Keq, Equilibrium Calculations and Given Initial Conditions

Question

A solution initially contains 0.350 mol/L of A and 0.750 mol/L of B. A reaction occurs according to the equation 

2 A(aq) + B(aq) --> 3 C(aq) + D(s). 

 

At equilibrium, [C] is found to be 0.100 mol/L. What is the value of Keq? 

Answer 1

I can help you get set up for this problem, but I will let you solve for Keq on your own.

So for a given reaction:

aA_(aq) + bB_(aq) -> cC_(aq) + dD_(aq) (lowercase letters are coefficients)

The equilibrium constant, K_eq of the reaction will be as follows:

K_eq = [C]^c [D]^d / [A]^a [B]^b

(where the lowercase letters are the coefficients from the balanced equation, and the upper case letters in brackets indicate the molar concentration of that reactant/ product)

An easy way to remember the formula formula is: K_eq = [products] / [reactants]

For your problem, you would write the equation as follows:

K_eq = [C]³ / [A]² [B]

(the reason why you do not include D in the equilibrium expression is because product D is a solid. The concentration of the solid remains constant in aqueous solution so it is negligible. You also would not include any products or reactants that are liquids)

We know the initial concentrations of A and B, and we know the equilibrium concentration of C. For this next part you need to construct an ICE (initial, change, equilibrium) table to determine how the concentrations actually change between the initial and equilibrium states, (I apologize if this table is weirdly formatted):

[A] [B] [C]

I 0.350 M 0.750 M 0

C -2x -x +3x

E 0.350 M - 2x 0.750 M - x 0.100 + 3x

Use the information from the equilibrium state to calculate Keq:

Keq = (0.100 +3x) / (0.350 - 2x) (0.750 - x)

In order to solve this, you need to figure out what x is.

This can be done using the data from C:

3x = 0.1000

x = 0.033

Substitute this x value into the equation and then solve for Keq.