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    (Original post by Don Pedro K.)
    Attachment 549117

    Somebody tell me why it's not B?

    Funky_Giraffe samb1234
    I got C, 3 moles of ester at equilibrium so there are also 3 moles of water. Minus 3 from initial moles of both reactants gives you 1mol of methanoic acid and 3 moles of ethanol, I put them into kc expression and got 3 (C)
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    1 to 1 ratio: therefore equilibrium moles of methanol acid would be 4-3= 1 mol and ethanol 6-3 = 3 mol therefore moles of ester and water must be 3. overall experession is 9/3 which is 3
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    (Original post by imnoteinstein)
    would it be right to say thermodynamically stable and thermodynamically feasible are are opposites?
    yes
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    (Original post by Don Pedro K.)
    This question really got me too haha. If you work out the moles of the acid and alkali, you'll find that the acid is in excess. When you subtract the moles of alkali from the moles of acid to see how many moles of acid you have left over, you'll find that it is the same as the moles of ethanoate ions formed. Thus, [HA] = [A-], which means that Ka = [H+] (think about the expression for Ka and you'll see why this is the case). So, to find [H+], you just use your value for Ka!
    Thanks for that, but now I am a little confused as to why ethanoate ions are also 0.05.... thanks again
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    (Original post by _H_V)
    I got C, 3 moles of ester at equilibrium so there are also 3 moles of water. Minus 3 from initial moles of both reactants gives you 1mol of methanoic acid and 3 moles of ethanol, I put them into kc expression and got 3 (C)
    (Original post by taer)
    1 to 1 ratio: therefore equilibrium moles of methanol acid would be 4-3= 1 mol and ethanol 6-3 = 3 mol therefore moles of ester and water must be 3. overall experession is 9/3 which is 3
    Ah crap yeah I forgot about the water haha Thanks guys!
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    (Original post by maria17xo)
    Thanks for that, but now I am a little confused as to why ethanoate ions are also 0.05.... thanks again
    Well, check this out:
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    (Original post by Don Pedro K.)
    Well, check this out:
    Name:  Untitled.jpg
Views: 122
Size:  29.4 KB
    OHHHHHHHHHHH!!!!!!!! I get it! Thank you. you superstar!
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    Do you leave water out of a Kc expression??
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    (Original post by maria17xo)
    OHHHHHHHHHHH!!!!!!!! I get it! Thank you. you superstar!
    Haha no problem xD!!!!
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    (Original post by CLGC98)
    Do you leave water out of a Kc expression??
    It'll probably be clear from the question and answers that you work out for equilibrium concs. etc. if you leave it out or not
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    (Original post by ayvaak)
    Attachment 549115

    Could someone please explain why there are different answers to similar questions?
    They look similar but they're not the same. The first one asks what happens if you add a catalyst, which doesn't affect the position of equilibrium but makes it be reached quicker. The second one asks what happens if you increase pressure, think about La Chateliers principle and how equilibrium shifts to the side with less moles when pressure increases, at A2 you need to know why it does that and it's because Kp stays constant so equilibrium moves to alter the partial pressures of the reactants and products to keep Kp constant.
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    (Original post by Whizbox)
    They look similar but they're not the same. The first one asks what happens if you add a catalyst, which doesn't affect the position of equilibrium but makes it be reached quicker. The second one asks what happens if you increase pressure, think about La Chateliers principle and how equilibrium shifts to the side with less moles when pressure increases, at A2 you need to know why it does that and it's because Kp stays constant so equilibrium moves to alter the partial pressures of the reactants and products to keep Kp constant.

    so to confirm an increase in pressure initially causes an increase in ratio of pH2 to pHI ^2 but because Kp must stay the same ratio must decrease to restore the the ratio to Kp?
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    Please could someone's explain the relationship between entropy change and the equilibrium constant?

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    (Original post by TeaAndTextbooks)
    Please could someone's explain the relationship between entropy change and the equilibrium constant?

    Posted from TSR Mobile
    Total entropy change = RlnK (where R is the gas constant 8.31kJmol-1)

    So basically, total entropy change is proportional to lnK.
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    (Original post by Don Pedro K.)
    It'll probably be clear from the question and answers that you work out for equilibrium concs. etc. if you leave it out or not
    Can you clarify , because i assumed you left it out due to the concentration of water being constant, but when i worked out the Kc, of the multiple choice question, i got the right answer when i left it in
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    (Original post by CLGC98)
    Can you clarify , because i assumed you left it out due to the concentration of water being constant, but when i worked out the Kc, of the multiple choice question, i got the right answer when i left it in
    I remember doing a question where we were told to assume that the moles of water = moles of ester.

    However, I also did a question where we were given an initial number of moles of water present, and then had to add on the number of moles lost by the reactants in order to get the equilibrium moles of water
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    (Original post by Don Pedro K.)
    Total entropy change = RlnK (where R is the gas constant 8.31kJmol-1)

    So basically, total entropy change is proportional to lnK.
    Thanks you.

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    Can anyone help with this question please? I'm kind of confused.


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    (Original post by TeaAndTextbooks)
    Can anyone help with this question please? I'm kind of confused.


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    Look back to post #641 on this page http://www.thestudentroom.co.uk/show...1#post65718631

    I did an explanation
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    Hey guys, can someone please explain what the heck this question is on about? How are we supposed to know to compare experimental and theoretical lattice energies?!

 
 
 
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