Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: HusamEddin on November 15, 2012, 12:25:31 PM
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Hi there ,,
Peace be upon U
can it be a psuedo-order reaction for a one-step reaction ???
if this hypothetical reaction is a one-step reaction:
A + B = C
could the rate of this reaction has an over all order which is a psuedo-order reaction ???
for example:
rate = K' [A] [ B]
change in [ B] is relativily so small
then:
rate = K* [A]
where K* = K' [ B](initial)
then the over all order is 1
could it be like this ?
or the concentrations exponents have the same as their coeffecients in the equation for a one-step reaction ?
because all the time my book says:
experimentally , is the only way to know for sure what the exponents are.
but if this statement is true ... this means it is also experimentally is the only way to determine the exponents in a one-step reaction
but I know previously that if I have a hypothetical one-step reaction:
A + B = C
then the rate would be as simple as:
rate = K [A]^1 [ B]^1
rate = K [A] [ B]
what is the right thing ??
help please
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waiting
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Didn't understand your question.
If "change in [ B] is relativily so small" then "change in [ A] is also relativily so small"
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I don't really understand your question either. If you have a process that is experimentally second-order overall (first order in A and first order in B), and >> [A], then you will observe pseudo-first order kinetics. rate = k[A], where k = k'. Of course, one would first have to determine that the rate is really first order in A first order in B.
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U really didn't understand me
look
my book says that the only way to determine the correct rate law for a certain reaction is by experiments
but !!
is it right that we can determine the rate law immediatly from the reaction equation ?? .. I mean like this:
A + 2B = C
if this is a one-step reaction ... then rate law is simply
rate = k [A] [ B]^2
If this is right ... then .. this contradict the sentence that my book mentioned !!!
can you explain please ?
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In general, one has to determine the kinetic order experimentally, and your book is correct. If you know that a reaction is elementary, then you can write the rate law. This sounds like an exception to the rule, but I am not at all certain that it is. How can you know that a reaction is elementary without doing experiments of some sort or another?
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if this is a one-step reaction...
Which you don't know without doing the experiment.
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ahaaaaaaaaaaaaaa
now I understood both of you
now the second question that you didn't understand from me:
if we are sure that this hypothetical reaction is elementary:
A + B = C
then the rate law is:
rate = K [A] [ B]
then the overall order is "2"
am I right ?
but ... I doubt
because some reactions are said to have psuedo order
psuedo-order means that if it appears to be a second-order ... then the right order is "first" not "second"
now my question is ...
could the elementary reactions have a psuedo-order .. or its order always can be determined immediatly from the reaction equation ?
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if we are sure that this hypothetical reaction is elementary:
A + B = C
then the rate law is:
rate = K [A] [ B]
then the overall order is "2"
am I right ?
Right
now my question is ...
could the elementary reactions have a psuedo-order .. or its order always can be determined immediatly from the reaction equation ?
Say, you had a large excess of B in your reaction. Now it will appear 1st order in A alone.
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if we are sure that this hypothetical reaction is elementary:
A + B = C
then the rate law is:
rate = K [A] [ B]
then the overall order is "2"
am I right ?
Right
now my question is ...
could the elementary reactions have a psuedo-order .. or its order always can be determined immediatly from the reaction equation ?
Say, you had a large excess of B in your reaction. Now it will appear 1st order in A alone.
yes ... that's it
it seams that it is second ... but in fact ... it is first
and this consequently lead me to say: we can't determine the overall order of the reaction eventhough we are sure it is an elementary reaction
Am I right ?
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?? :)
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A pseudo-first order reaction such as the one you described is really a second order reaction. The disappearance of the reactant that is low in concentration A follows an exponential decay (just like a true first order reaction), but that is only because the reactant in high concentration B does not change in concentration appreciably over the course of the reaction.
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A pseudo-first order reaction such as the one you described is really a second order reaction. The disappearance of the reactant that is low in concentration A follows an exponential decay (just like a true first order reaction), but that is only because the reactant in high concentration B does not change in concentration appreciably over the course of the reaction.
my frind ... this is totally OK
but eventhough it is really a second order ... we wrote it as:
rate = K* [A]
then .. this contradict with the fact that says for this elementary reaction "A + B = C" the rate equals K [A] [ B]
because
K [A] [ B] doesn't equal k* [ B]
!!!
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I am probably not following you. For any second-order reaction:
rate = k'[A][ B]
If [ B] >> [A], then
rate = k[A], where k = k'[ B]
and k is the pseudo-first order rate constant. I don't think we have to worry about whether the reaction is or is not elementary.
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To all: please note that [B] disappears and makes everything bold, use [ B] instead (put a single space after the opening parentheses). Otherwise you may be not able to understand each other.
Read more about post formatting (http://www.chemicalforums.com/index.php?topic=59314.0) if you want to know more fancy methods of avoiding this problem.
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I think you didn't got my question
for this elementary reaction:
A + B = C
rate law is simply:
rate = K [A] [ B]
and it can't be:
rate = K* [A]
where k* = K' [ B]
never
because it is elementary
Right ??
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I've the nagging feeling that we are flying in circles here. I give up.
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OK , don't worry ma friend ... you did your best
thank you very much :)