1. Chemical Equilibrium
Chapter 14.2 & 14.3

2. Dynamic Equilibrium
Is the point at which the forward and reverse rates of a reaction are equal to each other
Is represented in equations by a double arrow in a reaction
Ex. 2H2+O2 ⇌ 2H2O
In addition to equal rates concentrations become stable on both sides of the equation
Although rates are equal concentrations at equilibrium are not equal

3. How to look at a Reaction Considering Dynamic Equilibrium
Beginning
Only reactant molecules exist so only reactant molecules collide, there is only the forward reaction happening
Middle
Concentration of the product(s) is increasing so collisions increase causing reverse reaction
Equilibrium
Collisions are happening at the same rate on both sides

4. What Does This Look Like Graphically?

5. Equilibrium Constant (K)
The equilibrium constant (K) is the way to quantify the concentrations at equilibrium and how far the reaction proceeds after equilibrium
K>>1 means the forward reaction is favored and there are high concentration for products and low concentration for reactants, forward reaction proceeds fully
K<<1 means the reverse reaction is favored and there are low concentrations for products and high concentration for reactants, forward reaction doesn’t proceed far
K≈1 means neither is favored and the reaction proceeds about half way

6. How to find K To find K we use the Law of Mass Action
Using the base equation aA+bB⇌cC+dD where lowercase represent coefficients we divide the concentration of the products by that of the reactants while all values are raised to their respective coefficients as following
Note that this formula is in your reference tables at the top of the section labeled Equilibrium

7. Changes to the Equilibrium Constant
When the base equation you are working with needs to be modified in some way the equilibrium constant must be change to account for the modifications
There are three common modifications
Reversing the equation
Adjusting coefficients
Adding multiple equations together

8. Adjusting the Constant
When reversing the equation invert the equilibrium constant
When adjusting coefficients (multiplying the coefficients by a factor) raise the equilibrium constant to the same factor
When combining equations find the individual constants then multiply them together to obtain an overall equilibrium constant for the combined equation

9. What the college board says

10. Find the Equilibrium constant for the following equation
2H2S(g)⇌2H2(g) + S2(g)
K=[H2]² [S2] / [H2S]²

11. Consider the reaction A(g)⇌ B(g) where K=10
If initially [A]=1.1M and [B]=0.0M which of the following would be the result at equilibrium?
[A]=1.0M and [B]=0.1M
2. [A]=0.1M and [B]=1.0M
3. Concentrations will be equal

12. Consider the following reaction and solve for K
2A + B ⇌ C + 2D Where the concentrations at equilibrium are M , 0.600M, 0.200M and 0.400M respectively.
[0.200][0.400]²/[0.100]²[0.600] = K
K= 5.33
And if we reverse the reaction?
[0.100]²[0.600]/[0.200][0.400]²
K=.1875