difference between reaction rate and rate law??

[Graffiti]

for a hypothetical reaction: aA + bB --> cC + dD, where the lower case letters represent coefficients

i'm told the rate expressions are:

rate=-[A]/a x change in time
rate = -/b x change in time
rate = [C]/c x change in time
rate = [D]/d x change in time

minus signs represents decrease of reactants and no sign represents increase in products over some time period.

anyways, it seems like we can use that expression even when the reaction isnt elementary. isnt this somehow inaccurate. maybe im missing something conceptually here but how does the rate expression above relate to the "rate" determined by the rate law?? is the above just an approximation whereas the rate determined by the rate law the actual value determined by the rate determining step of the reaction???

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[Graffiti]

i feel like i just confused myself.

 

[Rabolisk]

Rate law relates the rate of a reaction to the concentration of the reactants. The rate expression just expresses or defines what rate is. In your example, rate can be defined in one of those four ways. They are equivalent because for every molecule of A consumed, b/a molecule of B has to be consumed, c/a of C has to be produced, and d/a of D has to be produced. This is based on stoichiometry, not reaction mechanism, so the rate expression is independent of whether or not the reaction is "elementary". The rate law is dependent on the mechanism of the reaction.

 

[Graffiti]

Rate law relates the rate of a reaction to the concentration of the reactants. The rate expression just expresses or defines what rate is. In your example, rate can be defined in one of those four ways. They are equivalent because for every molecule of A consumed, b/a molecule of B has to be consumed, c/a of C has to be produced, and d/a of D has to be produced. This is based on stoichiometry, not reaction mechanism, so the rate expression is independent of whether or not the reaction is "elementary". The rate law is dependent on the mechanism of the reaction.

i understand what rate expression is. what i dont understand is how the rate expression relates to the rate law. if you plugged numbers in the rate expression and also plugged in numbers for the rate law, would they be equal if this were the elementary reaction?

 

[GreenRabbit]

No difference between these rates.

In your example, rate = (-1/a) d[A]/dt is the definition of rate (definitely not an approximation).

Something like rate = k[A] tells us more about the order of the reaction.

The rates are the same, and in fact you set them equal to each other for some disgusting equations (those solving for concentration and time).


For example, you've probably seen for A --> B

rate = -d[A]/dt
rate = k[A]
​

Set -d[A]/dt = k[A], do some calc, and get

ln([A]/[A]i) = -k(delta t)
​

The example you gave with both A and B is really ugly, and definitely not on the MCAT.

 

[Graffiti]

No difference between these rates.

In your example, rate = (-1/a) d[A]/dt is the definition of rate (definitely not an approximation).

Something like rate = k[A] tells us more about the order of the reaction.

The rates are the same, and in fact you set them equal to each other for some disgusting equations (those solving for concentration and time).


For example, you've probably seen for A --> B

rate = -d[A]/dt
rate = k[A]
​

Set -d[A]/dt = k[A], do some calc, and get

ln([A]/[A]i) = -k(delta t)
​

The example you gave with both A and B is really ugly, and definitely not on the MCAT.

thank you!! this is exactly what i was trying to understand. no review book i've seen has really stated this explicitly.

 

[Graffiti]

the reason why i said approximately though is because ek said this, "although the rate equation above is strictly correct only for an elemntary reaction, it is a good approximation for multistep reaction if the concentration of intermediates is kept low."

 

[GreenRabbit]

the reason why i said approximately though is because ek said this, "although the rate equation above is strictly correct only for an elemntary reaction, it is a good approximation for multistep reaction if the concentration of intermediates is kept low."

Whether the reaction is elementary or not is important only in predicting the second equation (the order). The first equation, being the definition of rate, doesn't change.

So if the reaction is elementary or approximates an elementary reaction, it follows that for aA + bB ---> products,

rate = k[A]^a ^b
​

If the reaction isn't elementary, then you can't predict the exponents, so you'd have some generic equation

rate = k[A]^m ^n
​

where 'm' and 'n' are not related to 'a' and 'b'.

 

[Graffiti]

Whether the reaction is elementary or not is important only in predicting the second equation (the order). The first equation, being the definition of rate, doesn't change.

So if the reaction is elementary or approximates an elementary reaction, it follows that for aA + bB ---> products,

rate = k[A]^a ^b
​

If the reaction isn't elementary, then you can't predict the exponents, so you'd have some generic equation

rate = k(A)^m (B)^n
​

where 'm' and 'n' are not related to 'a' and 'b'.

what if its found that reactant B is zero order overall. that wouldn't affect the overall rate either? Im probably over thinking this. ill stick with your earlier statement and ignore what ek said. you're really smart thanks for your help.

 

[Graffiti]

what if its found that reactant B is zero order overall. that wouldn't affect the overall rate either? Im probably over thinking this. ill stick with your earlier statement and ignore what ek said. you're really smart thanks for your help.

i think i see what my confusion was. the rate determined by the rate law predetermines how much the concentration of reactants will change over a period of time. the change in concentration of reactants over a period of time is equal to the rate of the reaction determined by the rate law.

 

[GreenRabbit]

i think i see what my confusion was. the rate determined by the rate law predetermines how much the concentration of reactants will change over a period of time. the change in concentration of reactants over a period of time is equal to the rate of the reaction determined by the rate law.

Hmm, I think you may be mixing up "rate" with "change in concentration." The rate law only tells you the speed of the reaction at one particular instant. Because those concentrations change, so will the rate (unless it's zeroth order). The only way to predict how much the concentration will change over a long period of time is by using what's called "integrated rate laws", which might help you get a better understanding.

 

[Graffiti]

Hmm, I think you may be mixing up "rate" with "change in concentration." The rate law only tells you the speed of the reaction at one particular instant. Because those concentrations change, so will the rate (unless it's zeroth order). The only way to predict how much the concentration will change over a long period of time is by using what's called "integrated rate laws", which might help you get a better understanding.

so based on what you just said, would it make sense now that the rate expression is an average? i think this is what ek was trying to say.

rate at some initial point (from rate law)
rate at some final point (also from rate law)

the average of that is the same thing as...

rate expression=d[reactant concentration]/d[time]

 

[Rabolisk]

No, it's not an average. It is the instantaneous rate at a specific time. d[A]/dt is a derivative, which is instantaneous by definition. The rate from the rate law and the rate from the expression are the same, and both change with time, unless the reaction is zeroth order. The equations don't change, but the values do.