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Writing the Equilibrium Expression (mass action equation)

 

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

 

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