Question

For the equilibrium, \[AB(g) \rightleftharpoons A(g) + B(g)\], \[{K_P}\] is equal to four times the total pressure. Calculate the number of moles of A formed per mole of AB taken.

Answer 1

Hint: In order to find the number of moles of A formed from the initial moles of AB taken, we must first calculate the partial pressure of A, B and AB. Then we have to apply these values in the law of mass action.

Complete answer:
In order to calculate the total number of moles of A formed per moles of AB taken, we have to first consider the total equilibrium pressure as ‘P’ atm.
In the question it is given that
\[{K_P}\]= 4P
Initial number of moles of AB(g) is 1 and the degree of dissociation is taken as ‘x’.
\[AB(g) \rightleftharpoons A(g) + B(g)\]
At the equilibrium, AB(g) is 1-x, A(g) is x and B(g) is x.
Therefore, the total number of moles at equilibrium is
1-x+x+x=1+x
The partial pressure of A is taken as
\[{P_A} = \dfrac{{x + 1}}{x}P\]
The partial pressure of B is taken as
\[{P_B} = \dfrac{{x + 1}}{x}P\]
The partial pressure of AB is taken as
\[{P_{AB}} = \dfrac{{1 - x}}{{1 + x}}P\]
We have to apply the Law of mass action
\[{K_P} = \dfrac{{{P_A} \times {P_B}}}{{{P_{AB}}}}\]
So,
\[4P = \dfrac{{{x^2}}}{{1 - {x^2}}}P\]
\[4 = \dfrac{{{x^2}}}{{1 - {x^2}}}\]
\[4 - 4{x^2} = {x^2}\]
\[x = \dfrac{2}{{\sqrt 5 }}\]
Therefore, we can say that the number of moles of A formed is \[\dfrac{2}{{\sqrt 5 }}\] times the initial moles of AB taken.

Note:
We have to remember that the law of mass action is helpful for explaining and predicting the solutions which are in the dynamic equilibrium. It is applicable only when there will be existence of the chemical equilibrium.