Question

In 50 words or more explain how chemical kinetics and chemical equilibrium are related to one another. Give a real world example of chemical kinetics.

Answer 1

Chemical equilibrium is the state of constant composition attained when opposing reaction rates become equal. There is an essential relationship between reaction rates and chemical equilibrium, one that we can describe quantitatively. At first thought, the connection may seem obscure - do we not need to be far from equilibrium to properly measure reaction rates? The dynamic nature of chemical equilibrium means that both forward and reverse reactions can be taking place at significant - even high - rates, but since these rates are equal, no change in concentration is observed over time.

To further illustrate the relationship between reaction rate and chemical equilibrium, let's consider the reaction coordinate diagram for our simple, reversible reaction.

This diagram, at right, shows a graph (purple line) of chemical potential energy (vertical axis) versus a "reaction coordinate" (horizontal axis), which here represents the progress of a reaction in which one bond must break and a new bond forms. Examples of such reactions are provided by the many isomerization reactions, such as the interconversion of cis-2-butene and trans-2-butene.

Let us first consider the forward reaction, A → B. Reactant molecules (A) in the system with enough energy to reach the energetic peak, the transition state (symbolized by the double dagger, ‡) can continue forward along the reaction coordinate to conversion to product B.

reactants # products Ea, fwd A Ea, rev chemical potential energy change in chemical potential energy for the reaction reaction: A =B B reaction coordinate (progress of reaction)

This minimum energy required is the activation energy. For the forward reaction, this is symbolized as E_{a, fwd}, and is the energetic cost required to advance the bond breaking in a molecule of A to the point where formation of the new bond in B can begin to provide an energetic payback. The increase in potential energy must be provided by the kinetic energy of the molecules in the system, which varies by temperature, according to a Maxwell-Boltzmann distribution. For molecules, the kinetic energy can reside not only in their motion through space, but also in bond stretching, bending, and twisting. Kinetic energy in these forms is constantly being redistributed by collisions between the molecules. We can easily imagine a bond breaking in molecules where an excess of kinetic energy is momentarily transferred into these bond motions. The reverse reaction, B → A, proceeds in a similar manner, but more conversion of kinetic to potential energy at the molecular level is required. Molecules of B in the system with enough energy (the activation energy for the reverse reaction, E_{a, rev}) to reach the transition state can continue backward along the reaction coordinate to conversion back to A. Since the forward reaction activation energy is less than the reverse reaction activation energy (E_{a, fwd} < E_{a, rev} ), the reverse reaction will be slower than the forward reaction when the concentrations of A and B are equal. In order for this reaction to reach equilibrium, the concentration of B will have to increase beyond that of A up to the point where the rates of the forward and reverse reactions become equal. Qualitatively, since E_{a, fwd} < E_{a, rev} in this case, K_eq > 1.

We can conclude that the net change in chemical potential energy is related to the equilibrium constant (K_eq) for the reaction, whereas the magnitude of the activation energy must be related to the rate constants for the reaction. Let us specify now that the forward reaction, A → B, and the reverse reaction, B → A, both follow first-order rate laws. Then we can readily derive the relationship between rate constants and the equilibrium constant for our simple first-order case. The rate laws for the forward and reverse reactions in this case are

forward reaction rate = V_{0, fwd} = k₁[A]         

reverse reaction rate = V_{0, rev} = k₋₁[B]

These are just first-order rate laws, with k₁ as the forward first-order rate constant, and k₋₁ as the first-order rate constant for the reverse reaction. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. Setting these expressions equal when equilibrium concentrations are reached, we have

k₁[A]_eq  = k₋₁[B]_eq

which can be rearranged to

k₁/k₋₁  =  [B]_eq/[A]_eq

The right side expression is of course the equilibrium constant for this reaction, so we have shown in this case that

K_eq = k₁/k₋₁  

2)

our body system consists lots of biological reactions. If a biochemical reaction in our body is too fast or too slow, it can endanger our life. There are a lot of studies involve in studying the optimization of the rate of reaction in our body. For example, if a medicine is developed to counter a specific disease, it is crucial to optimize the rate of its effect in our body. If such medicine will release heat during reaction in our body, it is important to make the reaction is slower, therefore ensuring the patient will not experience high fewer during medication.

In other example, rate of reaction is obviously very important to the chemical industry. The rate of reaction dictates the rate of production of our daily products. In order to meet the demand and safety standards, optimization of rate of reaction is nonetheless the most important subject to control and study.

There are some other small things that involve rate of reaction in our life as well, such as cooking, cleaning and so on.