- #1
minivanhighwa
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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
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