Advertisement Remove all ads

A Sample of Pure Pcl5 Was Introduced into an Evacuated Vessel at 473 K If the Value Of Kc Is 8.3 × 10–3, What Are the Concentrations of Pcl3 And Cl2 At Equilibrium - Chemistry

A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, the concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If the value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5 (g) ⇌ PCl3 (g) + Cl2(g)

Advertisement Remove all ads

Solution 1

Let the concentrations of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:

                                    `PCl_(5(g))  "              ↔           " PCl_(3(g))    +    Cl_(2(g))`

At equilibrium   0.5xx10^(-1) mol L^(-1)              `xmol L^(-1)`   `xmol L^(-1)`

It is given that the value of equilibrium constant, `K_C`  is `8.3 xx 10^(-3)`

Now we can write the expression for equilibrium as:

`([PCl_2][Cl_2])/[PCl_5] = K_C`

`=> (x xx x)/(0.5 xx 10^(-1)) = 8.3 xx 10^(-1)`

`=> x^2 = 4.15 xx 10^(-4)`

`=> x = 2.04 xx 10^(-2)`

= 0.0204

= 0.02 (approximately)

Therefore at equilibrium.

`[PCl_3] = [Cl_2] = 0.02 mol L^(-1)`

Solution 2

Let the initial molar concentration of PCl5 per litre = x mol
Molar concentration of PCl5 at equilibrium = 0.05 mol
.’. Moles of PCl5 decomposed = (x – 0.05) mol
Moles of PCl3 formed = (x – 0.05) mol
Moles of Cl2 formed = (x – 0.05) mol
The molar conc./litre of reactants and products before the reaction and at the equilibrium point are

                                PCl5   ⇌    PCl3  + Cl2

Initial moles/litres                x                0              0

Moles/litre at eqm. point   0.05    (x - 0.05) (x - 0

.05)

Equilibrium constant (`K_c`) = 8.3 xx 10^(-3)  = 0.0083

Applying Law of chemical equilibrium,

`K_c = ([PCl_3][Cl_2])/[PCl_5]`

`0.0083 xx ((x - 0.05) xx (x - 0.05))/0.05`

`(x - 0.05)^2 = (0.0083) xx 0.05 = 4.15 xx 10^(-4)`

`(x - 0.05) = (4.15 xx 10^(-4))^(1/2) = 2.037 xx 10^(-2) = 0.02 moles`

x = 0.05 + 0.02 = 0.07 mol

The molar concentration per liter of `PCl_3` at eqm = 0.07 -  0.05 = 0.02 mol

The molar concentation per liter of `Cl_2` at eqm = 0.07 - 0.05 =0.02 mol

Concept: Homogeneous Equlibria - Equilibrium Constant in Gaseous Systems
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 19 | Page 226
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×