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Equilibrium Constant Expressions, and How to Calculate Kc and Kp

SECTION 13 - Chemical Equilibrium

Equilibrium Constant Expressions

For the general reaction:

jA  +  kB  ⇌  ℓC  +  mD

An expression for chemical equilibrium can be written:

How to Write Equilibrium Constant Expressions

As we can see above, the equilibrium constant (K) is equal to:

K = [C][D]m / [A]j[B]k  , where...

K (or K) is unitless and the [brackets] represent the concentrations at equilibrium.

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How to Calculate the Equilibrium Constant, Kc

Now that we know how to write an equilibrium expression, how exactly do we calculate the equilibrium constant, K ?

First, there's 2 NOTES I want to make.

Then, we'll do a detailed example 😊

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➞ NOTE #1:  If the original reaction above is written in reverse, then we have:

K' = [A]j[B]k / [C][D]m  =  1 / K

Or, in a more visually appealing view...

Manipulating the Equilibrium Constant
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➞ NOTE #2:  If the original reaction at the beginning of this post is multiplied by a coefficient (i.e. 4), then for...

4jA  +  4kB  ⇌  4ℓC  +  4mD

We have:

K'' = [C]4ℓ[D]4m / [A]4j[B]4k  =  (K)4  =  K4

Or, in a more visually appealing view...

Manipulating the Equilibrium Constant Expression

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Equilibrium Constant Calculations

Now, as promised, let's do some equilibrium constant calculation examples...

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ex:  Consider the following reaction and the following equilibrium concentrations:

Equilibrium Expression Example

a. Calculate the equilibrium constant, K.

How to Calculate the Equilibrium Constant

b.  Calculate K for the the reverse reaction.

Calculate Keq Equilibrium Constant

c. Calculate the equilibrium constant for the reaction:

Calculate Keq Manipulations

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So far today, we've seen how the value for K changes as we also change the reaction (reversing it, multiplying it through by 1/4th, etc).

However...

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Equilibrium Positions

➞ There's only one equilibrium constant (K) for a specific reaction at a particular temperature...

But, there's an infinite number of equilibrium positions.

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Equlibrium Position  =  a set of equilibrium concentrations.

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ex:  Consider the reaction below, where 4 separate experiments are run

-- each with a different amount of initial concentrations of reactants and product.

N2(g)  +  3H2(g)  ⇌  2NH3(g)

_________
answer:

➞  as you can see in the image below, all 4 experiments (at the same temperature of course), give rise to one equilibrium constant...

Equilibrium Position Examples

All three sets of equilibrium concentrations, known as equilibrium positions, give rise to one equilibrium constant (K).

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Equilibrium Partial Pressures and Kp

Besides being described via equilibrium concentrations, equilibria involving gases can also be described in terms of Equilibrium Partial Pressures, (instead of concentrations).

WHY ?

Because we can manipulate the Ideal Gas Law...

➞ Concentration (C) is moles / L , right?

➞ And so is "n / V" in the Ideal Gas Law (moles / L).

See here:

Kp and Kc Relationship

Both R (the ideal gas constant) and T (temperature) are constant, so for the reaction,

N2(g)  +  3H2(g)  ⇌  2NH3(g)

We can represent equilibrium with either K (~ Kc) or Kp ...

Kp vs. Kc Relationships

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ex:  Calculate the value of Kp for the following reaction at 25°C given these equilibrium partial pressures:

PNOCl  =  1.2 atm
PNO  =  0.05 atm
PCl2  =  0.30 atm

2NO(g)  +  Cl2(g)   ⇌   2NOCl(g)

How to Calculate the Equilibrium Partial Pressures

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Kc and Kp

The relationship between Kc and Kp is shown below:

R  =  universal gas constant (0.0821 L.atm / K.mol),
T  =  Kelvin temperature, and
Δn  =  the change in moles of gas particles (products − reactants).

Kp  =  Kc(RT)Δn

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Stick around for the 3rd set of video notes on SECTION 13 - Chemical Equilibrium where we'll discuss, 

Heterogeneous Equilibria.