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Writing the Equilibrium Expression (mass action equation)

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 (K_eq) 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. K_eq is what we want to find, so that's our "x."

Here is what we get:

K_eq= 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

K_C = 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 K_c for the reaction PCl_3(g) + Cl_2(g) <--> PCl_5(g)

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

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