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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?

Shoes
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    For the reaction $\ce{A=B}$ , write down $dA/dt$ then set this to zero at equilibrium and solve for $K_e=[B]/[A]$ – porphyrin Jul 12 '23 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 $K_\mathrm{eq,k} = \prod_j a_j^{\nu_{j,k}}$ – Metal Storm Jul 12 '23 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. – Shoes Jul 12 '23 at 13:54
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    To have equilibrium there must be a reverse reaction so $d[A]/dt=-k_f[A]+k_r[B]$. – porphyrin Jul 12 '23 at 14:09
  • So $-k_f[A]+k_r[B]=0\implies k_f[A]=k_r[B]\implies k_f/k_r=[B]/[A]=K_c$ But my question remains: How did we assume that the forward and reverse reactions are elementary so that the powers in the rate law (1 in this case) are the coefficients of the balanced chemical equation? – Shoes Jul 12 '23 at 15:34
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    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. – porphyrin Jul 12 '23 at 16:24

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