Chemical equilibrium vs pressure

[subhradeep mahata]

Homework Statement

Consider the reaction : A⇔2B (⇔ stands for reversible). 1 mol of A and 2 mol of B are taken in a closed container at equilibrium. Suppose the pressure is increased, how much of B (by moles) will decompose to restore equilibrium ?

Homework Equations

The Attempt at a Solution

Okay, i know that reaction will be in backward direction in this case if pressure is increased.
A ⇔ 2B
Eq initial 1mol , 2mol
Eq final 1+x mol , 2-2x mol (2x mol of B decomposes to form x mol of A)
equilibrium constant = 2² / 1 = 4
it will not change with pressure
So, (2-2x)² / 1+x = 4
x is comming out to be 3.
But if i see, at eq final the remaining moles of B is 2-2*3, -4 !
How is it possible ? Please help me rectify my mistake.

 

[Borek]

equilibrium constant = 2² / 1 = 4

This is not the equilibrium constant, this is just the initial reaction quotient. You need the equilibrium constant value from some other source.

 

[mjc123]

Even if it is at equilibrium initially, that is not the equilibrium constant. The equilibrium constant is not expressed using moles but pressures (or concentrations), and if the number of moles is not equal on both sides of the equation, pressure is a factor in the equilibrium constant. If partial pressure is P*mole fraction, then
K = (2P/3)²/(P/3) = 4P/3 where P is the initial equilibrium pressure
If P₁ is the final equilibrium pressure, can you calculate the mole fractions and partial pressures (as functions of x), and an expression for the equilibrium constant?
Suppose the question is "you change the pressure to P₂, and the system adjusts to a new equilibrium state." Can you find the new equilibrium pressure, then proceed as above?

 

[subhradeep mahata]

Yes, the value of K will be same, so 4P/3 = (mole fraction of B * P₂)² / (mole fraction of A* P₂)
But i think i need some help in calculating the final mole fraction.
Shall i proceed in the same way as i did in my original attempt ?

 

[Borek]

There is not enough data to solve the problem.

 

[subhradeep mahata]

Yes i too think so, as i myself made the problem ! ;-)