Chemistry

Chemical reactions can be written as

$$ A + B \stackrel{k}{\longrightarrow} C $$

meaning "$A$ plus $B$ yields $C$ with rate constant $k$"

The changes in the amounts of $A,B$ and $C$ can be expressed in change equation (a.k.a differential equations)

The state variables are concentrations: $[A]$ for the concentration of $A$, etc.

But for simplicity, we use $A$ for $[A]$, etc.

Let us see how to do it with a simple equation first

$$ A \stackrel{k}{\longrightarrow} C$$

In this case $A$ is depleting at a rate of $k$ so the change equation should be

$$ A' = -kA$$

1.4.17 What's the change equation for $C$?

Ans Since the A molecules all turns into B molecules, so

$$C' = kA$$

Next let's look at a slightly more complicate equation

$$ A + A \stackrel{k}{\longrightarrow} C \quad \text{or} \quad 2A \stackrel{k}{\longrightarrow} C$$

This time two molecules of $A$ must collide in order to form a molecule of $C$. How often will that happen?

The law of mass action tells us that it should be proportional to $A^2$ (how often an $A$ bumps into another $A$)

Thus, the equation should be

$$ A' = -2kA^2$$

Note the "2" on the right-hand side is there because two molecules of $A$ combine to form a molecule of $C$.

Exercise 1.4.18 What is the change equation for $C$ then?

Ans. $$C' = kA^2$$

Following the same logic, the change equation for $A$ in the reaction

$$ A + B \stackrel{k}{\longrightarrow} C$$

should be

$$ A' = - kAB$$

Exercise 1.4.19 Finish writing the change equations for the system

Ans. $$B' = -kAB$$ and $$C' = kAB$$

Same logic as in the logistic population growth model

Reversible Case and Equilibrium

Equations with reversible reaction can be written as

$$ A + B \stackrel{k_f}{\longrightarrow} C$$ $$ A+B \stackrel{k_b}{\longleftarrow}C$$

So $$A' = -k_fAB + k_bC$$

Exercise 1.4.20 Write the change equation for $B$ and $C$

Ans $$B' = -k_fAB + k_bC$$ and $$C' = k_fAB-k_bC$$

Question What happen when $A = B$? What are the change equations then?

Ans. $$A' = -2k_f A^2 + 2k_bC$$ and $$C' = k_fA^2 - k_bC$$

Equilibrium, by definition, is the state when there is no net change in the concentrations.

In other words, the state of equilibrium is characterized by

$$X' = 0$$

for every molecule concentration (state variable) $X$.

By setting $A' = 0$, we get

$$ 0 = -k_fAB + k_bC.$$

That is $$ \frac{k_f}{k_b} = \frac{C}{AB}$$

This means at equilibrium, the final concentrations in a chemical reaction stand in a certain ratio.

Remark Note that every equation $X'=0$ should give the same ratio. This is useful in detecting any mistake in the original system of change equations