# What does it mean for chemical reaction network to be in equilibrium?

What does being in "equilibrium" mean for a set of chemical reactions? If we have some number $$n$$ of $$\ce{A}$$ molecules and $$m$$ of $$\ce{B},$$ and the following reaction where $$\ce{A}$$ and $$\ce{B}$$ can reversibly bind:

\begin{align} \ce{A + B &->[k_\mathrm{f}] AB} \\ \ce{AB &->[k_\mathrm{r}] A + B} \end{align}

Suppose $$k_\mathrm{f} = 2,$$ $$k_\mathrm{r} = 1.$$ Can this system be "in equilibrium" and what would it mean? Does it mean all reactions have to be reversible and that as $$t \to \infty$$ go on at the same rate? In that case it seems that this system cannot be in equilibrium, since $$k_\mathrm{f} \neq k_\mathrm{r}.$$

• k$_f$ and k$_r$ are rate constants. You have to multiply them by concentrations to get rates. So the rates can be equal at equilibrium even if the rate constants are not. – Karsten Theis Jun 15 '19 at 14:17
• and so at equilibrium $\ce{ k_r [AB]=k_f[A][B] }$ making $\ce{K_e=k_f/k_r=[AB]/[A][B]}$ which means that species A, B and AB all exist. – porphyrin Jun 15 '19 at 16:06
• @porphyrin is it correct to say that any set of such chemical reactions will always have at least one equilibrium? if not what is an intuitive example of one without it, if we assume conservation of mass? – bal Jun 15 '19 at 19:52
• No it is not a rule, reactions do not need to have an equilibrium step, $A \to B$ or $A\to B \to C$ type reactions are common. – porphyrin Jun 16 '19 at 8:27