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# finding initial rate

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1. can someone pls explain how to find the rate here..
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i.e. tell me/us what you've tried and why you can't do it.
3. Did you get given a rate equation?
4. (Original post by Pigster)

i.e. tell me/us what you've tried and why you can't do it.
I dont understand how we do it without the rate equation, is it possible? Plus the conc of (H+) didnt double so that confused me even more, isnt it supposed to be like when one doubles the other doubles if its first order and so on...
Did you get given a rate equation?
Nope :/
6. (Original post by pondsteps)
Nope :/
The best way to figure it out is to look at a similar question on a mark scheme then try and work back from that. That way in the process of figuring it out it'll solidify your understanding
7. If I were to point out that this is (effectively) a clock reaction, would that help?
8. (Original post by Pigster)
If I were to point out that this is (effectively) a clock reaction, would that help?
What info would that give me? :///
9. [OH-] does not affect reaction rate, but it does affect reaction time.
10. (Original post by Pigster)
[OH-] does not affect reaction rate, but it does affect reaction time.
Can you pleaseee explain what a clock reaction is and how you knew it was a clock reaction, im so confused.. Bare with me im an only student and my teacher is so crap 😭
11. Clock reactions are those in which a certain amount of reaction takes place until something observable occurs.

The 'disappearing cross' reaction is one of the most famous - the reaction produces a sulfur precipitate until you can't see the cross. You don't know how much sulfur is made, but you can compare experiments as (roughly) the same amount of sulfur must be made in each experiment, so if the change is the same, but the time is different, you can work out the relative rates.

The 'iodine clock' reaction is more similar to your OP. There are various forms, but they involve the oxidation of I- to I2 by a slow reaction, then the rapid reduction back to I- using something like S2O32-. Once the S2O32- is used up, I2 accumulates and the starch indicator goes blue/black.

In your OP, it says in the Q that the reaction ends when all of the OH- is used up. Imagine if it were 1st or 2nd order WRT [OH-], as the OH- gets used up the rate would fall and, in a Xeno's paradox stylee, the reaction would never end. But end it does. I guess for me, the easy way I think about it is that [OH-] goes from a tiny number, OH- gets used up and [OH-] falls to zero. If [OH-] is higher, it means that there is more OH- that needs using up, so naturally will take longer. Twice as much OH- would take twice as long. It is this idea that a certain amount of reaction takes place before something happens which made me describe it as a clock reaction.

The other thing that jumps out to me is that the Q MUST be answerable with the information given, since it is taken from an AQA exam paper (or at least it is their layout). You are not told the order WRT [OH-] nor are given any way (data) of working it out, therefore it can't be important to working out the rate constant.
12. (Original post by Pigster)
Clock reactions are those in which a certain amount of reaction takes place until something observable occurs.

The 'disappearing cross' reaction is one of the most famous - the reaction produces a sulfur precipitate until you can't see the cross. You don't know how much sulfur is made, but you can compare experiments as (roughly) the same amount of sulfur must be made in each experiment, so if the change is the same, but the time is different, you can work out the relative rates.

The 'iodine clock' reaction is more similar to your OP. There are various forms, but they involve the oxidation of I- to I2 by a slow reaction, then the rapid reduction back to I- using something like S2O32-. Once the S2O32- is used up, I2 accumulates and the starch indicator goes blue/black.

In your OP, it says in the Q that the reaction ends when all of the OH- is used up. Imagine if it were 1st or 2nd order WRT [OH-], as the OH- gets used up the rate would fall and, in a Xeno's paradox stylee, the reaction would never end. But end it does. I guess for me, the easy way I think about it is that [OH-] goes from a tiny number, OH- gets used up and [OH-] falls to zero. If [OH-] is higher, it means that there is more OH- that needs using up, so naturally will take longer. Twice as much OH- would take twice as long. It is this idea that a certain amount of reaction takes place before something happens which made me describe it as a clock reaction.

The other thing that jumps out to me is that the Q MUST be answerable with the information given, since it is taken from an AQA exam paper (or at least it is their layout). You are not told the order WRT [OH-] nor are given any way (data) of working it out, therefore it can't be important to working out the rate constant.
Is the mechanism of the reaction of iodine and propane related to the iodine clock???
13. I guess you meant with propanone, but I guess you could turn it into a clock by adding starch and when the I2 has been used it, the colour would vanish, I've never thought to try it.

I'm also a little wary about your usage of the word mechanism, it suggests you aren't sure what it means.
14. (Original post by Pigster)
[OH-] does not affect reaction rate, but it does affect reaction time.
So a clock rxn is when one of the reactants doesnt effect the ratw of rxn but only the time???? And in the iodine clock rxn its the s2o3 -2 instead of H+??
15. In the iodine clock, the S2O32- doesn't even get involved in the reaction that determines the rate of the reaction.

Simplifying it as following:

Rxn 1 A -> B
Rxn 2 B + C -> A + D
(C = S2O32- and is in a fixed quantity, when C is all gone, B builds up and the indicator changes)

Doubling the conc. of A will double the rate of rxn 1. This will double the creation of B and therefore double the rate of rxn 2. Therefore C will get used up twice as quickly and the indicator changes quicker.

Doubling the conc. of C (and hence doubling the amount) has no effect on the rate of rxn 1, it simply means that rxn 1 has to produce twice the amount of B (in roughly twice the time) before all of C is used up.

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