Question

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 help

Answer 1

Answer: Expression for equilibrium constant kp is
`K_p=(p_B^2)/(p_A)` and equilibrium concentration of B is 0.141 atm.

Explanation: Equilibrium expression is written as:

`K_p=(products)/(reactants)`

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:

`K_p=(p_B^2)/(p_A)`

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:

`0.001=((2x)^2)/(20-x)`

`0.001=(4x^2)/(20-x)`

`4x^2=0.02-0.001x`

`4x^2+0.001x-0.02=0`

This is a quadratic equation. On solving this equation:

`x=0.0706`

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