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

Answer

A, B

Explanation

Let the total equilibrium pressure be=P atm
Given `K_(p)=4P`
Let the start be made with `1` mol of AB(g) and the degree if dissociation be x.
`AB(g) hArr A(g)+B(g)`
`{:("at equilibrium,",1-x,,x,,x):}`
Total "moles" at equilibrium `=1-x+x+x=1+x`
Thus, `p_(A)`=Partial pressure of A =`x/(1+x)P`
`p_(B)` =Partial pressure of B `=x/(1+x).P`
`p_(AB)`=Partial pressure of AB `=(1-x)/(1+x).P`
Applying the law of mass action
`K_(p)=(p_(A)xxp_(B))/p_(AB)=((x/(1+x).P)(x/(1+x).P))/(((1-x)/(1+x).P))`
So `4P=x^(2)/(1-x^(2)).P`
or `4-4x^(2)=x^(2)`
or `5x^(2)=4`
or `x=2/sqrt(5)`
Hence, number of "moles" of A formed `=2//sqrt(5)` times initial moles of AB taken.