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The reaction

$$\ce{2 A + B + C -> D + 2 E}$$

is found to follow the rate law as

$$r = k[\ce{A}][\ce{B}]^2[\ce{C}]^0$$

If the concentration of $\ce{A}$, $\ce{B}$ and $\ce{C}$ increases two times then the rate of reaction becomes eight times higher:

$$r' = k\cdot 2[\ce{A}]\left(2[\ce{B}]\right)^2\left(2[\ce{C}]\right)^0 = 8k[\ce{A}][\ce{B}]^2$$

Note: there is no change in the order of reaction. I want to change the order of this reaction by changing the concentration of reactants. Can I do this?

Because if we double the concentration of $\ce{A}$, $\ce{B}$ or $\ce{C}$, it does not change the order. But if we increase the concentration of $\ce{A}$ by 4 times, then we can change order by

$$\left(4[\ce{A}]\right)^2 = \left(2[\ce{A}]\right)^{2+2} = \left(2[\ce{A}]\right)^4$$

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You would change the reaction speed, but not the order.

The order of the reaction can be changed by reagent concentration, if it is result of multiple reaction steps with different orders. By variation of concentrations you can manipulate the relative reaction speed, as the order is often ruled by the slowest reaction.

Let suppose the 2 step reaction $$\begin{align} \ce{2A + B &-> C} \\ \ce{A + A &->[slow] D} \\ \ce{B + D &->[fast] C} \\ \end{align}$$

For small $c_A$, the reaction speed $$\frac{dc_C}{dt}=k_{low A}.{c_A}^2$$.

For high $c_A$, the first reaction becomes faster then the second.

$$\frac{dc_C}{dt}=k_{high A}.{c_A}^2.c_B$$.

You can also change the effective order of reaction, if some reactant is in about constant abundance and is not explicitly considered.

E.g. The speed of the reversed esterification $$\ce{R1COOR2 + H2O -> R1COOH + R2OH}$$ is the reaction kinetics of the 2nd order. But in abundance of water, it is effectively a reaction of the 1st order: $$\frac{dc_{\ce{R1COOR2}}}{dt}=-k_1.c_{\ce{R1COOR2}}$$.

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You shouldn't be able to change the order of the reaction or the powers of the concentrations in the rate equation (which lead to the overall order). Those powers correspond to the rate determining step, so changing the rate equation would mean you have a different mechanism and potentially a different reaction.

Your substitution of different concentrations into the rate equation is imprecise which leads to error. [A] corresponds to the concentration of A in general. You should only get [2A], or rather 2[A], if you have a different chemical species, e.g F, which is produced from two moles of A. If you double the concentration of A, what you have is double a specific concentration of A, e.g 2[initial A].

To answer the edited question, your maths seems flawed as you take the power which applies to 2 only and apply it to the whole term, which would only work if they were same base (but for most cases of [A], 2 =/= [A]). So as well as the chemical intuition supporting not being able to change a reaction's order, there doesn't seem to be any mathematical way of doing so.

$$ \left(2^{2}[\ce{A}]\right)^{2} =/= \left(2[\ce{A}]\right)^{2+2} $$

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Conceptually the easiest way to change the rate law in your example is to lower the concentration of C. In your rate law, the reaction is zero-order in C. However, if you lower the concentration of C to zero, you get a new rate law:

$$ \text{rate} = 0$$

instead of

$$ \text{rate} = k [A]^2 [B]$$

The reason the reaction rate did not depend on C in your rate law is that C does not come into play until after A and B have reacted to form some intermediate. If the reaction of the intermediate with C is much faster than the formation of the intermediate, the concentration of C has negligible effect on the overall rate (if there is sufficient C available). If you leave out C, the last step obviously can't happen, so now the concentration of C has a significant effect.

To summarize this argument, having C be at zero order is only true for a certain range of concentrations of C, and if you go below it, a different rate law will be appropriate.

But if we increase the concentration of A by 4 times, then we can change order [...]

That argument is flawed. The defining equation for the order in A is

$$ \frac{\text{rate}(f \times [A_0], [B_0], [C_0])}{\text{rate}([A_0], [B_0], [C_0])} = f^\text{order}$$

In your example, you change [A] by a factor four, and the rate increases by a factor 16. That is second order. It is true if you had changed [A] by a factor of two and the rate would have increased by a factor 16, you would conclude that it is fourth order in A. However, you said you increased the concentration by a factor of four, so you have to stick with that.

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  • $\begingroup$ If we take the concentration of C= 0 , then that will become the case of [0]^0 , the answer to this is controversial in maths I think @Karsten $\endgroup$ – Beyond Zero Apr 8 at 1:20
  • $\begingroup$ @Beyond_Zero : Yes, but not controversial in chemistry. Without C, there will be no reaction. But I guess you are right, that this gives a hint that under certain circumstances, the rate of reaction might dependent on the concentration of C after all. $\endgroup$ – Karsten Theis Apr 8 at 2:39

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