Question

The dissociation of CO2 can be expressed as
2Co <=> 2Co +O2 . lf the 2 mol of CO2 is taken
initially and 40% of the CO2 is dissociated
completely. What is the total number of moles at
equilibrium:-
(1) 2.4 (2) 2.0 (3) 1.2 (4) 5
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Answer 1

Answer:

option 1 is correct

Explanation:

2CO2 ------> 2CO + O2

Initial moles of CO2 are two

Initial moles of CO and O2 are zero

At equilibrium degree of dissociation a = 40%=0.4

for the given reaction

equilibrium moles of CO2 = 2—2a = 2—2(0.4)= 1.2

equilibrium moles of CO = 2a = 2(0.4)= 0.8

equilibrium moles of O2 = a = 0.4

therefore total moles at equilibrium are

1.6+0.8+0.4=2.8

Answer 2

Answer:

2.4 moles

Explanation:

Number of moles dissociated= 40%×initial no. of moles

                                = (40/100)×2

                                      =0.8

             2CO₂ ⇄ 2CO + O₂

@ t=0        2           0        0

@ t=t      2-2x        2x       x

 no. of moles dissociated = 2x= 0.8⇒ x=0.4

total no. of moles at equilibrium= (2-2x)+ (2x) + x

                                                       = 2-0.8 + 0.8 + 0.4

                                                       = 2.4 moles