3
$\begingroup$

I have read in a general chemistry textbook that rate laws can only be determined by experiment. This seems intuitive because if a mechanism is not elucidated and tabulated reaction data would be one way of bypassing that info to reach the rate law. However, I have seen instances where the rate law is simply determined by using the reactants and raising them to their respective stoichiometric power...which I thought contradicted the idea of needing an experiment? I highly speculate that this is only acceptable if it a singular reaction that the reaction order can be determined experimentally? For example:

 

$$\ce{2A -> B_2}$$

Can be surmised as:

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

Is this true? My speculation is based on the notion that if one reactant is splitting to a product and that stoichiometric equivalent is needed for the conversion of reactant to product, then the rate law is determined by stoichiometric equivalent number of molecules hence the justification for the "raise to the power" method. Or is there another underlying factor when the exponents have been raised and no detail is given to rate laws determine by experiment?

$\endgroup$
2
$\begingroup$

Your reasoning is correct. When the reaction you are investigating is an elementary reaction step, you can write down the rate law using the stoichiometry of the reaction. Mathematically, this is done as follows.

Consider a reaction as

$$ \sum_{i} \nu_{i} R_{i} \rightarrow \sum_{j} \nu_{j} P_{j} $$

Where $R$ and $P$ are reactants and products respectively and $\nu_{i}$ and $\nu_{j}$ are the stoichiometric coefficients.

If this reaction is an elementary reaction step, than the rate law in the forward direction is given by:

$$ r = k \prod_{i} [R_{i}]^{\nu_{i}} $$

where $[R_{i}]$ is the concentration of reactant $R_{i}$ and $k$ is the rate constant.

For example, if you consider the reaction

$$ A + 2B \rightarrow 3C, $$ then the rate in the forward direction is

$$ r = k[A][B]^{2}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.