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 The Equilibrium Constant and the  Mass Action Expression

What does K imply?

Large K > 1 products are "favored", the equilibrium lies to the right

K = 1 neither reactants nor products are favored

Small K < 1 reactants are "favored", the equilibrium lies to the left

How do we calculate K when given concentrations?

Calculate the equilibrium constant (Keq) for the following reaction:

H2 + I2 <==> 2 HI

[H2] = 0.0505 M
[I2] = 0.0498 M
[HI] = 0.389 M

The first thing to do is write the equilibrium expression for the reaction as written in the problem. This is what to write:

Now, all you have to do is substitute numbers into the problem. Keq is what we want to find, so that's our "x."

Here is what we get:

 Keq = (0.389)2 ____________________ (0.0505) (0.0498)

Keq= 60.2

Solving this and rounding to the correct number of sig figs (remember those??), we get 60.2

How do we calculate K when given Moles and Liters?

What is the equilibrium constant for the gaseous reaction

2 N2 + O2 <--> 2 N2O

if, at equilibrium, 0.00670 mole N2, 0.000928 mol O2 and 1.090 mol N2O are in a 2.00-L container.

Solution:

The mass action expression for this reaction is

KC

=

 [N2O]2 [N2]2[O2]

Calculate Molarity for each species

nN2

=

 0.00670 mole N2 =0.00335M 2.00L

nO2

=

 0.000928 mol O2 =0.000464M 2.00L

nN2O

=

 1.090 mol N2O =0.545M 2.00L

Using the given values in the equation,

KC

=

 (0.545)2 (0.00335)2(0.000464)

KC

=

 0.297025 (5.20724) x 10–9

KC = 5.7 x 107

****note no units for K

Another Calculation

At a certain temperature, the equilibrium mixture of PCl5 , PCl3 , and Cl2 has the following concentrations:

[PCl3] = 0.035 M, [PCl5] = 0.017 M, [Cl2] = 0.074 M

Calculate Kc for the reaction PCl3(g) + Cl2(g) <--> PCl5(g)

 ANSWER Keq=[PCl5] /[PCl3][Cl2]=[0.017 M]/[0.035 M][0.074 M]=6.56