For this gas phase equilibrium shown below, 6 mol of C is placed in a 2.5 L flask and allowed to equilibrate. If the equilibrium constant for this reaction is 2.50, what are the final concentrations for all the three gases?
A + B <-----> C
We place 6 mol of C in a 2.5 L flask. So its molarity is:
molarity of C = 6 mol / 2.5 L
molarity of C = 2.4 M
The equilibrium constant for that reaction can be expressed like this:
Keq = [C] / ( [A] * [B] )
To solve the problem we have to set up an ICE table:
A + B <-----> C
I 0 0 2.4 M
C x x - x
E x x 2.4 M - x
Replacing those values in the Keq (Keq = 2.50):
Keq = [C] / ( [A] * [B] )
2.50 = (2.4 - x) / (x * x )
2.50 = (2.4 - x) / (x²)
And we have to solve that equation for X:
2.50 * x² = 2.4 - x
2.50 *x² + x - 2.4 = 0
This equation has 2 roots:
x₁ = -1.2 and x₂= 0.8
We can't have a negative concentration, so -1.2 is not our answer, then we use 0.8 M.
From the ICE table we said that the equilbrium concentrations are:
[A] = x = 0.8 M [B] = x = 0.8 M [C] = 2.4 M - x = 2.4 M - 0.8 M = 1.6 M
Answer: the final concentrations are [A] = 0.8 M [B] = 0.8 M [C] = 1.6 M