Equilibrium constants

Question

Consider the reaction below.  1.44 mol of A and 8.20 mol of B are added to a 6.0 L container.  At equilibrium, the concentration of A is 0.178 M.  What is the value of the equilibrium constant?  

 3 A + 4 B   2 C

I know the equilibrium constant is the fraction of concentrations in the reaction but how do i find c's concentration? 

Answer 1

You need to set up an ice table, and the concentration of A at equilibrium will allow you to solve for x.

First, find the concentration in moles/L of everything:

initial concentration of A: 1.44 mol/6 L = 0.24 M

initial concentration of B: 8.2 mol/6 L = 1.37 M

equilibrium concentration of A: 0.178/6 = 0.03 M

3A + 4B <----> 2C

0.24 1.37 0

-3x -4x +2x

0.24-3x 1.37-4x 2x

You know from the concentration of A that 0.24 - 3x = 0.03

We can use that to solve for x:

0.24 - 3x = 0.03

0.24 - 0.03 = 3x

.21 = 3x

x = 0.07

Now, you can find the equilibrium concentrations of everything using the ICE table:

[A] = 0.03 M

[B] = 1.37 - 4x = 1.37 - 4(0.07) = 1.09 M

[C] = 2x = 2(0.07) = 0.14

Write an expression for Keq:

Keq = [C]²/ ([A]³*[B]⁴)

Keq = 0.14²/(0.03³*1.09⁴)

Keq = 514.26 M^-5

Answer 2