# Dynamic Equilibrium

### Examples of Equilibria

- If water is in a closed container, after a while, water vapor molecules condense to water at the same rate as they vaporize.
- When a solute is added to a solvent, the system reaches a point where the rate of dissolution matches the rate at which the solute crystallizes.

- Those examples above are describing a quantity known as an
*equilibrium constant.*

# Equilibrium Constant

- A reaction can either be one-way or two-way, called a reversible reaction
- Eg: aA + bB -> cC + dD (a moles of A, b moles of B -> ...)

- When a reaction is reversible, some molecules could be going from (aA + bB -> cC + dD), and others could be going from (cC + dD -> aA + bB)
- Later, there would be less and less of the left hand side, and the other reaction would begin to occur (cC + dD -> aA + bB)
- Eventually, these reactions would be happening at the same rate, called the equilibrium.

- At the equilibrium condition, the amounts of the reactants and products become constant
- Regardless of what quantities of each molecule you start with, there is a ratio that describes every reversible reaction, called the
**equilibrium constant expression.**

**Equilibrium Constant Expression:**

**Equilibrium Constant Expression:**

## [C]^{c}[D]^{d}/([A]^{a}[B]^{b})

[C] denotes the molarity of C, expressed in moles/liter, raised to their respective coefficients. The result of the expression is called the equilibrium constant. So, it's the molarity of the products to the coefficients divided by the reactant molarities to their coefficients.

## Relationships involving Equilibrium Constants

**Relationships of K to the Balanced Chemical Equation**

- When we reverse an equation, we invert the value of K.
- When we multiple the coefficients in a balanced equation by a common factor (2,3, ...), we raise the equilibrium constant to the
*corresponding power*(2, 3, ...) - When one
*divides*by a common factor (2,3, ...) the*corresponding root*of the equilibrium constant is taken.

### Example:

Find the equilbrium constant of the following chemical equation:

CH_{3}OH(g) = CO(g) + 2 H_{2}(g)

The original is:

CO(g) + 2 H_{2}(g) = CH_{3}OH(g), K_{c} = 14.5

So, the equilibrium value is just 1/K. So, 1/14.5 = 0.0690

## Combining Equilibrium Constant Expressions

**RULE: **When two equations are combined, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction

## Example

Suppose we want to find the equilibrium constant for this reaction:

N_{2}O(g) + 1/2 O_{2}(g) = 2 NO (g)

However, we only know the K values of these two equilibria

N_{2}(g) + 1/2 O_{2}(g) = N_{2}O (g) K = 2.7 * 10 ^ -18

N_{2}(g) + O_{2}(g) = 2 NO (g) **k** = 4.7 * 10^-31

So, we should get the first equation by adding a form of the other two. So, we can flip the first KNOWN equation around, making it N2O = N2(g) + 1/2 O2(g). That makes the equilibrium constant 1/(2.7e-18), making it 3.7e17.

The second equation stays the same, being 4.7e-31.

So, we take the product of those two, to get the overall K value, 4.7e-31 * 3.7e17 = 1.7e-13, which is our final answer.

## Equilibria Involving Gases: The Equilibrium Constants, Kc and Kp

Mixtures of gases are as much solutions as are mixtures in a liquid solvent. They could be written as moles/liter, but it is common to express them as their partial pressures.

The formula is the same but instead of the molarity, it's the partial pressure.

K_{p} = K_{c}(RT)^(change in n)

n = sum of the coefficients of the products - sum of the coefficients of the reactants

## Equilibria Involving Pure Liquids and Solids

When reactions in a gas or in an aqueous solution occur, they are *homogeneous *reactions: They occur within a single phase. When there are reactions which include solids, liquids, and gases, those are called *heterogeneous *reactions. One of the most important ideas about heterogeneous reactions are that:

Equilibrium constant expressionsdo notcontain concentration terms for solid or liquid phases of a single component (that is, for pure solids and liquids).

So, to summarize, the pure solids and liquids are ignored when calculating the equilibrium constant.

For example, give this equation for decomposing of limestone.

CaCO_{3}(s) = CaO(s) + CO_{2}(g). So, you ignore the solids, and K_{c} = [CO_{2}]

Now, to calculate K_{p} = K_{c}(RT)^n, and you calculate n by ignoring the coefficients of the solids too.

## The Reaction Quotient, Q: Predicting the Direction of Net Change

Let's say we have the reaction: CO(g) + 2H_{2}(g) = CH_{3}OH(g), and we start with only CO and H_{2}, so obviously the equation is a forward reaction, meaning CH_{3}OH is produced. Also, let's consider the scenario when we only have CH_{3}OH, obviously 2H_{2} and CO(g) would be produced, a reverse reaction.

Now, let's consider a case where all of the three molecules are present, which way does it go? Predicting the direction of net change is very useful because:

- Sometimes, you need to know the direction of net change as a first step
- At times we just want a qualitative description of the changes as opposed to a quantitative.

The ratio of **activities**, or molarity/1 M with the same form as the equilibrium constant is called the **reaction quotient **and is designated

**Q.**

## [C]^{c}_{init}[D]^{d}_{init}/([A]^{a}_{init}[B]^{b}_{init})

So, that's the ratio at the current time, so if it equals the equilibrium constant, it's at the equilibrium. If it's less than the equilibrium constant, it's a forward reaction, or occurs to the right. If it's greater than the equilibrium constant, it's a reverse reaction, or occurs to the left.

### Example:

Let's say that you have CO, H2, and CH3OH all at 0.100 M, which direction does the equation go in?

Qc = (0.100)/((0.100)(0.100)^2) = 100. (recall that the equation is CO + 2H2 = CH3OH)

Since, Qc > Kc which is 14.5, the reaction is a reverse reaction.