# rate equation

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Thread starter 5 years ago
#1
in the rate equation im confused in what the power for a certain reactant means...
rate = k(NO)^2
my teacher said the two has nothing to do with the number of molecules/mols and its just the order of the reactant to the reaction but when are writing the equation we're using it as the number of molecules reacted (2 moles)
i guess im not exactly sure what order of reactant is.. can someone please expalin in details.. thanks x
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5 years ago
#2
The order of the reaction is how many molecules of the reactant are used in the rate determining step. This is the step of the reaction which is the slowest.
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Thread starter 5 years ago
#3
(Original post by alow)
The order of the reaction is how many molecules of the reactant are used in the rate determining step. This is the step of the reaction which is the slowest.
But i thought when the order of a certain reactant is lets say zero it means that no matter how much we increased the concentration there will be no effect on the reaction.. But it will be used up anyway since its a reactant . is this right or wrong..?
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5 years ago
#4
(Original post by pondsteps)
But i thought when the order of a certain reactant is lets say zero it means that no matter how much we increased the concentration there will be no effect on the reaction.. But it will be used up anyway since its a reactant . is this right or wrong..?
In a reaction, there is usually one step which occurs much much slower than the others. This step does not need to use all of the different reactants.

Yes, if a reactant is zero order, no matter how much you increase the concentration, the reaction will not occur any faster. This is because it is not involved in the rate determining step. However it is used in a different, faster, step so will be used as the reaction proceeds.

No matter how fast the other steps are, a reaction can never proceed faster than the slowest step.
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5 years ago
#5
(Original post by alow)
In a reaction, there is usually one step which occurs much much slower than the others. This step does not need to use all of the different reactants.

Yes, if a reactant is zero order, no matter how much you increase the concentration, the reaction will not occur any faster. This is because it is not involved in the rate determining step. However it is used in a different, slower, step so will be used as the reaction proceeds.

No matter how fast the other steps are, a reaction can never proceed faster than the slowest step.
Typo "faster".
1
5 years ago
#6
(Original post by charco)
Typo "faster".
Thanks
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Thread starter 4 years ago
#7
(Original post by charco)
Typo "faster".
(Original post by alow)
In a reaction, there is usually one step which occurs much much slower than the others. This step does not need to use all of the different reactants.

Yes, if a reactant is zero order, no matter how much you increase the concentration, the reaction will not occur any faster. This is because it is not involved in the rate determining step. However it is used in a different, faster, step so will be used as the reaction proceeds.

No matter how fast the other steps are, a reaction can never proceed faster than the slowest step.
i got confused here (pic attached) .. no2 has an order of 2 therefore two molecules of NO2 were used in the rate determining step (also means that if NO2 increased by *2 the rate increases by *4 ?????? is this right?) and zero molecules of CO were used... but the mol ratio shows that its a one to one ratio therfore dont two mols of CO also have to be also used when two mols of NO2 are used???
(sorry im all over the place here but im struggling to understand this :/)
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4 years ago
#8
(Original post by pondsteps)
i got confused here (pic attached) .. no2 has an order of 2 therefore two molecules of NO2 were used in the rate determining step (also means that if NO2 increased by *2 the rate increases by *4 ?????? is this right?) and zero molecules of CO were used... but the mol ratio shows that its a one to one ratio therfore dont two mols of CO also have to be also used when two mols of NO2 are used???
(sorry im all over the place here but im struggling to understand this :/)
the bottom of the image shows that the overall reaction happens in two steps, notice that the second step can't take place before the first one because NO3 is a product in step 1 and a reactant instep 2. This means the overall rate of step 2 (and the rate of the reaction) is limited by how fast step 1 can produce NO3. Step 1 happens slowly, so it is the rate determining step. Step 1 only involves NO2 and so only NO2 appears in the rate equation.

So yes! 2 molecules of NO2 in the rate determining step and yes! doubling concentration of NO2 ought to quadruple the rate!

From these two equations it may look like 2 moles of NO2 have reacted and only 1 mole of CO, however, step 2 produces a mole of NO2 so overall the 1:1 ratio is what you see from this pair of reactions.

Prepare yourself for a bit of a wacky analogy for rate determining steps:

Imagine that you and a friend are making a cake with 2 layers, your friend is making the first layer (think of this like the first step in the reaction) and you are making the second layer (like the second step in the reaction).

Every cake has to have the first layer made by your friend and the second layer made by you.

So you set off, making one cake at a time, but it turns out your friend is very slow at making cakes and you are quite fast.

That would mean you would finish your first cake but you'd have to wait for your friend to finish theirs for the first layer, before you can put the second layer on top.

IE the overall rate of cake making (overall reaction rate) is limited by your friends cake making (by step 1) so making step 2 quicker (increasing the concentration of CO to speed up step 2 of the reaction) has no effect on the overall rate at which cakes can be made.

Hopefully that explains why CO is zeroeth order, because speeding up step 2 doesn't make any difference if step 1 isn't supplying the NO3 any faster!

So yes the overall reaction uses 1 mole of each, but the rate determining step (the slowest step) only uses 2 molecules of NO2.
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4 years ago
#9
Could anyone help my friend finish this please he doesn't know how to and doesn't know if the ones he has done are right or not.
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Thread starter 4 years ago
#10
(Original post by MexicanKeith)
the bottom of the image shows that the overall reaction happens in two steps, notice that the second step can't take place before the first one because NO3 is a product in step 1 and a reactant instep 2. This means the overall rate of step 2 (and the rate of the reaction) is limited by how fast step 1 can produce NO3. Step 1 happens slowly, so it is the rate determining step. Step 1 only involves NO2 and so only NO2 appears in the rate equation.

So yes! 2 molecules of NO2 in the rate determining step and yes! doubling concentration of NO2 ought to quadruple the rate!

From these two equations it may look like 2 moles of NO2 have reacted and only 1 mole of CO, however, step 2 produces a mole of NO2 so overall the 1:1 ratio is what you see from this pair of reactions.

Prepare yourself for a bit of a wacky analogy for rate determining steps:

Imagine that you and a friend are making a cake with 2 layers, your friend is making the first layer (think of this like the first step in the reaction) and you are making the second layer (like the second step in the reaction).

Every cake has to have the first layer made by your friend and the second layer made by you.

So you set off, making one cake at a time, but it turns out your friend is very slow at making cakes and you are quite fast.

That would mean you would finish your first cake but you'd have to wait for your friend to finish theirs for the first layer, before you can put the second layer on top.

IE the overall rate of cake making (overall reaction rate) is limited by your friends cake making (by step 1) so making step 2 quicker (increasing the concentration of CO to speed up step 2 of the reaction) has no effect on the over rate at which cakes can be made.

Hopefully that explains why CO is zeroeth order, because speeding up step 2 doesn't make any difference if step 1 isn't supplying the NO3 any faster!

So yes the overall reaction uses 1 mole of each, but the rate determining step (the slowest step) only uses 2 molecules of NO2.
OMG THANK YOU THANK YOU THANK YOU. I was very confused and you helped so much!!!!!! xxxx
0
4 years ago
#11
(Original post by pondsteps)
OMG THANK YOU THANK YOU THANK YOU. I was very confused and you helped so much!!!!!! xxxx
No problem
I hope thinking about cakes helped
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Thread starter 4 years ago
#12
(Original post by MexicanKeith)
No problem
I hope thinking about cakes helped
yes deffo did
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