# What is the relationship between the order of a reaction and rate determining step?

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#1
What is the relationship between the order of a reaction and the rate determining step?

I need to explain this relationship for part of my coursework - yet I am unaware of such a relationship.

It would be much appreciated if someone could shed some light on this relationship! tia
0
7 years ago
#2
You know that the species involved in the rate determining step (RDS) appear in the rate equation.

So take a general reaction A + 2B ---> C + D

RDS is A + B ---> X

then X + B ----> C + D

the rate equation is expressed as rate = k[A][B] since one A and one B take place in the RDS
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#3
(Original post by EierVonSatan)
You know that the species involved in the rate determining step (RDS) appear in the rate equation.

So take a general reaction A + 2B ---> C + D

RDS is A + B ---> X

then X + B ----> C + D

the rate equation is expressed as rate = k[A][B] since one A and one B take place in the RDS
I think I understand, so are you saying that only reactants involved in the RDS, will appear in the rate equation.
If this is so, does that mean that any reactants that are first or second order, must be involved in the RDS, because they are present in the rate equation. Whereas, reactants of zero order will not be involved in the RDS because they do not appear in the rate equation (or at least because they are to the power of 0, they are effectively not there - because anything to the power of 0 is just 1).

An answer to my original question: "The relationship between the order of a reaction and the rate determining step, is that the order of the reaction depends on the order of the reaction with respect to each reactant involved in the rate determining step."

Would this be correct?
0
7 years ago
#4
(Original post by miles christisum)
I think I understand, so are you saying that only reactants involved in the RDS, will appear in the rate equation.
Exactly

If this is so, does that mean that any reactants that are first or second order, must be involved in the RDS, because they are present in the rate equation. Whereas, reactants of zero order will not be involved in the RDS because they do not appear in the rate equation (or at least because they are to the power of 0, they are effectively not there - because anything to the power of 0 is just 1).
Yep

An answer to my original question: "The relationship between the order of a reaction and the rate determining step, is that the order of the reaction depends on the order of the reaction with respect to each reactant involved in the rate determining step."

Would this be correct?
Close. The bold bit is technically correct, but there is a simpler way to put it. The order of the reactants in the RDS is the stoichiometry of the RDS itself because it is only one step.

Meaning that if the RDS was A + B ----> stuff then rate = k[A][B] (with those orders)

if the RDS was 2A ---> stuff then rate = k[A]2

if you have a single step reaction that you know is only one step, then that is the RDS by default and so you can easily get it's rate equation by looking at the stoichiometry alone.
1
#5
(Original post by EierVonSatan)
Exactly
(Original post by EierVonSatan)

Yep

Close. The bold bit is technically correct, but there is a simpler way to put it. The order of the reactants in the RDS is the stoichiometry of the RDS itself because it is only one step.

Meaning that if the RDS was A + B ----> stuff then rate = k[A][B] (with those orders)

if the RDS was 2A ---> stuff then rate = k[A]2

if you have a single step reaction that you know is only one step, then that is the RDS by default and so you can easily get it's rate equation by looking at the stoichiometry alone.
Brilliant. So in the simple iodine clock reaction......

[which I investigated for my coursework which involves these two reactions:

A) S2O82- (aq) + 2I- (aq) ➝ SO42- (aq) + I2 (aq)

B) 2S2O32- (aq) + I2 (aq) ➝ S4O62- (aq) + 2I- (aq)

(Where the thiosulfate (VI) ions in reaction B, transform the iodine produced by reaction A, back into iodide ions - until the thiosulfate (VI) ions run out. So reaction A is the RDS, because reaction B just postpones the colour change to make the test measurable).]

......the rate equation would just be:

Rate = k x [S2O82-] x [I -]

because the stoichiometry of the RDS is 1:1 (peroxodisulfate (VI) : Iodide)

Thinking about what you said about the stoichiometry of the RDS determining the rate equation. I am sure my chemistry teacher explained to the class before that this does not always hold true; that sometimes the order of the reaction with respect to a specific reactant is not always the same as that reactants stoichiometry in the reaction equation?
I am not saying you are wrong haha, just double checking because I distinctly remember my chemistry teacher telling me something along those lines.
0
7 years ago
#6
(Original post by miles christisum)

Brilliant. So in the simple iodine clock reaction......

[which I investigated for my coursework which involves these two reactions:

A) S2O82- (aq) + 2I- (aq) ➝ SO42- (aq) + I2 (aq)

B) 2S2O32- (aq) + I2 (aq) ➝ S4O62- (aq) + 2I- (aq)

(Where the thiosulfate (VI) ions in reaction B, transform the iodine produced by reaction A, back into iodide ions - until the thiosulfate (VI) ions run out. So reaction A is the RDS, because reaction B just postpones the colour change to make the test measurable).]

......the rate equation would just be:

Rate = k x [S2O82-] x [I -]

because the stoichiometry of the RDS is 1:1 (peroxodisulfate (VI) : Iodide)

The most important thing to remember about rates/orders is that they can only be determined through experiment. You can't assume which step would be faster, unless of course you have been told that it is Also, using the RDS of a multistep reaction to form the rate equation only works when one step is significantly faster than every other step, if they're at all similar then you can't use the RDS alone as the basis for the rate equation.

Thinking about what you said about the stoichiometry of the RDS determining the rate equation. I am sure my chemistry teacher explained to the class before that this does not always hold true; that sometimes the order of the reaction with respect to a specific reactant is not always the same as that reactants stoichiometry in the reaction equation?
I am not saying you are wrong haha, just double checking because I distinctly remember my chemistry teacher telling me something along those lines.
Your teacher is quite correct, you can't just tell the orders from the stoichiometry of any old chemical reaction. The exception is when the reaction takes place in only one step or you know that one step in the reaction is rate determining and then you can use its stoichiometry for the rate equation. In other words you can't assume it is - you need more information
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#7
(Original post by EierVonSatan)
The most important thing to remember about rates/orders is that they can only be determined through experiment. You can't assume which step would be faster, unless of course you have been told that it is Also, using the RDS of a multistep reaction to form the rate equation only works when one step is significantly faster than every other step, if they're at all similar then you can't use the RDS alone as the basis for the rate equation.

Your teacher is quite correct, you can't just tell the orders from the stoichiometry of any old chemical reaction. The exception is when the reaction takes place in only one step or you know that one step in the reaction is rate determining and then you can use its stoichiometry for the rate equation. In other words you can't assume it is - you need more information
Alright cool, cheers for the help man
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