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Rate constant

I'm reading about an experiment involving k and it says for a reaction between HCl and Mg(s) the HCl needs to be in excess to ensure that the rate of reaction is constant. But surely as the Mg is used up the rate of reaction will decrease anyway as there is less chance of collisions etc?


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Reply 1
Imagine you had 1 mol of Mg + 2 mol of HCl, as the reaction proceeded, the rate would decrease since the surface area of the Mg decreases AND because the [HCl] decreases.

Now imagine repeating it with 1000 mol of HCl. The rate would decrease as the S.A. decreases, but the [HCl] would barely change and would barely have an effect.

How much does the S.A. change up to the point that 90% remains? Not by much* and hence rate is barely affected.

Now, you should be asking, what if you had 2 mol of HCl, how much would [HCl] change by up to the point that 90% of the Mg remains, you'd work out 1.8 mol remains which you might conclude is only a small change, but since it is second order WRT [H+] is quite a big difference, in fact rate = 0.81 x initial rate.

Perhaps a better Q would have been to say "to ensure that the rate of reaction is as close to constant as possible"

*a quick and dirty calc suggests that it is in the region of 0.93 x initial rate.
Original post by Pigster
Imagine you had 1 mol of Mg + 2 mol of HCl, as the reaction proceeded, the rate would decrease since the surface area of the Mg decreases AND because the [HCl] decreases.

Now imagine repeating it with 1000 mol of HCl. The rate would decrease as the S.A. decreases, but the [HCl] would barely change and would barely have an effect.

How much does the S.A. change up to the point that 90% remains? Not by much* and hence rate is barely affected.

Now, you should be asking, what if you had 2 mol of HCl, how much would [HCl] change by up to the point that 90% of the Mg remains, you'd work out 1.8 mol remains which you might conclude is only a small change, but since it is second order WRT [H+] is quite a big difference, in fact rate = 0.81 x initial rate.

Perhaps a better Q would have been to say "to ensure that the rate of reaction is as close to constant as possible"

*a quick and dirty calc suggests that it is in the region of 0.93 x initial rate.


Oh ok so is this for calculating initial rate rather than overall rate?
Is the calculation always initial rate?


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Reply 3
If you have a large excess of acid, the conc of said acid will be more or less the same at the start as at the end, so the change in conc will hardly affect the rate. There's nothing you can do about the decrease in rate due to surface area, but at least will be controlled when you do fair tests.

Initial rates are better as they allow you to work out k, but you can compare reactions using average rate to see the effects of changing temp/conc etc.
Original post by Pigster
If you have a large excess of acid, the conc of said acid will be more or less the same at the start as at the end, so the change in conc will hardly affect the rate. There's nothing you can do about the decrease in rate due to surface area, but at least will be controlled when you do fair tests.

Initial rates are better as they allow you to work out k, but you can compare reactions using average rate to see the effects of changing temp/conc etc.


Is that why you talked about up to the point where 90% of the mg remains? Because you were using initial rate? Sorry I'm just freaking out now I feel like I know nothing.


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