Chemistry A2 Equilibirum question help
Hi, would anyone be able to tell me where I can access the mark scheme to these questions? https://gofile.io/?c=78mjOF
Or if not would anyone be willing to help me do some of the questions since I'm really struggling with a few of them. Thanks in advance.
Or if not would anyone be willing to help me do some of the questions since I'm really struggling with a few of them. Thanks in advance.
Original post by janejones659
Hi, would anyone be able to tell me where I can access the mark scheme to these questions? https://gofile.io/?c=78mjOF
Or if not would anyone be willing to help me do some of the questions since I'm really struggling with a few of them. Thanks in advance.
Or if not would anyone be willing to help me do some of the questions since I'm really struggling with a few of them. Thanks in advance.
Which ones are you struggling with?
Thanks for replying! Question 3b particularly...I don't understand the relevance of 1g/cm^3 or the "30cm^3 sulfuric acid contains 30cm^3 water"?
Original post by charco
Which ones are you struggling with?
A student is investigating the acid-catalysed hydrolysis of the ester methyl propanoate, CH3CH2COOCH3, to form propanoic acid, CH3CH2COOH, and methanol.
The equation for this process is:
CH3CH2COOCH3(l) + H2O(l) ⇌ CH3CH2COOH(l) + CH3OH(l)
0.200 mol of methyl propanoate are mixed with 30.0 cm3 of 5.0 mol dm3 sulfuric acid. The student allowed the mixture to reach equilibrium at 50 °C.
b 0.12 moles of propanoic acid are present at equilibrium. Calculate a value for the equilibrium constant, Kc, at this temperature.
You may assume that the total volume of the equilibrium mixture remains constant at 50 cm3 and that 30 cm3 sulfuric acid contains 30 cm3 water. AND 1g/cm3
Kc =
(5 marks)
Is this the question?
The equation for this process is:
CH3CH2COOCH3(l) + H2O(l) ⇌ CH3CH2COOH(l) + CH3OH(l)
0.200 mol of methyl propanoate are mixed with 30.0 cm3 of 5.0 mol dm3 sulfuric acid. The student allowed the mixture to reach equilibrium at 50 °C.
b 0.12 moles of propanoic acid are present at equilibrium. Calculate a value for the equilibrium constant, Kc, at this temperature.
You may assume that the total volume of the equilibrium mixture remains constant at 50 cm3 and that 30 cm3 sulfuric acid contains 30 cm3 water. AND 1g/cm3
Kc =
(5 marks)
Is this the question?
Yes this is the one
Original post by p_helena
A student is investigating the acid-catalysed hydrolysis of the ester methyl propanoate, CH3CH2COOCH3, to form propanoic acid, CH3CH2COOH, and methanol.
The equation for this process is:
CH3CH2COOCH3(l) + H2O(l) ⇌ CH3CH2COOH(l) + CH3OH(l)
0.200 mol of methyl propanoate are mixed with 30.0 cm3 of 5.0 mol dm3 sulfuric acid. The student allowed the mixture to reach equilibrium at 50 °C.
b 0.12 moles of propanoic acid are present at equilibrium. Calculate a value for the equilibrium constant, Kc, at this temperature.
You may assume that the total volume of the equilibrium mixture remains constant at 50 cm3 and that 30 cm3 sulfuric acid contains 30 cm3 water. AND 1g/cm3
Kc =
(5 marks)
Is this the question?
The equation for this process is:
CH3CH2COOCH3(l) + H2O(l) ⇌ CH3CH2COOH(l) + CH3OH(l)
0.200 mol of methyl propanoate are mixed with 30.0 cm3 of 5.0 mol dm3 sulfuric acid. The student allowed the mixture to reach equilibrium at 50 °C.
b 0.12 moles of propanoic acid are present at equilibrium. Calculate a value for the equilibrium constant, Kc, at this temperature.
You may assume that the total volume of the equilibrium mixture remains constant at 50 cm3 and that 30 cm3 sulfuric acid contains 30 cm3 water. AND 1g/cm3
Kc =
(5 marks)
Is this the question?
okay gimme a min to work it out 
Original post by janejones659
Yes this is the one
thank you
Original post by p_helena
okay gimme a min to work it out
Aight so with these questions it's really important that you establish a process which works for you.
For all the Ks (Ka, Kp, Kc, Kw and k) I made individual posters which I still refer to as they are laid out better than my textbook.
What you must remember is that Kc is calculated from equilibrium concentrations, not initial concentrations nor equilibrium/initial AMOUNTS.
So at the beginning you have 0.2 mol of methyl propanoate. It reacts in a 1+1--> 1+1 ratio with water to produce propanoic acid and methanol.
