chem alevel question

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helloo
pls help
i’ve been stuck on this question for like 30 minutes trying to figure it out

the question is to find the order of (CO)
and in the first question worked out that (Hb) is 1st order

The answer is that CO is also 1st order but i don’t get why,
i thought you’d have to apply Hb, to a rate and then from that intermediate rate work out the effect of CO but that didn’t work 🫡

how would you work this out?
thank you in advance

(edited 11 months ago)

Reply 1

George H.

The attachment doesn't seem to work. Do you want to reattach it, or copy it in text or something?

Reply 2

hihihihihuhh

Original post by George H.

The attachment doesn't seem to work. Do you want to reattach it, or copy it in text or something?

yeahh my bad,
A68396E9-2EFE-4A07-AA1F-A0A14E8D82A6.jpg.jpeg

i’ll do both in case it doesn’t work again (:

all in mol/dm-3
exp1: HB: 2.09x10-6. CO: 1.4x10-6 Rate: 8.20x10-7

exp2: HB: 4.18x10-6. CO: 1.4x10-6 Rate: 1.64x10-6

exp3: HB:3.26x10-6. CO:2.8x10-6 Rate: 2.56 x10-6

(edited 11 months ago)

Reply 3

user23184

Hi!

So you've been able to deduce that Hb is first order.

For the third experiment, Hb is multiplied by a factor of 1.56 (you can do this by simply dividing 3.26 by 2.09). If you divide the rate by this factor, you would get 2.56x10^-6 / 1.56x10^-6, which is equal to 1.64x10^6. As you can see in the second experiment, this value is double the rate of the initial reaction, and hence CO alone has doubled the rate and doubled in concentration, making CO first order as well.

Sorry I know its a strange way to get that answer and I'm not sure if there's another way to solve this, but I hope that makes sense as to why it would be first order for CO

Reply 4

xivst

Original post by hihihihihuhh

yeahh my bad,
A68396E9-2EFE-4A07-AA1F-A0A14E8D82A6.jpg.jpeg

I have no idea if this is the right way of working it out but I did:
3.26e-6/4.18e-6=163/209
then multiplying that by 1.64e-6 to get 1.279e-6
then 2.56e-6/1.279e-6 to get 2.001 for the multiplying factor of the rate as co doubles giving first order

Sorry if it's a very wrong way of explaining it but I did a similar question to this and used the thought process above

Reply 5

TypicalNerd

Original post by hihihihihuhh

yeahh my bad,
A68396E9-2EFE-4A07-AA1F-A0A14E8D82A6.jpg.jpeg

Calculate the following:

-How many times larger [Hb] is in experiment 3 than it is in experiment 2. Let’s call this number a.

-How many times larger [CO] is in experiment 3 than it is in experiment 2. Let’s call this number b.

-How many times larger the rate is in experiment 3 than it is in experiment 2. Let’s call this number c.

Using typical rate law rules, a^(order of Hb) x b^(order of CO) = c

Now work out the order of CO.