Question

2CO2(g)
2CO(g) + O2(g), the dissociation of
2 moles of Co, is carried out in container with
volume 5 L. If at equilibrium 60% CO2 is
dissociated then number of moles at equilibrium
will be :-
(1) 2.3
(2) 2.6 (3) 3.0
(4) 3.2
7 Y ir heated at​

Answer 1

Answer:

Moles at equilibrium will be 3.0

Answer 2

Answer:

If 60% CO₂ is dissociated at equilibrium, then the total number of moles at equilibrium is 2.6 moles.

Therefore, option b) 2.6 moles is the correct option.

Explanation:

Given,

• Number of moles of CO₂ dissociated in a container = 2 moles

• At equilibrium, the amount of CO₂ dissociated, α = 60%

The total number of moles left after dissociation of 60% CO₂ at equilibrium =?

Let x = dissociated moles

Initial moles, a = 2 moles

The reaction of CO₂ dissociation given is,

•             2CO₂ (g)  ⇄  2CO (g) + O₂ (g)

At initial:               2                    0             0

At equilibrium:   2-x                   $\frac{2x}{2}$             $\frac{x}{2}$

At equilibrium:    2-x                  x              $\frac{x}{2}$

Now, after 60% dissociation occurs

Then,

• $\alpha = \frac{x}{a} \times 100$
• $60 = \frac{x}{2} \times 100$
• x = 1.2 moles

Therefore,

At equilibrium:      

• Number of moles of CO₂ = 2- x = 2- 1.2 = 0.8 moles
• Number of moles of CO = x = 1.2 moles
• Number of moles of O₂ = $\frac{x}{2}$ = $\frac{1.2}{2}$ = 0.6 moles

Hence,

The total number of moles at equilibrium = 0.8 + 1.2 + 0.6 = 2.6 moles.