General Chemistry/Reaction Rates
Introduction
Reaction rates of a chemical system provide the underpinnings of many theories in thermodynamics and chemical equilibria.
Elementary reactions are onestep processes in which the reactants become the products without any intermediate steps. The reactions are unimolecular (A → products) or bimolecular (A + B → products). Very rarely, they could be trimolecular (A + B + C → products), but this is not common due to the rarity of three molecules colliding at the same time.
A complex reaction is made up of several elementary reactions, with the products of one reaction becoming the reactants of the next until the overall reaction is complete.
Rate Equation
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle m\hbox{A} + n\hbox{B} \rightarrow p\hbox{C} + q\hbox{D}}  Consider an arbitrary chemical reaction. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r = k[\hbox{A}]^m[\hbox{B}]^n}  The rate at which the products will form from the reactants is given by this rate equation. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r = k[\hbox{C}]^p[\hbox{D}]^q}  The rate of the reverse reaction (which also occurs to a lesser extent) has its own rate equation. 
Note [A] is raised to the power of m, its coefficient, just like an equilibrium expression. The rate of the reaction may rely on the molar coefficients of the reactant species, but it might not. However, for an elementary reaction, the concentrations of the species A and B are always raised to their molar coefficients. This only applies to elementary reactions, which is a very important distinction to make.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k} is the rate reaction coefficient, which is reactionspecific. It can be considered a constant, although it does change with temperature (and possible other factors). 
Order
The order of an equation is what the concentration of a substance is raised to in the rate equation. The greater the number, the greater effect it will have on rate. For example, zero order equations do not effect the rate. To find the order, you must alter one concentration and keep the rest the same, Dividing gives an equation which can be used to solve for the order. To find overall order, simply add all orders together.
ZeroOrder Equations
Zeroorder equations do not depend on the concentrations of the reactants.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r = k \,}  There is only a rate coefficient with no concentrations. The rate probably depends on temperature, and possibly other factors like surface area, sunlight intensity, or anything else except for concentration. These reactions usually occur when a substance is reacting with some sort of catalyst or solid surface. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle [\hbox{A}] = [\hbox{A}]_0  kt \,}  The integrated rate law tells us how much reactant will remain after a given amount of time. Integrated rate laws can be found using calculus, but that isn't necessary. In this zeroorder integrated rate law, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k} is the rate coefficient from the rate equation, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t} is time, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle [\hbox{A}]_0} is the starting concentration. 
If you make a graph of concentration vs. time, you will see a straight line. The slope of that line is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k} . This is how you can identify a zeroorder rate. 
FirstOrder Equations
Firstorder equations depend on the concentration of a unimolecular reaction.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r = k [\hbox{A}] \,}  There is a rate coefficient multiplied by the concentration of the reactant. As with a zeroorder equation, the coefficient can be though of as a constant, but it actually varies by the other factors like temperature. There can be other reactants present in the reaction, but their concentrations do not effect the rate. Firstorder equations are often seen in decomposition reactions. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \ln [\hbox{A}] = kt + \ln [\hbox{A}]_0 \,}  This is the integrated rate law. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t_{1/2} = \frac{\ln 2}{k} \,}  The halflife of a reaction is the amount of time it takes for one half of the reactants to become products. One halflife is 50% completion, two halflives would lead to 75% completion, three halflives 88%, and so on. The reaction never quite reaches 100%, but it does come close enough. To find the halflife, you can algebraically manipulate the integrated rate law. 
If you make a graph of the logarithm of concentration vs. time, you will see a straight line with a slope of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k} . This is how you can identify a firstorder rate. 
SecondOrder Equations
Secondorder equations depend on the two concentrations of a bimolecular reaction.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r = k [\hbox{A}] [\hbox{B}] \,}  This is the rate law for a secondorder equation. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r = k [\hbox{A}]^2 \,}  If there are two molecules of the same kind reacting together, the rate law can be simplified. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \frac{1}{[\hbox{A}]} = \frac{1}{[\hbox{A}]_0} + kt \,}  In that case, this is the integrated rate law. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle t_{1/2} = \frac{1}{k[\hbox{A}]_0} \,}  This is the halflife for a secondorder reaction (with only one reactant). 
To see a graph with a straight line of slope Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k} , graph the reciprocal of concentration vs. time. 
Equilibrium
Equilibrium will occur when the forward and reverse rates are equal. As you may have already noticed, the equilibrium expression of a reaction is equal to the rate equations divided.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 2 \hbox{NO}_2 \to \hbox{N}_2\hbox{O}_4}  Consider this reaction, the dimerization of nitrogen dioxide into dinitrogen tetraoxide. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle r_f = k_f[\hbox{NO}_2]^2 \,}

The forward reaction rate is secondorder, and the reverse reaction rate is firstorder. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle k_f[\hbox{NO}_2]^2 = k_r[\hbox{N}_2\hbox{O}_4] \,}  The rate coefficients may be different for the two reactions. If the reaction is in equilibrium, the forward and reverse rates must be equal. 
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle K_{eq} = \frac{k_f}{k_r} = \frac{[\hbox{N}_2\hbox{O}_4]}{[\hbox{NO}_2]^2}}  Rearranging the equation gives the equilibrium expression. 
Understanding kinetics explains various concepts of equilibrium. Now it should make sense why increasing the reactant concentration will make more products. The forward rate increases, which uses up reactants, which decreases the forward rate. At the same time, products are made, which increases the reverse reaction, until both reaction rates are equal again.
Arrhenius Equation
The Arrhenius equation determines a rate coefficient based on temperature and activation energy. It is surprisingly accurate and very useful. The Arrhenius equation is:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle E_a} is the activation energy for the reaction, in joules per mole. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle R} is the Universal Gas Constant, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle T} is the temperature (in kelvin), and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle A} is the prefactor. Prefactors are usually determined experimentally.