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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

Respuesta :

Answer:  Expression for equilibrium constant kp is [tex]K_p=\frac{p_B^2}{p_A}[/tex]   and equilibrium concentration of B is 0.141 atm.

Explanation: Equilibrium expression is written as:

[tex]K_p=\frac{products}{reactants}[/tex]

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:

[tex]K_p=\frac{p_B^2}{p_A}[/tex]

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:

[tex]0.001=\frac{(2x)^2}{20-x}[/tex]

[tex]0.001=\frac{4x^2}{20-x}[/tex]

[tex]4x^2=0.02-0.001x[/tex]

[tex]4x^2+0.001x-0.02=0[/tex]

This is a quadratic equation. On solving this equation:

[tex]x=0.0706[/tex]

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

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