The whole thing about sulfuric acid is just something to throw you off. Because sulfuric acid is acting as a catalyst, not a reactant, it doesn't go into the expression. Look at the equation; the methyl propanoate reacts with water. The solution is aqueous; therefore it contains 30cm3 water as the volume of the actual sulfuric acid in the solution is probably negligible, hence you are given that assumption
Because you have 30cm3 water in a 50cm3 solution, the concentration of H20 in the entire solution at the start is 33.33 moldm3.
This is because 30cm3=30g water, so 30/18 (molar mass of water)= 1.67 moles. Divide this by 0.05 (dm3 volume of whole solution- says it's 50cm3 in question) and you get 33.33 moles per dm3.
Finding the initial conc of methyl propanoate is less complicated. Just divide the moles by the volume (0.2/0.05) and you get 4 moldm3.
Then for the products- you have 0.12 moles of propanoic acid at equilibrium. This means there are 2.4 mol/dm3 because 0.12/ 0.05=2.4.
Because the ratio of the products in the equation are the same, we can assume that 2.4 moldm3 of methanol has also been produced at equilibrium.
Initially there would have been no propanoic acid or methanol- so the change in moles/dm3 is +2.4.
Because the two reactants are also in the same stoichiometric quantity as the products, we can be sure that 2.4 mol/dm3 of each reactant has been lost.
so, as 33.33-2.4= 30.93 and 4-2.4= 1.6, we now have the four equilibrium concentrations.
The equation for Kc is
Kc= ([methanol] x [propanoic acid])/([water] x [methyl propanoate]).
as such, you have (2.4 x 2.4) / (30.93 x 1.6) which should give a value shown in the spoiler
I'm not entirely sure my answer is right as I haven't done A-Level chem for a while so I'm a bit rusty. Either way I hope it helps.
For all the Ks (Ka, Kp, Kc, Kw and k) I made individual posters which I still refer to as they are laid out better than my textbook.
What you must remember is that Kc is calculated from equilibrium concentrations, not initial concentrations nor equilibrium/initial AMOUNTS.
So at the beginning you have 0.2 mol of methyl propanoate. It reacts in a 1+1--> 1+1 ratio with water to produce propanoic acid and methanol.
The whole thing about sulfuric acid is just something to throw you off. Because sulfuric acid is acting as a catalyst, not a reactant, it doesn't go into the expression. Look at the equation; the methyl propanoate reacts with water. The solution is aqueous; therefore it contains 30cm3 water as the volume of the actual sulfuric acid in the solution is probably negligible, hence you are given that assumption
Because you have 30cm3 water in a 50cm3 solution, the concentration of H20 in the entire solution at the start is 33.33 moldm3.
This is because 30cm3=30g water, so 30/18 (molar mass of water)= 1.67 moles. Divide this by 0.05 (dm3 volume of whole solution- says it's 50cm3 in question) and you get 33.33 moles per dm3.
Finding the initial conc of methyl propanoate is less complicated. Just divide the moles by the volume (0.2/0.05) and you get 4 moldm3.
Then for the products- you have 0.12 moles of propanoic acid at equilibrium. This means there are 2.4 mol/dm3 because 0.12/ 0.05=2.4.
Because the ratio of the products in the equation are the same, we can assume that 2.4 moldm3 of methanol has also been produced at equilibrium.
Initially there would have been no propanoic acid or methanol- so the change in moles/dm3 is +2.4.
Because the two reactants are also in the same stoichiometric quantity as the products, we can be sure that 2.4 mol/dm3 of each reactant has been lost.
so, as 33.33-2.4= 30.93 and 4-2.4= 1.6, we now have the four equilibrium concentrations.
The equation for Kc is
Kc= ([methanol] x [propanoic acid])/([water] x [methyl propanoate]).
as such, you have (2.4 x 2.4) / (30.93 x 1.6) which should give a value shown in the spoiler
I'm not entirely sure my answer is right as I haven't done A-Level chem for a while so I'm a bit rusty. Either way I hope it helps.
Spoiler
Original post by p_helena
Aight so with these questions it's really important that you establish a process which works for you.
For all the Ks (Ka, Kp, Kc, Kw and k) I made individual posters which I still refer to as they are laid out better than my textbook.
What you must remember is that Kc is calculated from equilibrium concentrations, not initial concentrations nor equilibrium/initial AMOUNTS.
So at the beginning you have 0.2 mol of methyl propanoate. It reacts in a 1+1--> 1+1 ratio with water to produce propanoic acid and methanol.
The whole thing about sulfuric acid is just something to throw you off. Because sulfuric acid is acting as a catalyst, not a reactant, it doesn't go into the expression. Look at the equation; the methyl propanoate reacts with water. The solution is aqueous; therefore it contains 30cm3 water as the volume of the actual sulfuric acid in the solution is probably negligible, hence you are given that assumption
Because you have 30cm3 water in a 50cm3 solution, the concentration of H20 in the entire solution at the start is 33.33 moldm3.
