# Question on equilibrium

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#1
Browsing through the section B of a practical paper + mark scheme and stumbled on this question.

Its 3(B) that gets me, i was always under the impression that a dynamic equilibrium only applies to reversible reactions (Which this one clearly is not given the answer to Part A).

Anyone care to explain this as it has really irked me.
0
6 years ago
#2
(Original post by Mr.bob)
Browsing through the section B of a practical paper + mark scheme and stumbled on this question.

Its 3(B) that gets me, i was always under the impression that a dynamic equilibrium only applies to reversible reactions (Which this one clearly is not given the answer to Part A).

Anyone care to explain this as it has really irked me.
It's one of the principles of equilibrium. Add to the right get more of the left ...

You can explain it using the euqilibrium law expression, kc.

If you increase the water then the quotient no longer equals the equilibrium constant and the system must move to the LHS to restore the value of kc.
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#3
(Original post by charco)
It's one of the principles of equilibrium. Add to the right get more of the left ...

You can explain it using the euqilibrium law expression, kc.

If you increase the water then the quotient no longer equals the equilibrium constant and the system must move to the LHS to restore the value of kc.
silly question in bound, but what is kc

incidentally, is this reaction reversible?
0
6 years ago
#4
(Original post by Mr.bob)
silly question in bound, but what is kc

incidentally, is this reaction reversible?
Difficult to understand equilibrium without knowing the equilibrium law (law of mass action)

kc = [products]coefficient/[reactants]coefficient

eg for a system at equilibrium:

2A <==> 3B

kc = B3/A2

The only thing that affects the value of kc for a given equilibrium is temperature.
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#5
(Original post by charco)
Difficult to understand equilibrium without knowing the equilibrium law (law of mass action)

kc = [products]coefficient/[reactants]coefficient

eg for a system at equilibrium:

2A <==> 3B

kc = B3/A2

The only thing that affects the value of kc for a given equilibrium is temperature.
sorry to bother you, but now im even more more worried for the exams as everything youve mentioned may as well be an alien language

i dont recognise any of that as being part if the ccea AS syllabus
0
6 years ago
#6
(Original post by Mr.bob)
sorry to bother you, but now im even more more worried for the exams as everything youve mentioned may as well be an alien language

i dont recognise any of that as being part if the ccea AS syllabus
Look, I don't know the spec for your exam, so you probably don't need it.

In which case you have probably just used Le Chatelier's principle to work out equilibrium changes of condition ...
1
6 years ago
#7
(Original post by Mr.bob)
Browsing through the section B of a practical paper + mark scheme and stumbled on this question.

Its 3(B) that gets me, i was always under the impression that a dynamic equilibrium only applies to reversible reactions (Which this one clearly is not given the answer to Part A).

Anyone care to explain this as it has really irked me.
We don't have to know the equilibrium constant and stuff for AS as far as I know (if we did I'm effed lol).

Water is a product of this reaction and if there's water to begin with that essentially means the concentration of product has increased. The equilibrium now favours backward reaction to counteract the increase in concentration of product according to Le Chatelier's principle.

According to my teacher, every reaction is reversible as in there's always some backward reaction going on but many reactions equilibrium is farrrr to the right so we don't call it reversible (backward reaction is rare to occur naturally so neglected). My GUESS here is normally there isn't significant backward reaction going on with pure reactants and no disturbance in system (which is the case in this process), but addition of water causes disturbance in the system as water is part of the reaction, encouraging backward reaction more that we can now call it reversible

this is just an educated guess by assuming what my teacher said was correct and using my AS knowledge. I may well be totally wrong

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0
6 years ago
#8
This is an esterification reaction and background knowledge is required that it is poor yielding (reversible) reaction. The reversed direction is the hydrolysis of esters.

If your concern is why wasn't a double arrow used in the suggested answer, I suspect both a single headed and double headed arrows are accepted for marking purposes.
0
6 years ago
#9
(Original post by Metanoia)
This is an esterification reaction and background knowledge is required that it is poor yielding (reversible) reaction. The reversed direction is the hydrolysis of esters.
How would we know this though
It looks as though there's no background info at the beginning of the question and as far as my AS syllabus is concerned ester is not covered at all (I don't know about OP's syllabus but I assume it's not that much different from my board, Edexcel)

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0
6 years ago
#10
Esters are covered in the AS OCR Syllabus which i assume the OP is doing
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6 years ago
#11
(Original post by mattcandy3)
Esters are covered in the AS OCR Syllabus which i assume the OP is doing
Oh is it lol
We only do ester for IUPAC naming exercise.

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0
6 years ago
#12
Yes, its very basic though, making of esters etc, however with that knowledge and knowledge of Le Chatelier's principle, the question above should be able to be answered relatively easy

Hope OP, now understands it

(Original post by C0balt)
Oh is it lol
We only do ester for IUPAC naming exercise.

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