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Rate equation question? thanks :)

2009 7 writ rate eqn.PNG

the answer given is below the question but I don't understand it

I know it is the slow rate determining step that relates to the order of the reaction but since H+ ions are not actually reactants in the reaction (ie. not in the overall balanced equation) I didn't think they'd be in the rate equation :s

also would you get the mark for using [H+] instead of [H3O+] ?
thnx
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ah4p
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2009 7 writ rate eqn MORE.PNG

ALSO just seen that the rest of the queston says the H+ ions are a catalyst
i didnt think a catalyst would be included in rate equation i thought it would just be reactants

thanks again
Original post by ah4p
2009 7 writ rate eqn.PNG

the answer given is below the question but I don't understand it

I know it is the slow rate determining step that relates to the order of the reaction but since H+ ions are not actually reactants in the reaction (ie. not in the overall balanced equation) I didn't think they'd be in the rate equation :s

also would you get the mark for using [H+] instead of [H3O+] ?
thnx


As you can see the RDS contains propanone and H+. This means that H+ should also be in the rate equation.

Why does H+ appear in it, you ask?
Catalysts can appear in rate equations because they affect the rate. They simply don't appear in chemical/stoichiometric equations because they remain chemically unchanged.

Similarly, I2 despite being in the chemical equation is not in the rate equation, as it is not in the slow step. The reaction is in first order with respect to propanone and H+ but zero order in respect to iodine.
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ah4p
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Original post by thymolphthalein
As you can see the RDS contains propanone and H+. This means that H+ should also be in the rate equation.

Why does H+ appear in it, you ask?
Catalysts can appear in rate equations because they affect the rate. They simply don't appear in chemical/stoichiometric equations because they remain chemically unchanged.

Similarly, I2 despite being in the chemical equation is not in the rate equation, as it is not in the slow step. The reaction is in first order with respect to propanone and H+ but zero order in respect to iodine.


ok thnku vvvv much :smile:

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