Conceptual question about rate laws

Click For Summary
SUMMARY

The discussion centers on the application of integrated rate laws in chemical kinetics, specifically addressing first-order reactions with stoichiometric coefficients. The participant questions whether the integrated rate law for a first-order reaction, expressed as ln[A]/[A]i = -kt, remains valid when the reaction involves stoichiometric coefficients, such as in the reaction 2A -> B. The conclusion drawn is that while the form of the integrated rate law does not change substantively with stoichiometry, the interpretation of the rate law must account for these coefficients to avoid confusion between reaction order and stoichiometry.

PREREQUISITES
  • Understanding of first-order reaction kinetics
  • Familiarity with integrated rate laws
  • Knowledge of stoichiometry in chemical reactions
  • Basic principles of chemical kinetics
NEXT STEPS
  • Study the derivation of integrated rate laws for zero-order and first-order reactions
  • Explore the impact of stoichiometric coefficients on rate laws in complex reactions
  • Learn about the distinction between reaction order and stoichiometry in kinetics
  • Investigate experimental methods for measuring reaction rates and their implications on rate law analysis
USEFUL FOR

Chemistry students, educators in chemical kinetics, and researchers analyzing reaction mechanisms and kinetics in laboratory settings.

minivanhighwa
Messages
1
Reaction score
0
So anytime I've seen textbooks explain integrated rate laws, they usually start with a reaction of the form A -> B and then from there say, if we know the reaction is first order with respect to [A] then:

v = -d[A]/dt = k[A]

And then subsequently integrate this to find ln[A]/[A]i = -kt

I get that. My issue is this. From what I always thought, the integrated rate laws are universal, and now that I have this equation I can use it any time I know that a reaction is first order with respect to A. But what if I have an equation of the form:

aA -> bB

where there is a stoichiometric coefficient in front of A. I was always under the impression that v = - (1/a) d[A]/dt = (1/b) d/dt... If experimental data still tells us that the rate law is first order with respect to A, can I just use the same integrated rate law that I found above? or would I have to say the following:

v = - (1/a) d[A]/dt = k[A]

in which case you would get:
ln[A]/[A]i = -(a)kt

So for a reaction 2A -> B

You would find ln[A]/[A]i = -2kt

Assuming the rate was always first order with respect to [A], wouldn't you get a different integrated rate law for every instance that you have a different stoichiometric factor (a) in front of your reactant? If so why isn't the general form of the equation given as

ln[A]/[A]i = -akt

Hope this makes sense - not sure where my reasoning is off
 
Chemistry news on Phys.org
I suspect this is a matter of definitions and pedagogy.

When I think of a "different rate law," I view the zero-order rate law

[A] = [A]0 - kt

to be different than the first-order rate law

[A] = [A]0*exp(-kt).

Adjusting a rate law by a constant to correct for stoichiometry does not substantively change the form of the rate law.

I suspect the reason that a more general rate law is not introduced for a few reasons, most of which has to do with avoiding confusion between order of a reaction and stoichiometry (I used to notice this happening quite a bit while a teaching assistant for introductory general chemistry) and since - if feasible - one can pick what to measure in an experiment to make your subsequent analysis of the reaction kinetics & mechanism easier.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 131 ·
5
Replies
131
Views
11K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
3K