InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» Correct Answer - A::B 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. |
|