Titration

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

In that case then yes, it never mentioned the volume as 1, that's why I was unsure

Posted from TSR Mobile

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.

Posted from TSR Mobile

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

Posted from TSR Mobile

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.

Posted from TSR Mobile

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)

b)How, if any, would addition of extra carbon monoxide affect the following? Justify your answers.

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)
ii)Use your answers to aii) and bi) to decide whether the reaction was exothermic or endothermic. Explain your answer.
(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

Posted from TSR Mobile

Well from the whole question, for the 77% one, the volume isnt 1 it's 5 so you need to recalculate Kc correctly

Posted from TSR Mobile

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?

Posted from TSR Mobile

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
Posted from TSR Mobile

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

Posted from TSR Mobile

oh I see what you are saying now?! So we assume its also 5 dm3