Iodine Clock | AQA Required Practical 7 | A level Chemistry | Question Walkthrough

[@AvaSun-yo2wy]

the fact I'm learning this the day before the exam

[@DrWaluigiii]

the morning of*

[@Ar-wu5lf]

@@DrWaluigiii 30 mins before

[@chemistrytutor]

Hope the exams went well

[@revision7685]

wow now that's me this year😭

[@saml5395]

​@@revision7685sammmmeeee

[@andreinecula9492]

Grear video! Wish us good luck tomorrow😅

[@chemistrytutor]

You got this! Good luck 😃

[@katheriner6456]

Just wondering if it’s possible to come up questions from topics / sub topics that has already came up? Also Thank you so much for the video this helped me a lot❤You are so kind.

[@chemistrytutor]

You're very welcome. No, I think it's unlikely topics would come up again (certainly in anything other than multiple choice). There's so much on the course, they'll never test it all so they wouldn't duplicate by asking something twice

[@J5gaming]

for the last rate of reaction question dont you have to half the rate as in the reaction it is 2H+ , the rate of reaction for a reaction eg. of A + 2B = 3C would be rate = -(d[A]/dt) = -(1/2)(d[B]/dt) = 1/3(d[C]/dt) because rate is always with respect to 1 mol not 2 or 3

[@chemistrytutor]

You can't make that kind of connection between the chemical equation and the order What we do know is that the gradient will show us the rate at which the H+ is being used up. So in this question the rate is being expressed as rate of change of concentration of H+. From that we deduce that [H+] is changing at a constant rate and so is zero order in spite of the 2 in the equation

[@cristinaax]

thank you soo muchhh, i really didnt understand this before

[@chemistrytutor]

You're very welcome 🙏

[@joshuahavenhand9570]

Thank you very much for taking the time to make these videos! Just a quick question on the analysis of the last question, would you get marks for talking about calculating initial rates and then how you could calculate orders by the change of the initial rate when the concentration of A is changed, Rather than looking at the shape of the graph?

[@chemistrytutor]

Depending on what you say you could get the first of the points, maybe getting you that 5th mark. For the 6 you really need to talk about the graph here. Remember in clock reactions particularly iodine Clock, you're only taking 1 measurement, which is the time take for it to go blue black. And so, this value itself (which we use to plot a graph) is actually an initial rate (we don't usually call it that though). The reaction actually continues after we stop the timer. Hope that helps

[@joshrufc6522]

@@chemistrytutor that’s great thank you very much I’ll make sure to talk about the graph!

[@ubxyy]

legend

[@chemistrytutor]

Thanks

[@AM-hn7gj]

Could you also make a video like this with the practical 7b, the one with the gas syringe

[@chemistrytutor]

I'll do my best, but not likely I will have time. I've an RP2 video lined up to do, but your idea is good too

[@AM-hn7gj]

@@chemistrytutor that’s ok, your videos have been so helpful! thanks 😊

[@chemistrytutor]

@@AM-hn7gj thanks 😊

[@michaelmuzusa698]

for part d, shouldnt the value of k be a negative value, since its going down?

[@abbabbcbbcbb]

the gradient is negative however its a rate of reaction so it's usually given as positive, the mark scheme would likely say *ignore negative signs* so you wouldn't be penalised for it, however because we know it's the rate with respect to the reactants, the negative is kinda implied

[@michaelmuzusa698]

@@abbabbcbbcbb Thank you🙌 makes sense

[@chemistrytutor]

Well explained dexometh!

[@abbabbcbbcbb]

@@chemistrytutor why thank you very much :)

[@chemistrytutor]

@@abbabbcbbcbb good luck tomorrow

[@shandana4661]

Hello Sir, does the concentration of X need to be the same as the concentration of A everytime you repeat with a new concentration of A so that we can see the colour change when all of A has reacted?

[@chemistrytutor]

I dont think you need to go into that much depth for the question. You definitely want the same conc and vol of X each time you do a repeat. Really, you don't want to be anywhere near the same conc as A as you need to make sure you use up all your X in good time

[@ojj7401]

Thanks again! Reckon we should know the method?

[@chemistrytutor]

I think there's a good chance a similar question could come up yes. 👍

[@user-jp6lq3vh6v]

Hi sir, a bit confused at 5:17 - I thought that a straight line graph of concentration against time shows that the reactant is first order, not 0 order? I thought that a 0 order graph would give a straight horizontal graph

[@divineifashe3452]

It’s a concentration time graph not a rate concentration graph. The line is only horizontal In a rate concentration graph!!

[@divineifashe3452]

If you plot the gradient (rate) against concentration it would be a horizontal line!!

[@user-jp6lq3vh6v]

@@divineifashe3452 great i thought so thank you so much :)

[@chemistrytutor]

Well explained!

[@primeaccount4094]

a bit confused as to when you calculate the gradient - for e.g. in part d) you took away 0.0 - 0.5 but in part e) u took away 0.46 - 0.0 how do when to take away the bigger number away from smaller or the other way around. thanks

[@AB-dn6fv]

Same confusion

[@chemistrytutor]

Worth clarifying 👌 d) was to find the gradient of a straight line so can work it out for the whole line e) was the gradient of a tangent to a curve. So I added a straight line, made it into a triangle and that intercepted the y axis at 0.46 😃

[@primeaccount4094]

@@chemistrytutor thanks! and this applies the same way across all graphs?

[@chemistrytutor]

@@primeaccount4094 yes. When youre tracking a reactant concentration its a curve 2 of the 3 times, but linear for zero order

[@m_za6878]

hey you know for the last question, you measure vol on x and vol of c but what do you add x and c to?

[@chemistrytutor]

You add them all together. So you've got A+B+X in one beaker, no rush as reaction won't start, THEN you add C (pre measured in a separate container) to the beaker and start the timer

[@diomisisalisandratos8362]

is it possible to apply this particular reaction on paper so that you can alternate printed shapes?

[@chemistrytutor]

Hi, I'm not sure I understand your question, sorry 🤔

[@stephanie4548]

thanks sir :)

[@chemistrytutor]

You're welcome 😊

[@mikaielshuaib1307]

love u bro

[@chemistrytutor]

Thanks 😊

[@maira_ae]

thank uuu

[@chemistrytutor]

Very welcome! Glad it was useful