{-----------[1]}\\ \text{The equilibrium pressure values are substituted above the equation (1)}\\ \text{Kc=}\frac{{(x)}^2}{(1.03-x)}\\ Kp=1.05atm\\ Hence1.05=\frac{{(x)}^2}{(1.03-x)}------[2]\\ Solvedaboveequation(2)\\ x^2+1.05x-1.08=0\\ x=0.639\end{array}$

The balanced equilibrium equations there are one mole of PCl5 dissociated to produce two moles pf products, if the pressure of the container will not affect above calculation method of (Kp) values. Hence the derived equilibrium constant in terms of showed above.

To find: Calculate the total pressure values and degree of dissociation values for given the statement of PCl5 equilibrium reaction.

Calculate and analyze the total pressure and degree of dissociation values with respective statement (c and d).

The obtained statement (a and d) values are substituted below method, we get the partial pressure values and total pressure values.

$\begin{array}{l}\text{Statement (c)}\\ {\text{PCl}}_{\text{5}}\text{(g)}\rightleftharpoons {\text{PCl}}_{\text{3}}\text{+}{\text{Cl}}_{\text{2}}\\ {\text{PCl}}_{\text{5}}\text{(g)}=\text{Reactant-Product}\text{values}\\ {\text{PCl}}_{\text{5}}\text{(g)}=1.03-0.603=0.39\text{atm}\\ {\text{P}}_{\text{total}}=\text{Product-Reactant}\text{(According}\text{to}ICEtablevalues)\\ {\text{P}}_{\text{total}}=(1.03-x)+x+x\\ {\text{P}}_{\text{total}}=1.03+0.639=1.67atm\\ \text{Statement (d)}\text{(We}\text{derived}\text{degreee of dessociation)}\\ \text{The }\frac{ICETablevalues}{molarmassvalues}=\frac{0.639atm}{1.03atm}=0.620\\ \text{The Degreee}\text{of}\text{dessociation}\text{=62}\text{.0\%}\end{array}$

The given dissociation reaction the respective reactant to give products  (two products) all exists in the different phase  and this equilibrium reaction expression contains different conditions like solid converted into gases phase, so this equilibrium reactions has heterogeneous.   The equilibrium constant can also be represented by Kp, were the “P” partial pressure. The each partial pressure, total and degree of dissociation values are derived given ${\text{PCl}}_{\text{5}}$ equilibrium reaction equation at ${250}^{\text{0}}\text{C}$ as showed above.

Conclusion

The different terms of equilibrium values are derived given the PCl₅ dissociation reaction.