The equilibrium constant for the following reaction is 1.20*10 (-2) at 500K.


Screenshot 2024 01 16 195036

Screenshot 2024 01 16 195036

Step 1 of 4

Given is the value of equilibrium constant for the reactions;

  1. PCL5(g)↽−−⇀PCl3(g)+Cl2(g) Keq=1.2×10-2
  2. 2NOBr(g)↽−−⇀2NO+Br2(g) Keq=6.50×10-3

3.SO2Cl2(g)↽−−⇀SO2(g)+Cl2(g) Keq=4.50×10-2

We found out the equilibrium concentration of Cl2,Br2 and Cl2 respectively⋅

Explanation:

We use the concept of law of chemical equilibrium.

Step 2 of 4

For 1st problem,

PCl5(g) −−⇀PCl3(g)+Cl2(g)

2.17×10-2M 1.71×10-2M ?

[PCl5] =2.17×10-2M

[PCl3] =1.71×10-2M

[Cl2] =?

Keq=1.20×10-2

Temperature =500K is constant.

On applying law of chemical equilibrium,

Keq=[Product]/[Reactant]

or Keq=[PCl3][Cl2]/[PCl5]

or 1.20×10-2=1.71×10-2 [Cl2] / 2.17×10-2

or [Cl2] = 1.20×10-2 ×2.17 / 1.71

or [Cl2]=1.5228×10-2M

or [Cl2] = 1.52×10-2M is concentration of [CL2] at equilibrium.

Step 3 of 4

For 2nd problem,

2NOBr(g)↽−−⇀2NO(g)+Br2(g)

[NOBr]=2.15×10-2M

[NO]= 2.00×10-2M

[Br2] =?

Keq= 6.50×10-3

T=298K is constant.

On applying law of chemical equilibrium;

Keq=[Product]/[Reactant]

or Keq=[NO]2 [Br2]/ [NOBr]2

or 6.50×10-3= [2.00×10-2]2 [Br2] / [2.15×10-2]2

or [Br2] = 6.50×10-3 (2.15)2/(2.00)2

or [Br2] = 30.04625 ×10-3/4

or [Br2] = 7.51 ×10-3M

or [Br2]= 7.51 ×10-3M is the equilibrium concentration of [Br2].

Step 4 of 4

For 3rd problem,

SO2Cl2(g)↽−−⇀SO2(g)+Cl2(g)

[SO2Cl2]=2.20×10-2M

[SO2]=1.67×10-2M

[Cl2]=?

Keq=4.50×10-2

Temperature=T=648K is constant.

On applying law of chemical equilibrium,

Keq= [SO2] [Cl2]/[SO2Cl2]

or 4.50×10 -2= (1.67×10-2) [Cl2] / 2.20× 10-2

or [Cl2] = 4.50×10-2 (2.20× 10-2) / 1.67×10-2

or [Cl2]= 5.9281×10-2 M

or [Cl2] = 5.93 ×10-2 M is the equilibrium concentration of [Cl2]

Final solution

So, as we seen that the equilibrium concentration of the given equations are following,

for 1st equation is, 1.52×10-2M

for 2nd equation is, 7.51×10-3M

for 3rd equation is, 5.93×10-2M

 

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