1. Reaction Rates and Equilibrium

2. Reaction Rates Indicate how quick the reaction takes place
Unit is moles per liter in a second (M/s)
Zero-order reactions:
A zero-order reaction has a constant rate, which is independent of the reactants’ concentrations.
y = x0
First-order reactions:
A first-order reaction has a rate which is proportional to the concentration of a reactant. Thus if a reactant’s concentration is doubled, then the rate of the reaction is also doubled.
y = x1
Second-order reactions:
A second-order reaction has a rate proportional to the product of the concentration of two reactants or the square of one reactant
y= x2

3. Factors Affecting Reaction Rate
The rate of a chemical reaction can depend heavily on the environment of where the reaction is taking place.
The rate of a reaction will increase if there is an increase in the collisions of the reactants
1. Reactant Concentrations - The greater the concentrations of the reactants (the more particles per volume), the greater the chance will be for collisions of the reactants. Therefore the reaction rate will increase for all but zero order reactions.
2. Pressure (gasses only) - Increasing the pressure of a gas is exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume. If you have the same mass in a smaller volume, then its concentration is higher. Changing the pressure on a reaction which involves only solids or liquids has no effect on the rate.
3. Temperature – For nearly all reactions, the reaction rate will increase as the temperature of a system/reaction increases. Since the temperature is a measure of the particles’ average kinetic energy, increasing the temperature increases how fast the molecules are moving around and providing enough energy to form products.
4. Catalysts – Catalysts are substances that increase reaction rates without being used up or consumed. They help the reactants collide in the right way, stabilize the transition states and/or the intermediates, and generally makes it easier to form the products. This increases the overall reaction rate if the rate-determining step is the formation and breaking of bonds. This is especially important in biology and biological reactions.

4. Equilibrium
So far, reaction rates have been discussed under the assumption that the reactions were irreversible (only goes from reactants to products).
A reversible reaction is one where products can react/breakdown to form the reactants.
When there is no net or overall change in the concentrations of the products and reactants after reacting for a while, it is said to be in equilibrium.
During equilibrium, products are still being formed and broken down but at an equal rate.
Equilibrium can be thought of as a balance between two reaction directions.
Consider the following: A <==> B
At equilibrium, the concentrations of A and B stay the same, but the concentration of A isn’t equal to the concentration of B
Only the rate is equal (speed, not concentration!)

5. Equilibrium Constant Keq
Keq is an expression used to describe the ratio between the concentration of products and reactants in a chemical reaction
The only condition that affects the ratio (Keq) is temperature
Since Keq is the ratio between the concentrationof products vs. reactants, pure solids and liquids are not included in the expression

6. Finding Keq When a reaction is in the form: aA + bB  cC + dD
Keq can be calculated as follows:
Example
Write the Keq expression for the following reaction and use the concentrations to find its value:
O2(g) + 2H2(g) 2H2O(l)
the concentrations are as follows:
[O2] = 3.45 M, [H2] = 5.67 M

7. Interpreting Keq
Since the expression is concentration of products OVER reactants:
A Keq value greater than 1 means the concentration of products is greater than the concentration of reactants
A small Keq value means the concentration of products is less than the concentration of reactants
A Keq value close to one means the concentration of products is equal to the concentration of reactants

8. Interpreting Keq
Example: Ag+(aq) + Cl-(aq) → AgCl(s)  This reaction at equilibrium has a Keq of 6x109 which heavily favors the product. In this case, you would expect almost all solid silver chloride precipitate and almost no silver or chloride ions in solution.