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# Titration Watch

1. (Original post by ZakRob)
The answer if calculated with moles, is however though 0.583. so your method is of working is correct. You just need to technically calculate concentration, then use the Kc equation.
but the volume is 1 so the values would be the same as the moles
2. (Original post by otrivine)
but the volume is 1 so the values would be the same as the moles
In that case then yes, it never mentioned the volume as 1, that's why I was unsure

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3. (Original post by ZakRob)
In that case then yes, it never mentioned the volume as 1, that's why I was unsure

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what would you said for part c)i)ii) ?

Kc do you write an expression with liquid or solid or is it only gas and aqueous conditions
4. (Original post by otrivine)
what would you said for part c)i)ii) ?

Kc do you write an expression with liquid or solid or is it only gas and aqueous conditions
What are the answers for aii and bi?so I can make a decision

As far as my knowledge stretches being taught off the ocr syllabus, I've only dealt with either gases and aqueous conditions for Kc.

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5. (Original post by ZakRob)
What are the answers for aii and bi?so I can make a decision

As far as my knowledge stretches being taught off the ocr syllabus, I've only dealt with either gases and aqueous conditions for Kc.

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is the first one that Kc is only affected by temperature?

part ii) concentration of product increases and equilibrium shifts from left to right/
6. (Original post by otrivine)
is the first one that Kc is only affected by temperature?

part ii) concentration of product increases and equilibrium shifts from left to right/
So if temperature only affect Kc, and when at the higher temperature the equilibrium shifted to the right increasing amount of product. This means the right side is endothermic, and the reaction is endothermic
This is because, according to la chatliers principle, the system will try to minimise changes, and in this instance when temp is increased the system will try to reduce that, by moving to the endothermic end
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7. (Original post by ZakRob)
So if temperature only affect Kc, and when at the higher temperature the equilibrium shifted to the right increasing amount of product. This means the right side is endothermic, and the reaction is endothermic

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i think its exothermic, because there was this part question also a) A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed. and this was at higher temp and the previous one was at lower temp. The value of Kc at this was 1.53 so it decreased which means its exo?
8. (Original post by otrivine)
i think its exothermic, because there was this part question also a)A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed. and this was at higher temp and the previous one was at lower temp. The value of Kc at this was 1.53 so it decreased which means its exo?
Wheres this question from? So I can have a look at the whole thing. Then I'll be able to give you a confirmative answer.

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9. (Original post by ZakRob)
Wheres this question from? So I can have a look at the whole thing. Then I'll be able to give you a confirmative answer.

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Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

a) A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

i) Write down an expression for Kc for this reaction (1 mark)
ii) Calculate the concentrations of methanol and carbon monoxide present at equilibrium. Show your working.
(4 marks)
iii) Hence calculate Kc (to 3 significant figures) including its units
(2 marks)

i) The value of the equilibrium constant for formation of ethanoic acid
(2 marks)
ii) The equilibrium yield of ethanoic acid.
(2 marks)

c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted.
i) Calculate the value of Kc at the lower temperature. Give your answer to 3 significant figures.
(6 marks)
(2 marks)

this is whole question
10. (Original post by otrivine)
Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

a)A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

i)Write down an expression for Kc for this reaction (1 mark)
ii)Calculate the concentrations of methanol and carbon monoxide present at equilibrium. Show your working.
(4 marks)
iii)Hence calculate Kc (to 3 significant figures) including its units
(2 marks)

i)The value of the equilibrium constant for formation of ethanoic acid
(2 marks)
ii)The equilibrium yield of ethanoic acid.
(2 marks)

c)Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted.
i)Calculate the value of Kc at the lower temperature. Give your answer to 3 significant figures.
(6 marks)
(2 marks)

this is whole question
It is exothermic. I say this now because I worked out the second Kc value using the volume of 5dm3, which gave a Kc value of 2.91 this compared to the Kc value of 1.52 at a higher temp shows that the exothermic side is favored at lower temperature, and this occurred with the increase of product

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11. (Original post by ZakRob)
It is exothermic. I say this now because I worked out the second Kc value using the volume of 5dm3, which gave a Kc value of 2.91 this compared to the Kc value of 1.52 at a higher temp shows that the exothermic side is favored at lower temperature, and this occurred with the increase of product

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No you mean higher temperature is 1.52 and the lower temperture one which said 77.6% that was 0.583.

where did u get 2.91 from?
12. Well from the whole question, for the 77% one, the volume isnt 1 it's 5 so you need to recalculate Kc correctly

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13. (Original post by ZakRob)
Well from the whole question, for the 77% one, the volume isnt 1 it's 5 so you need to recalculate Kc correctly

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You are confusing me now , what did you get for the 77, you said 0.853
14. (Original post by otrivine)
You are confusing me now , what did you get for the 77, you said 0.853
I said that on the presumption of the volume being 1, but after reading the whole question im thinking the volume is actually 5dm3, so then I had to recalculate kc and now I got 2.91. Do you see what I mean?

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15. (Original post by ZakRob)
I said that on the presumption of the volume being 1, but after reading the whole question im thinking the volume is actually 5dm3, so then I had to recalculate kc and now I got 2.91. Do you see what I mean?

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so are 1.53 for the first calculation and 0.583 for the second calculation wrong/yes or no
16. (Original post by otrivine)
so are 1.53 for the first calculation and 0.583 for the second calculation wrong/yes or no
1.53 is correct because we used 5dm3 as a volume. But 0.583 is wrong as we used 1 as the volume. For the second one the kc value is actually 2.91 when we use 5dm3 as a
Volume
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17. (Original post by ZakRob)
1.53 is correct because we used 5dm3 as a volume. But 0.583 is wrong as we used 1 as the volume. For the second one the kc value is actually 2.91 when we use 5dm3 as a
Volume
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the first one uses 5dm3 the second one does not or else they would have put it down?
18. (Original post by otrivine)
the first one uses 5dm3 the second one does not or else they would have put it down?
Not necessarily, my understanding is the same sized container is being used, same moles of reactants just the temperature is different

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19. (Original post by ZakRob)
Not necessarily, my understanding is the same sized container is being used, same moles of reactants just the temperature is different

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oh I see what you are saying now?! So we assume its also 5 dm3
20. (Original post by otrivine)
oh I see what you are saying now?! So we assume its also 5 dm3
Yes, since it's the same question and they havent said anything otherwise.

Does that clear everything up now?

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