# Why can we use stoichiometric coefficients in the equilibrium-constant expression? [duplicate]

I just studied the chapter on chemical kinetics on Coursera, wherein I was repeatedly admonished not to use the stoichiometric coefficients in the rate law formula. I was told that this formula has to be experimentally determined instead. The only exceptions are elementary reactions where one can indeed use the stoichiometric coefficients in the rate law formula.

The equilibrium constant is $$\displaystyle K_c=\frac{k_f}{k_r}$$ However, for some reason, one can use the stoichiometric coefficients in the equilibrium constant expression regardless of the reaction mechanism. Why is this?

• For the reaction $\ce{A=B}$ , write down $dA/dt$ then set this to zero at equilibrium and solve for $K_e=[B]/[A]$ Commented Jul 12, 2023 at 12:34
• Typically you write that the rate of change of species $j$ is $\mathrm{d}C_j/\mathrm{d}t = \sum_k \nu_{j,k} r_k$, where $\nu_{j,k}$ is the stoichiometric coefficient of species $j$ in reaction $k$. This formalism is independent of how you write $r_k$, it can be thermodynamically reversible, a catalyst-type rate law, etc. Only if it is thermodynamically reversible, then you recuperate that $K_\mathrm{eq,k} = \prod_j a_j^{\nu_{j,k}}$ Commented Jul 12, 2023 at 13:20
• @porphyrin If it's a first-order reaction, $-dA/dt=k[A]$. So if, at equilibrium, $dA/dt=0$, would it mean that $[A]=0$? I'm getting confused. Commented Jul 12, 2023 at 13:54
• To have equilibrium there must be a reverse reaction so $d[A]/dt=-k_f[A]+k_r[B]$. Commented Jul 12, 2023 at 14:09
• When molecules react you have to assume a scheme then test it against experiment. If it is simple, such as I gave, then you get that result, if not then you have to work it through. Commented Jul 12, 2023 at 16:24