Consider the following gas phase chemical reaction:
a(g) -- > 2b(g)
write down the expression for the equilibrium constant of this reaction.
if the initial concentration of a is 20 atm pressure, the initial concentration of b is 0 atm and the equilibrium constant kp for the reaction is .001 atm-1, calculate the equilibrium concentration of b.
i know the first part of this would be kc = [a] / [b]2 i need the second part

 Expression for equilibrium constant kp is   and equilibrium concentration of B is 0.141 atm.

Explanation: Equilibrium expression is written as:

Note: for Kp, we use the partial pressures where as for Kc, we use the concentrations.

If we look at the given reaction then, the reactant is A and the product is B. Coefficient of B is 2 so we will do the square of B and the equilibrium expression will be:

small p stands for partial pressures.

If change in pressure is x then the equilibrium pressure of A will be (20-x) atm and the equilibrium pressure of B will be 2x atm.

Let's plug in the values in the equilibrium expression:

This is a quadratic equation. On solving this equation:

So, equilibrium pressure of B = 2(0.0706) atm   = 0.141 atm

u i forgot the answer?