**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}= K^{4}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.

**N _{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(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}

_{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,

**N _{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(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) + Cl _{2}(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,