# General Chemistry/Reaction Rates

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## Introduction

Reaction rates of a chemical system provide the underpinnings of many theories in thermodynamics and chemical equilibria.
Elementary reactions are one-step 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

| $m{\hbox{A}}+n{\hbox{B}}\rightarrow p{\hbox{C}}+q{\hbox{D}}$ | Consider an arbitrary chemical reaction. |
| $r=k[{\hbox{A}}]^{m}[{\hbox{B}}]^{n}$ | The rate at which the products will form from the reactants is given by this rate equation. |
| $r=k[{\hbox{C}}]^{p}[{\hbox{D}}]^{q}$ | 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.

### 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.

### Zero-Order Equations

Zero-order equations do not depend on the concentrations of the reactants.
| $r=k\,$ | 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. |
| $[{\hbox{A}}]=[{\hbox{A}}]_{0}-kt\,$ | 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 zero-order integrated rate law, $k$ is the rate coefficient from the rate equation, $t$ is time, and $[{\hbox{A}}]_{0}$ is the starting concentration. |

### First-Order Equations

First-order equations depend on the concentration of a unimolecular reaction.
| $r=k[{\hbox{A}}]\,$ | There is a rate coefficient multiplied by the concentration of the reactant. As with a zero-order 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. First-order equations are often seen in decomposition reactions. |
| $\ln[{\hbox{A}}]=-kt+\ln[{\hbox{A}}]_{0}\,$ | This is the integrated rate law. |
| $t_{1/2}={\frac {\ln 2}{k}}\,$ | The half-life of a reaction is the amount of time it takes for one half of the reactants to become products. One half-life is 50% completion, two half-lives would lead to 75% completion, three half-lives 88%, and so on. The reaction never quite reaches 100%, but it does come close enough. To find the half-life, you can algebraically manipulate the integrated rate law. |

### Second-Order Equations

Second-order equations depend on the two concentrations of a bimolecular reaction.
| $r=k[{\hbox{A}}][{\hbox{B}}]\,$ | This is the rate law for a second-order equation. |
| $r=k[{\hbox{A}}]^{2}\,$ | If there are two molecules of the same kind reacting together, the rate law can be simplified. |
| ${\frac {1}{[{\hbox{A}}]}}={\frac {1}{[{\hbox{A}}]_{0}}}+kt\,$ | In that case, this is the integrated rate law. |
| $t_{1/2}={\frac {1}{k[{\hbox{A}}]_{0}}}\,$ | This is the half-life for a second-order reaction (with only one reactant). |

## 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.
| $2{\hbox{NO}}_{2}\to {\hbox{N}}_{2}{\hbox{O}}_{4}$ | Consider this reaction, the dimerization of nitrogen dioxide into dinitrogen tetraoxide. |
| $r_{f}=k_{f}[{\hbox{NO}}_{2}]^{2}\,$ $r_{r}=k_{r}[{\hbox{N}}_{2}{\hbox{O}}_{4}]\,$ | The forward reaction rate is second-order, and the reverse reaction rate is first-order. |
| $k_{f}[{\hbox{NO}}_{2}]^{2}=k_{r}[{\hbox{N}}_{2}{\hbox{O}}_{4}]\,$ | The rate coefficients may be different for the two reactions. If the reaction is in equilibrium, the forward and reverse rates must be equal. |
| $K_{eq}={\frac {k_{f}}{k_{r}}}={\frac {[{\hbox{N}}_{2}{\hbox{O}}_{4}]}{[{\hbox{NO}}_{2}]^{2}}}$ | 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:
$k=Ae^{-E_{a}/RT}$
$E_{a}$ is the activation energy for the reaction, in joules per mole. $R$ is the Universal Gas Constant, $T$ is the temperature (in kelvin), and $A$ is the prefactor. Prefactors are usually determined experimentally.