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2000 - Multiple Choice - Q23 - HSN forum

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2000 - Multiple Choice - Q23


10 replies to this topic

#1 DecadentParadise

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Posted 27 May 2005 - 02:48 PM

Is it right that the pressure should be high as the product has overall less moles than reactants? Also, i have no idea how to work out whether the reaction is exo/endothermic and therefore what temperature it should be at.. could someone please explain this to me? blink.gif blink.gif

#2 dondon

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Posted 27 May 2005 - 02:52 PM

Exothermic is when the reaction gives out energy and endothermic is when it takes in energy for use in a reaction but i dont know what the question is i'm assuming its something about equlibrium??

#3 DecadentParadise

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Posted 27 May 2005 - 02:56 PM

The question IS about equilibrium.. and i know the definition you've given but its something specific about the question. I was meaning, How do you work out whether its exo/endo from the information given. so you'll have to see the question to help me.

#4 dondon

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Posted 27 May 2005 - 03:02 PM

could u not just copy it out if its an equlibrium equation co i dont have past papers

#5 dfx

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Posted 27 May 2005 - 03:07 PM

"Also, i have no idea how to work out whether the reaction is exo/endothermic "

Strange, normally they specify its for the forward or reverse reaction, but they haven't.

It IS high pressure because there's 2 moles of gasses on the right hand side (reactants) and 1 mole on the left (products). Increasing the pressure shifts the equilibrium to the side with less moles of gases - which is the products, and so high pressure.

I think you have to know the process is exothermic off by heart from the chemical industry or something... cause they've quoted the DeltaH as +/- 46.

#6 DecadentParadise

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Posted 27 May 2005 - 03:11 PM

Lol, i'll give it my best.


Under the conditions used industrially, ethene and steam react as follows.


C2H2(g) + H2O(g) ->/<- C2H5OH(g) /_\H= +/- 46kJmol^+/-1

(Ethene + Steam -> Ethanol)

Which set of conditions would give the best yield of ethanol at equilibrium?

A - High temperature, low pressure
B - High temperature, high pressure
C - Low temperature, high pressure
D - Low temerature, low pressure

As i said, i can work out the pressure part. But am clueless as how to assert the endothermic/exothermic directions in the reaction.

#7 dondon

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Posted 27 May 2005 - 03:22 PM

QUOTE(DecadentParadise @ May 27 2005, 03:11 PM)
Lol, i'll give it my best.


Under the conditions used industrially, ethene and steam react as follows.

                           
C2H2(g) + H2O(g)    ->/<-    C2H5OH(g)  /_\H= +/- 46kJmol^+/-1
                           
    (Ethene + Steam -> Ethanol)   

Which set of conditions would give the best yield of ethanol at equilibrium?

A - High temperature, low pressure
B - High temperature, high pressure
C - Low temperature, high pressure
D - Low temerature, low pressure

As i said, i can work out the pressure part. But am clueless as how to assert the endothermic/exothermic directions in the reaction.

View Post




Bonds are forming in the forward reaction and bonds forming take in energy whereas bonds breaking release energy so the forward reaction is endothermic so its high temperature.


#8 e-unit

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Posted 27 May 2005 - 05:08 PM

QUOTE(dondon @ May 27 2005, 04:22 PM)
QUOTE(DecadentParadise @ May 27 2005, 03:11 PM)
Lol, i'll give it my best.


Under the conditions used industrially, ethene and steam react as follows.

                            
C2H2(g) + H2O(g)    ->/<-    C2H5OH(g)   /_\H= +/- 46kJmol^+/-1
                           
     (Ethene + Steam -> Ethanol)   

Which set of conditions would give the best yield of ethanol at equilibrium?

A - High temperature, low pressure
B - High temperature, high pressure
C - Low temperature, high pressure
D - Low temerature, low pressure

As i said, i can work out the pressure part. But am clueless as how to assert the endothermic/exothermic directions in the reaction.

View Post




Bonds are forming in the forward reaction and bonds forming take in energy whereas bonds breaking release energy so the forward reaction is endothermic so its high temperature.

View Post



Its a typo. They would never say mol l^+/- . It doesnt make sense


#9 Ally

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Posted 27 May 2005 - 10:19 PM

That questions has a typo. It meant to be -ve.

#10 dfx

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Posted 27 May 2005 - 11:01 PM

wow how convenient for leckie and leckie to completely ignore their stupid typos mad.gif

#11 Ally

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Posted 28 May 2005 - 07:00 PM

There is also a typo in question 12 (c ) in the 2000 paper.

The equation is meant to read as:

CH3NHNH2(l) + 2.5 O2 ----> CO2(g) + 3H2O(l) + N2(g)

delta.gifH = -1305KJmol-1





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