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Am hoping that someone can help me with the below question, and that I am asking in the right place. Am new here, so please forgive any missteps.

Given a reaction like this:

$$\ce{A + B <=> C + D},$$

is the differential/initial rates method of determining a reaction order valid? That is, the method which determines $b$ in $r = k C_\ce{B}^b$ as the slope in a $\ln({\mathrm{d}C_\ce{C}}/{\mathrm{d}t})$ vs $\ln(C_\ce{B})$ plot.

Basically, I wonder about three points here:

  1. I have only seen examples for this method on single forward reactions, so is it valid when there is a reverse reaction?

  2. Am I correct in thinking that I can use measurements of $C$ for $\ln(\text{rate})$, even though I'm interested in the reaction order of $\ce{B}$, because the rate can be written in terms of any of the reactants or products?

  3. How crucial is it that the concentration (rate) measurements must come from initial times? If the change in all concentrations is very small, is it then not valid to use even a later values?

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closed as too broad by A.K., Mithoron, Melanie Shebel, Todd Minehardt, airhuff Feb 28 at 3:23

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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The difficulty with problems in kinetics is that you can envision a pool of intermediates such that, while B is consumed, you don't see significant formation of C, so that C is a poor reporter of the consumption rate of B. Concepts such as intermediates, autocatalysis, feedback, steady-states, lag times can be important here. Things can get complicated pretty quickly.

The initial forward rate, in the absence of a significant accumulation of reaction intermediates, can for simplicity be written as

$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b$$

Again, this assumes a negligible accumulation of intermediates whose reverse reaction may lead back to formation of B, as in a process under steady state.

For this to be useful in the determination of $k_\mathrm{f}$ it helps to satisfy two conditions:

  1. $k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b \gg k_\mathrm{rev}[intermediates]^m$
  2. $[\ce{A}]^a\approx \text{const}$

These conditions are essential to guarantee a decay of the concentration of $\ce{B}$ with a rate constant $k_\mathrm{fwd}[\ce{A}]^a$. The answer to your second question is no unless the mechanism is simple and conditions 1 and 2 hold (small amount of product), that is, if the reverse reactions from intermediates or products are not important.

So, it is not essential that the measurements come from the beginning as long as the mechanism is elementary (no rate limiting steps leading to accumulation of intermediates), the reverse reaction is insignificant in rate and the concentration of A greatly exceeds that of B throughout your measurement time.

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  • $\begingroup$ I edited some notations used in physical chemistry: note that "f" refers to formation, "r" – to reaction. In kinetics commonly used notations for the forward and reverse reactions are "fwd" and "rev", respectively. $\endgroup$ – andselisk Feb 26 at 0:22
  • $\begingroup$ RE: "the initial forward rate, in the absence of product, will always be given by the rate law $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}][\ce{B}]$$" This just isn't true. You're assuming an elementary reaction, which may or may not be the case. For instance if [A] << [B] then the reaction may be pseudo zero order in B. $\endgroup$ – MaxW Feb 26 at 1:46
  • $\begingroup$ In the most general case: $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b + k_\mathrm{rev}[\ce{C}]^c[\ce{D}]^d $$ where a, b, c, and d are exponents to be determined. $\endgroup$ – MaxW Feb 26 at 1:53
  • $\begingroup$ @NightWriter - At t=0 there is no product. For any t>0 there is. $\endgroup$ – MaxW Feb 26 at 7:34
  • $\begingroup$ @NightWriter - Do understand that if the coefficients in the kinetic equation are equal to the coefficients in the chemical reaction, then the kinetic equation is an elementary reaction. A reaction in actuality doesn't have to be elementary. $\endgroup$ – MaxW Feb 26 at 8:43

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