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November 01, 2025, 12:34:02 pm

Author Topic: question from stav 2008 exam  (Read 1210 times)  Share 

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cama23

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question from stav 2008 exam
« on: September 14, 2010, 11:04:15 pm »
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struggling to understand question 2bii) in the short answer, any help would be greatly appreciated  :P
the question asks what whould happen to the value K (up, down or no effect) if the volume is increased at constant temperature and to give an ecplanation to your answer.
the equation is CH4(g) + H20(g) <===> CO(g) + 3H2(g)
i presumed since volume increased, pressure would subsequently decrease causing a net forward reaction, meaning more products meaining a higher K value.
but in the suggested solutions it says no change in K and gives the reason that the system returns to the original CF value at the same temperature.
CONFUSED AS  :idiot2:

kakar0t

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Re: question from stav 2008 exam
« Reply #1 on: September 14, 2010, 11:15:52 pm »
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THE ONLY THING THAT CHANGES K VALUE IS TEMPERATURE CHANGE

Sorry for caps.

Your logic is absolutely correct, except for the fact that after the -shift in equilibrium- occurs the concentration fraction will be absolultely equivalent to the K value you started with. The only thing that will change the proportion of the fracation (and hence the K equilibrium constant) is temperature.

I'm sure I read something mathematical on why that is, i'll try and find it.

cama23

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Re: question from stav 2008 exam
« Reply #2 on: September 14, 2010, 11:24:05 pm »
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ohh i do remember doing that when we did equilibrium so long ago!
thanks for that, wont make that mistake again :P

sajib_mostofa

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Re: question from stav 2008 exam
« Reply #3 on: September 14, 2010, 11:29:06 pm »
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So I am right in saying pressure affects the concentration fraction but not the value of k?

taiga

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Re: question from stav 2008 exam
« Reply #4 on: September 14, 2010, 11:30:41 pm »
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yes
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kakar0t

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Re: question from stav 2008 exam
« Reply #5 on: September 14, 2010, 11:56:11 pm »
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So I am right in saying pressure affects the concentration fraction but not the value of k?

Well, the concentration fraction is composed of the product (multiplication) of products over reactants. Thus any change in either numerator or denominator would be followed by a proportional change in the other in order to maintain the value of K. Damn, this is intriguing stuff

sajib_mostofa

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Re: question from stav 2008 exam
« Reply #6 on: September 15, 2010, 12:12:37 am »
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It's got me in a bit of a spin  :)

taiga

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Re: question from stav 2008 exam
« Reply #7 on: September 15, 2010, 06:26:40 pm »
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Just avoid being trapped by questions like that.

IF the temperature doesn't change, as far as I know, K does not change :P
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Greggler

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Re: question from stav 2008 exam
« Reply #8 on: September 19, 2010, 01:44:19 pm »
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i just did tssm 2008 and a similar concern arose. theres a table asking you for the effects of changing concentration, pressure, temp etc.
I assumed that K values were unchanged with a change in pressure but it stated otherwise..

fady_22

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Re: question from stav 2008 exam
« Reply #9 on: September 19, 2010, 01:58:47 pm »
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i just did tssm 2008 and a similar concern arose. theres a table asking you for the effects of changing concentration, pressure, temp etc.
I assumed that K values were unchanged with a change in pressure but it stated otherwise..

That was a mistake.
K values only change with a change in temperature.
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