First order reactions - For the irreversible reaction, A -> B, the reaction rate (velocity, V) is given by

V = d[B]/dT

for the rate of appearance of the product, B, or by

V = -d[A]/dT

for the rate of disappearance of the substrate, A. These equations are equally valid for this reaction, so we can say

V = d[B]/dT = -d[A]/dT = k1[A],

where k1 is called the rate constant and for this reaction has units of (seconds)^-1. This type of a reaction is called a first-order reaction, because its rate depends on the first power of the reactant concentration. If k1 is large, the reaction is fast and if k1 is small, the reaction is slow. Integrating the above equation yields

[A]/[A]0 = e^-(k-1t),

where [A]0 is the starting concentration of A when t = 0.

A plot of this equation (Figure 11.1a) shows that the concentration of A decreases exponentially with time. The amount of time it takes for half of A to be lost is called the half-life and is given by t1/2. The half-life is inversely proportional to k1.

A plot of ln[A] vs t, as shown in Figure 11.1b, will always yield a straight line with slope = -k1 if the reaction is first order. If one plots the initial rate of the reaction versus varying starting concentrations ([A]0), a straight line plot with slope of k1 will be produced for a first order reaction. The most common example of a first order reaction is the decay of radioactive elements.

This approach to reactions in biological cells is too simple to explain them properly. One reason is that many of the reactions and processes in cells are reversible-that is, the equilibrium does not lie far to one side and, as product accumulates, the reverse reaction becomes important.

So, for the reaction A < = > B, a forward reaction constant, k1, can be used to define the reaction moving rightward and a reverse reaction constant, k--1 can be used to define the reaction moving leftward.

Now the molecule A is being consumed in the reaction to the right and formed by the reaction to the left, so the corresponding rate equation is

-V = d[A]/dt = -k1[A] + k-1[B]

Here, k1 and k-1 are the rate constants for the first-order forward and reverse reactions. Such a reaction approaches a state of equilbrium, at which point the rates of the forward and reverse reactions become equal. At the same time, the overall rate becomes zero, so

0 = -k1[A]eq + k-1[B]eq

or [B]eq/[A]eq = k1/k-1 = K,

where K is the equilibrium constant. For a reversible reaction that is first-order in both directions, the equilibrium constant is always the ratio of the forward and reverse rate constants.

Second-Order Reactions - A reaction of this type typically occurs when two molecules come together to form products. A simple example is

2A -> A2

with a rate constant given by k2. The rate of such a reaction is proportional to the second power of the concentration of the reactant. Therefore, V = -d[A]/dt = -k2[A]²

Here k2 is the second-order rate constant. It has dimensions of (mol/L)^-1s^-1