S13E2 - Equilibrium Constant Expressions, and How to Calculate Kc and Kp

Equilibrium Constant Expressions

For the general reaction:

jA  +  kB  ⇌  ℓC  +  mD

An expression for chemical equilibrium can be written:

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

K = [C]^ℓ[D]^m / [A]^j[B]^k  , where...

K (or K_c ) 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]^ℓ[D]^m  =  1 / K

Or, in a more visually appealing view...

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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...

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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:

a. Calculate the equilibrium constant, K.

b.  Calculate K for the the reverse reaction.

c. Calculate the equilibrium constant for the reaction:

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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 1/4^th, 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...

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 K_p

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:

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 (~ K_c) or K_p ...

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

P_NOCl  =  1.2 atm
P_NO  =  0.05 atm
P_Cl2  =  0.30 atm

2NO(g)  +  Cl₂(g)   ⇌   2NOCl(g)

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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).

K_p  =  K_c(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.