Question

For the equilibrium AB(g)= A(g) + B(g), K, is
equal to four times the total pressure. The number
of moles of 'A' formed is​

Answer 1

Answer:

camera-retroopen sidebar

Watch 1 minute video

Question from Class 11 Chapter Chemical Equilibrium

For the equilibrium AB(g)⇔A(g)+B(g). Kp is equal to four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially.

check-circle

Text Solution

Answer :

A::B

Solution :

Let the total equilibrium pressure be=P atm

Given Kp=4P

Let the start be made with 1 mol of AB(g) and the degree if dissociation be x.

AB(g)⇔A(g)+B(g)

at equilibrium,1−xxx

Total "moles" at equilibrium =1−x+x+x=1+x

Thus, pA=Partial pressure of A =x1+xP

pB =Partial pressure of B =x1+x.P

pAB=Partial pressure of AB =1−x1+x.P

Applying the law of mass action

Kp=pA×pBpAB=(x1+x.P)(x1+x.P)(1−x1+x.P)

So 4P=x21−x2.P

or 4−4x2=x2

or 5x2=4

or x=25–√

Hence, number of "moles" of A formed =2/5–√ times initial moles of AB taken.

PLZZ MARKS BRAINLIST