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
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
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.
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
I've not yet, no. I'll be making one soon... it will have something in common with the first few sections of this video...ruclips.net/video/0iKtr3xXEj8/видео.htmlsi=7h4Q7-FTJRxKGl-Y
Good thought, but it wouldn't affect it if you always added the same concentration and volume of it, and if you kept the total volume of the reaction mixture constant. That way you'd always have the same H+ moles spread over the same volume
@@chemistrytutorby diluting do you mean the addition of water? Wouldn’t that increase the volume and overall decrease the concentration of the H+ ions? Or am I missing something?
@John-dx3fw when you do a titration, the major reason why you do it is to determine an amount in moles of something. Then maybe you'd work out a concentration perhaps, but not always. So if you stop the reaction by adding lots of water then the concentration will drop very low, which is why the reaction will effectively stop. But the amount of acid in the flask in moles will still be the same
@@chemistrytutorIs the amount of water added measured so you could account for it when calculating the concentration at a particular time? Also thank you for responding, it’s greatly appreciated.
@John-dx3fw you don't need the concentration of the contents of the conical flask. You just need to know the volume of sample you withdraw, and the moles of H+ that there are. That's the thing you work out the concentration of
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?
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
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
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
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
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
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 😃
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?
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
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the fact I'm learning this the day before the exam
the morning of*
@@DrWaluigiii 30 mins before
Hope the exams went well
wow now that's me this year😭
@@revision7685sammmmeeee
Grear video! Wish us good luck tomorrow😅
You got this!
Good luck 😃
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
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
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.
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
thank you soo muchhh, i really didnt understand this before
You're very welcome 🙏
hi sir, do you have any videos on required practical 5?
I've not yet, no. I'll be making one soon... it will have something in common with the first few sections of this video...ruclips.net/video/0iKtr3xXEj8/видео.htmlsi=7h4Q7-FTJRxKGl-Y
3:50, if you diluted it, wouldn’t that affect H+ concentration
Good thought, but it wouldn't affect it if you always added the same concentration and volume of it, and if you kept the total volume of the reaction mixture constant. That way you'd always have the same H+ moles spread over the same volume
@@chemistrytutorby diluting do you mean the addition of water? Wouldn’t that increase the volume and overall decrease the concentration of the H+ ions? Or am I missing something?
@John-dx3fw when you do a titration, the major reason why you do it is to determine an amount in moles of something. Then maybe you'd work out a concentration perhaps, but not always. So if you stop the reaction by adding lots of water then the concentration will drop very low, which is why the reaction will effectively stop. But the amount of acid in the flask in moles will still be the same
@@chemistrytutorIs the amount of water added measured so you could account for it when calculating the concentration at a particular time? Also thank you for responding, it’s greatly appreciated.
@John-dx3fw you don't need the concentration of the contents of the conical flask. You just need to know the volume of sample you withdraw, and the moles of H+ that there are. That's the thing you work out the concentration of
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?
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
@@chemistrytutor that’s great thank you very much I’ll make sure to talk about the graph!
for part d, shouldnt the value of k be a negative value, since its going down?
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
@@abbabbcbbcbb Thank you🙌 makes sense
Well explained dexometh!
@@chemistrytutor why thank you very much :)
@@abbabbcbbcbb good luck tomorrow
Could you also make a video like this with the practical 7b, the one with the gas syringe
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
@@chemistrytutor that’s ok, your videos have been so helpful! thanks 😊
@@AM-hn7gj thanks 😊
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
It’s a concentration time graph not a rate concentration graph. The line is only horizontal In a rate concentration graph!!
If you plot the gradient (rate) against concentration it would be a horizontal line!!
@@divineifashe3452 great i thought so thank you so much :)
Well explained!
legend
Thanks
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?
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
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
Same confusion
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
😃
@@chemistrytutor thanks! and this applies the same way across all graphs?
@@primeaccount4094 yes. When youre tracking a reactant concentration its a curve 2 of the 3 times, but linear for zero order
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?
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
love u bro
Thanks 😊
is it possible to apply this particular reaction on paper so that you can alternate printed shapes?
Hi, I'm not sure I understand your question, sorry 🤔
thanks sir :)
You're welcome 😊
thank uuu
Very welcome! Glad it was useful