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Consider the reaction

$$\ce{A + B + C -> D + E}$$ and $\mathrm{rate} = k[\ce{A}][\ce{B}]^2.$ My book said if we double the amount of $\ce{A},$ then the reaction rate will double.

I understand this, but what about the law of definite proportion?

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    $\begingroup$ "…but what about the law of definite proportion?" — what about it? Could you please elaborate more? $\endgroup$ – andselisk May 15 at 10:32
  • $\begingroup$ we cant simply put more of a substance to get more of the product, the reactants must have a fixed ratio. $\endgroup$ – 12345bird May 15 at 10:43
  • $\begingroup$ Ehhm, and what is has to do with the law of definite proportions? $\endgroup$ – andselisk May 15 at 10:48
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    $\begingroup$ Either you interpreted your book incorrectly or your book in wrong. If you double the concentration of $\ce{A}$, then the rate of reaction will increase $\ce{A}$. This has nothing to do with the Law of Definite Proportions because the reaction will proceed to the make the product in the correct ratio and will stop when we run out of the limiting reactant. $\endgroup$ – Zhe May 15 at 12:26
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    $\begingroup$ No ! let's take a numerical example. If you start with all concentrations equal to $0.1$ mol/L, and the rate is $\ce{10^{-4}}$ mol/L/s, it means that after $10$ s, the concentration of A, B and C has decreased by $0.001$ mol/L, and becomes [A] = [B] = [C] = $0.1 - 0.001 = 0.099$ mol/L. Now if you double the amount of A, the rate will become equal to $\ce{2·10^{-4}}$ mol/L/s. It means that after the same delay ($10$ s), the amount of A, B and C will be respectively : [A] = $0.2 - 0.002 = 0.198$ mol/L, [B] = [C] = $0.1 - 0.002 = 0.098$ mol/L. $\endgroup$ – Maurice May 15 at 13:58
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[OP, in a comment] oh ok, so if we double the amount of only one substance, that substance will get left?

Yes. The rate law tells you how fast product is being made. The change in the different species is still given by the chemical reaction equation (which tells you in which proportion the amount of species will change, not in which proportion they are present at the beginning or at the end of the reaction).

You can see that when you look at the definition of rate:

$$\mathrm{rate} = \frac{d[X_i]}{\nu_i dt}$$

where $\nu_x$ is the (signed, i.e. negative for reactants) stoichiometric factor of the $i$-th species. Dividing by the stoichiometric factor ensures that the law of definite proportions is followed.

[OP in another comment] we cant simply put more of a substance to get more of the product, the reactants must have a fixed ratio.

The law of definite proportions is not about the amounts present, but about the change in those amounts. I can set up a reaction with 1000 times more elemental oxygen than hydrogen, but they will still react in a 1:2 ratio if the reaction yields water.

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  • $\begingroup$ thank for clearing it out! my book never said anything about there being residue of the limiting reactant. $\endgroup$ – 12345bird May 16 at 4:09

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