# 4.8: Rates and Concentrations

- Page ID
- 35141

\(rate = k\) (a horizontal line)

rate| / | /rate = k[A] | / |- -/- - - - -rate = k| / | / |/_________________ 0 1 2 3 4 [A]

For a first order reaction, the plot is a straight line (linear), as shown above, because

\(rate = k \ce{[A]}\) (a straight line)

Note that \(rate = k\) when \(\mathrm{[A] = 1}\).

For a second order reaction, the plot is a branch of a parabola, because

\(rate = k \ce{[A]}^2\)

rate| . |rate = k[A]^{2}| . (a branch of | a parabola) | . | - . - - - - - - | . |._________________ 0 1 2 3 4 [A]

For a reaction with an infinite order, the plot is a step function. The *rate* is small, almost zero, when \(\ce{[A]}\) is less than 1. When \(\ce{[A]}\) is greater than or equal to 1, then the reaction rate is very large. This model applies to nuclear explosion, except that \(\mathrm{[A] = 1}\) is actually *the critical mass* of the fission material.

\(rate = k \ce{[A]}^{\infty}\)

rate| (order = infinity) | |rate = k[A]^{00}| | (a vertical line) | | | | | | | | |...|_________________ 0 1 2 3 4 [A]

Is there a chemical process like this? Well, we all know that one of the key conditions in an atomic bomb is to have a critical mass of the fission material, \(\ce{^235U}\) or \(\ce{^239Pu}\). When such a mass is put together, the reaction rate increases dramatically, leading to an explosion. Thus, this model seems to apply; however, the mechanism for the fission reaction is not described by the order of the fission material.

## Contributors and Attributions

Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)