Does integrated rate law have stoichiometric coefficient?

  • Thread starter khoivu
  • Start date
  • #1
6
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 -> B

where there is a stoichiometric coefficient in front of A. 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
 

Answers and Replies

  • #2
member 392791
I'm interested in this as well...the textbook says that order of the reactant is not related to the stoichiometric coefficient.

Off-topic..but I knew a guy named Khoi Vu :P
 
  • #3
6
0
maybe it's me :D
 
  • #4
member 392791
do u go to cypress? cuz if so then it is you lol
 
  • #5
1,082
25
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 -> B

where there is a stoichiometric coefficient in front of A. 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
You pretty much have it. The general form is dA/dt=-kA, where k=ak' for a first order reaction. The general form does not change, only the k value effectively changes.

I'm interested in this as well...the textbook says that order of the reactant is not related to the stoichiometric coefficient.
If I recall correctly, for the order of the reaction is the stoichiometric coefficient for an elementary reaction but does not need to be so far a composite reaction.
 
  • #6
6
0
I understand. A constant is constant, no matter how we says it k or k'. But if the question asks to find a value of rate constant based on integrated rate law, then it would be important to know what kind of equation we should use, the one that has stoichiometric coefficient or the one that does not since they will produce different rate constants. Then how can you deal with such situation?
 
  • #7
6
0
i go to occ
 
  • #8
epenguin
Homework Helper
Gold Member
3,864
895
Stoichiometry does not predict rate law, reaction mechanism does. Conversely, finding rate laws experimentally gives insight into mechanism of the reaction (or at least limits the possible ones).

So if your mechanism was that two molecules of A had to collide (with a certain energy and orientation) and when they did they turned fast into B then your rate would = k[A]2.

But if the mechanism was that A turned itself into a reactive state A* and A* rapidly and efficiently reacted with any A molecule around

A → A* followed by A* + A → A2 , second reaction fast, then rate would have the form k[A].

Yours is just a preliminary doubt, I think as soon as you get into some detailed examples of mechanisms or kinetics you won't have any problems.
 
  • #9
member 392791
haha khoi I tried getting into that class but when she did the hat pull out names I didn't get in =(

atleast I got in at fullerton though

I liked that dr. gonzales she seemed funny =/
 

Related Threads on Does integrated rate law have stoichiometric coefficient?

Replies
3
Views
2K
Replies
2
Views
2K
Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
27K
  • Last Post
Replies
1
Views
4K
Top