This is because 30cm3=30g water, so 30/18 (molar mass of water)= 1.67 moles. Divide this by 0.05 (dm3 volume of whole solution- says it's 50cm3 in question) and you get 33.33 moles per dm3.
Finding the initial conc of methyl propanoate is less complicated. Just divide the moles by the volume (0.2/0.05) and you get 4 moldm3.
Then for the products- you have 0.12 moles of propanoic acid at equilibrium. This means there are 2.4 mol/dm3 because 0.12/ 0.05=2.4.
Because the ratio of the products in the equation are the same, we can assume that 2.4 moldm3 of methanol has also been produced at equilibrium.
Initially there would have been no propanoic acid or methanol- so the change in moles/dm3 is +2.4.
Because the two reactants are also in the same stoichiometric quantity as the products, we can be sure that 2.4 mol/dm3 of each reactant has been lost.
so, as 33.33-2.4= 30.93 and 4-2.4= 1.6, we now have the four equilibrium concentrations.
The equation for Kc is
Kc= ([methanol] x [propanoic acid])/([water] x [methyl propanoate]).
as such, you have (2.4 x 2.4) / (30.93 x 1.6) which should give a value shown in the spoiler
I'm not entirely sure my answer is right as I haven't done A-Level chem for a while so I'm a bit rusty. Either way I hope it helps.
For all the Ks (Ka, Kp, Kc, Kw and k) I made individual posters which I still refer to as they are laid out better than my textbook.
What you must remember is that Kc is calculated from equilibrium concentrations, not initial concentrations nor equilibrium/initial AMOUNTS.
So at the beginning you have 0.2 mol of methyl propanoate. It reacts in a 1+1--> 1+1 ratio with water to produce propanoic acid and methanol.
The whole thing about sulfuric acid is just something to throw you off. Because sulfuric acid is acting as a catalyst, not a reactant, it doesn't go into the expression. Look at the equation; the methyl propanoate reacts with water. The solution is aqueous; therefore it contains 30cm3 water as the volume of the actual sulfuric acid in the solution is probably negligible, hence you are given that assumption
Because you have 30cm3 water in a 50cm3 solution, the concentration of H20 in the entire solution at the start is 33.33 moldm3.
This is because 30cm3=30g water, so 30/18 (molar mass of water)= 1.67 moles. Divide this by 0.05 (dm3 volume of whole solution- says it's 50cm3 in question) and you get 33.33 moles per dm3.
Finding the initial conc of methyl propanoate is less complicated. Just divide the moles by the volume (0.2/0.05) and you get 4 moldm3.
Then for the products- you have 0.12 moles of propanoic acid at equilibrium. This means there are 2.4 mol/dm3 because 0.12/ 0.05=2.4.
Because the ratio of the products in the equation are the same, we can assume that 2.4 moldm3 of methanol has also been produced at equilibrium.
Initially there would have been no propanoic acid or methanol- so the change in moles/dm3 is +2.4.
Because the two reactants are also in the same stoichiometric quantity as the products, we can be sure that 2.4 mol/dm3 of each reactant has been lost.
so, as 33.33-2.4= 30.93 and 4-2.4= 1.6, we now have the four equilibrium concentrations.
The equation for Kc is
Kc= ([methanol] x [propanoic acid])/([water] x [methyl propanoate]).
as such, you have (2.4 x 2.4) / (30.93 x 1.6) which should give a value shown in the spoiler
I'm not entirely sure my answer is right as I haven't done A-Level chem for a while so I'm a bit rusty. Either way I hope it helps.
Spoiler
As all of the components are in the same homogeneous liquid solution and in a 1 to 1 to 1 to 1 ratio, you can forget about concentrations and work in moles
wait wth! this is the end of chapter test for chapter 19 that i did in october 2018! the answers are only held by the teachers lol!!!
i got a big fat U in this paper. i didnt attempt any of the questions. thanks, i'll do them again and get them marked by my teachers >:]
Original post by anonymoussse
wait wth! this is the end of chapter test for chapter 19 that i did in october 2018! the answers are only held by the teachers lol!!!
Bless your heart I know how it feels :/
Try not to think about it too much right now; if you overanalyse these things it makes you more confused. I would suggest you ask your teacher to go over it either in class or with you so that you can build your system of doing it.
Honestly with A-Level, it's not always a case of if you can finish the question and get an answer, but how many bits you get done. Even if you figure out the equilibrium amounts (or in some cases the units) you can get a couple of marks.
I'm assuming you just started learning this section? If you are in year 12 don't worry you have time to perfect this. I didn't learn it until January and had my exams in May/June and I managed to get it down, I'm sure that with time you will too.
Try not to think about it too much right now; if you overanalyse these things it makes you more confused. I would suggest you ask your teacher to go over it either in class or with you so that you can build your system of doing it.
