I was solving a question and the reaction $\ce{3A -> B}$ was written as a form of a zero order reaction. Why? It’s depending of $\ce{3A}$. So why is $[\ce{A}]^3$ not the rate equation?
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1$\begingroup$ Sometimes the order of a reaction is the same number as the stoichiometric coefficient in the corresponding equation. But it is not always true. The order of a reaction is experimentally obtained, with no systematic connection with the stoichiometric coefficients. $\endgroup$– MauriceCommented Sep 22 at 20:32
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1$\begingroup$ The rate of a reaction depends on the mechanism of the reaction not on the overall stoichiometry. So you need to know the mechanism before making any deductions about the rate equation. $\endgroup$– matt_blackCommented Sep 24 at 12:18
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$\begingroup$ See also chemistry.stackexchange.com/questions/32304/law-of-mass-action and maybe also chemistry.stackexchange.com/questions/38167/… for additional answers on highly related questions. $\endgroup$– Curt F.Commented Sep 27 at 19:21
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3 Answers
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The reaction rate order reflects (unconditionally) the stoichiometric coefficients only for elementary reactions without more complex reaction mechanism. For complex reactions with a reaction mechanism schema, the order may or may not reflect the stoichiometric coefficients.
Imagine the reaction
$$\ce{3 A -> B}$$
has the reaction mechanism:
$$\ce{2 A -> C}$$ $$\ce{2 C -> B + A}$$
For a steady state:
$$\text{d}[\ce{C}]/\text{d}t = 0 = k_1[\ce{A}]^2 - k_2[C]^2$$ $$[\ce{C}] = \sqrt{\frac{k_1}{k_2}}[\ce{A}]$$
Then for the product $\ce{B}$:
$$\text{d}[\ce{B}]/\text{d}t = k_3[C]^2=\frac{k_3}{k_1k_2}[\ce{A}]^2$$
therefore $\ce{3 A -> B}$ has in the steady state the reaction kinetics of the second order.
If not in the steady state yet, the reaction rate equation would be more complicated.
Different case: the reaction $\ce{3 A -> B}$ is triggered by UV/Vis light absorption:
$$\ce{A + A ->[h \nu] C}$$ $$\ce{A + C ->[h \nu] B}$$
The reaction rate is in wide range of [A] controlled by the incident photon rate, independent on the [$\ce{A}$], so the reaction kinetics wrt $\ce{A}$ would be of the zeroth order.
If [$\ce{A}$] is too low, part of light can transmit through and it would become [A] dependent (but not with an integer order).
If [$\ce{A}$] is too high, the light could be absorbed in a surface layer and the reaction rate would be partially controlled by diffusion. It would become [A] dependent again (not with an integer order).
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First, the chemical equation $$\ce{3A -> B}$$ tells you nothing about the kinetics. It is a mass balance equation: each time the reaction occurs, three molecules of $\ce{A}$ transform to a molecule of $\ce{B}.$
If you consider the reaction mechanism, you could broadly talk about two kinds of reactions:
- Simple reactions. The reaction is produced in one step, the stoichiometric coefficient coincides with the molecularity, and the reaction order coincides with the molecularity. For example, $$\ce{A + A + A -> B} \qquad v = k[\ce{A}]^3$$
- Complex reactions. The reaction mechanism corresponds to a sequence of simple reactions. The mechanism must be consistent with the stoichiometry of the reaction, but the coefficients tell you nothing about the reaction orders. Derivation of the rate expression requires you to know the reaction mechanism. Unless you know the mechanism, you can not obtain the reaction orders.
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I think you got confused between molecularity of a reaction and its order.
The molecularity of a reaction is simply the number of molecules involved in the reaction.
For example, $\ce{A + 2B -> C + D}$ has a molecularity of 3.
Molecularity is the number of molecules that come together to react in a reaction and is equal to the sum of stoichiometric coefficients of reactants in the elementary reaction with effective collision (sufficient energy) and correct orientation.
The order of a reaction on the other hand, involves experimentally finding out the dependence of the rate on the concentration of the reactants. It is a experimentally derived value.
$\ce{A + 2B -> C + D}$; this reaction CAN be a 3th order reaction but it could be a 2nd order too. So nothing can be said about the order just from the stoichiometric equation.
Another thing, for simple reactions molecularity = order but for complex (multi-step) reactions order maybe not be equal to molecularity. In conclusion, unless it is stated that the reaction is a simple reaction, you cannot write the rate equation with the orders from the stoichiometric equation of the reaction ie molecularity should not be assumed to be the same as the order.
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2$\begingroup$ molecularity = The number of reactant molecular entities that are involved in the 'microscopic chemical event ' constituting an elementary reaction . (For reactions in solution this number is always taken to exclude molecular entities that form part of the medium and which are involved solely by virtue of their solvation of solutes.) A reaction with a molecularity of one is called 'unimolecular', one with a molecularity of two 'bimolecular' and of three 'termolecular'. ( IUPAC GoldBook Molecularity) $\endgroup$– PoutnikCommented Sep 23 at 18:01
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1$\begingroup$ Note that even true trimolecular reactions are extremely rare. Many are just pseudo-trimolecular, forming extremely short-lived intermediate/activated complex, that still lasts much longer than elastic collisions of molecules. Both cases can have the kinetics of the 3rd order. // Typical case is the triple alpha process $\ce{3 ^{4}He -> ^{12}C}$ in stars fusing helium to carbon. The intermediate $\ce{^{8}Be}$ has the mean life $\pu{\approx e-16 s}$ what is still $\pu{\approx e6}$ time longer than typical nuclei collision. $\endgroup$– PoutnikCommented Sep 27 at 9:08