Honestly with A-Level, it's not always a case of if you can finish the question and get an answer, but how many bits you get done. Even if you figure out the equilibrium amounts (or in some cases the units) you can get a couple of marks.
I'm assuming you just started learning this section? If you are in year 12 don't worry you have time to perfect this. I didn't learn it until January and had my exams in May/June and I managed to get it down, I'm sure that with time you will too.
Original post by anonymoussse
i got a big fat U in this paper. i didnt attempt any of the questions. thanks, i'll do them again and get them marked by my teachers >:]
thank you for the help
ty 
no i finished year 13 a month ago haha but ur advice is still true & relevant for year 12s who come across this thread
i think at a level , if u fail a chapter badly, ur confidence in the content becomes so low which affects ur performance in the exam sooo badly. even if uve mastered the chapter by now...& it's so easy to u.....u just have to make sure u learn it well the FIRST time round. in my OCR chem exam paper 1 i messed up the Kp question and it's because those failure feelings resurfaced...despite having done sooo many questions on kp. kp isnt hard at all as well. getting a U on a chapter plummets ur confidence during the a level exam. i was thinking "how did i get a U? do i really not know this stuff even now? was it that hard?" and poof the time was up lol
no i finished year 13 a month ago haha but ur advice is still true & relevant for year 12s who come across this thread i think at a level , if u fail a chapter badly, ur confidence in the content becomes so low which affects ur performance in the exam sooo badly. even if uve mastered the chapter by now...& it's so easy to u.....u just have to make sure u learn it well the FIRST time round. in my OCR chem exam paper 1 i messed up the Kp question and it's because those failure feelings resurfaced...despite having done sooo many questions on kp. kp isnt hard at all as well. getting a U on a chapter plummets ur confidence during the a level exam. i was thinking "how did i get a U? do i really not know this stuff even now? was it that hard?" and poof the time was up lol
Original post by p_helena
Bless your heart I know how it feels :/
Try not to think about it too much right now; if you overanalyse these things it makes you more confused. I would suggest you ask your teacher to go over it either in class or with you so that you can build your system of doing it.
Honestly with A-Level, it's not always a case of if you can finish the question and get an answer, but how many bits you get done. Even if you figure out the equilibrium amounts (or in some cases the units) you can get a couple of marks.
I'm assuming you just started learning this section? If you are in year 12 don't worry you have time to perfect this. I didn't learn it until January and had my exams in May/June and I managed to get it down, I'm sure that with time you will too.
Try not to think about it too much right now; if you overanalyse these things it makes you more confused. I would suggest you ask your teacher to go over it either in class or with you so that you can build your system of doing it.
Honestly with A-Level, it's not always a case of if you can finish the question and get an answer, but how many bits you get done. Even if you figure out the equilibrium amounts (or in some cases the units) you can get a couple of marks.
I'm assuming you just started learning this section? If you are in year 12 don't worry you have time to perfect this. I didn't learn it until January and had my exams in May/June and I managed to get it down, I'm sure that with time you will too.
(edited 4 years ago)
Oh I agree with you so much on this, we sensitive folks get roasted by OCR 
I think theres a lot of others who feel the same- this year's OCR Chem A Paper 1 was pure evil in general so dw, the grade boundaries *should* be lower.
I think theres a lot of others who feel the same- this year's OCR Chem A Paper 1 was pure evil in general so dw, the grade boundaries *should* be lower.
Original post by anonymoussse
ty no i finished year 13 a month ago haha but ur advice is still true & relevant for year 12s who come across this thread
i think at a level , if u fail a chapter badly, ur confidence in the content becomes so low which affects ur performance in the exam sooo badly. even if uve mastered the chapter by now...& it's so easy to u.....u just have to make sure u learn it well the FIRST time round. in my OCR chem exam paper 1 i messed up the Kp question and it's because those failure feelings resurfaced...despite having done sooo many questions on kp. kp isnt hard at all as well. getting a U on a chapter plummets ur confidence during the a level exam. i was thinking "how did i get a U? do i really not know this stuff even now? was it that hard?" and poof the time was up lol
no i finished year 13 a month ago haha but ur advice is still true & relevant for year 12s who come across this thread i think at a level , if u fail a chapter badly, ur confidence in the content becomes so low which affects ur performance in the exam sooo badly. even if uve mastered the chapter by now...& it's so easy to u.....u just have to make sure u learn it well the FIRST time round. in my OCR chem exam paper 1 i messed up the Kp question and it's because those failure feelings resurfaced...despite having done sooo many questions on kp. kp isnt hard at all as well. getting a U on a chapter plummets ur confidence during the a level exam. i was thinking "how did i get a U? do i really not know this stuff even now? was it that hard?" and poof the time was up lol
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