Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on April 03, 2012, 12:26:28 PM
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How many percents of iodine will dissociate on 1273K if the density of iodine vapors toward hydrogen in this temperature is 92?
How can I calculate it without the Keq? How to use the density towards hydrogen? 92*2=184g of I2?
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can you get density out of PV=nRT?
what's relative density?
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p=mRT/(MV)
m/V=d
p=dRT/V
What to do now? What is that relative density?
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relative density here is 184.
since we are chemists we want to deal with moles and molar masses.
what exactly does density(I2)/density(H2) mean in this case?
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Don't have a clue. I only know that it is 92. Could you explain it more clearly?
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i am sorry. i meant the relative density is 92 as per your problem.
you need to plug that in and focus on mollar mass ratios.
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density(I2)/density(H2)=92
m(I2)/M(H2)=92
m(I2)=92*2=184
So, what to do now?
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what's the molar mass of I2. what's the molar mass of your mixture? why are they different?
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Hard for me to understand because I never done similar problems before and I there is no explanation in my book.Is this right:
The mass of the mixture is 184g, and the molar mass of I2 is 254g, so 70g dissociated?
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write the equation of dissociation, come up with the formula for the average weight: we started with 1 mole, x moles dissociated, etc.
the average weight of the mixture is dependent on x, so is Keq.
what's your formula for the average weight of your mixture?
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I2 ::equil:: 2I
1-x 2x
[m(I2)+m(2I)]/2=... is this 184?
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we have 6 apples 80 g. each and 8 apples 60 g. each, what's the average weight of an apple?
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10.42g?
How to use it here?
I2 ::equil:: 2I
1-x 2x
[m(I2)+m(2I)]/2=... is this 184?
Why is this wrong?
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how can it be 10.42 g.? the smallest apple weights 60 g.
it is a good idea to check if the answer makes sense.
>>>[m(I2)+m(2I)]/2=... is this 184?
why do you divide by 2? you will get the same value regardless of the degree of dissociation. how is it useful for anything?
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The average weight is of an apple is 70g. Could you please write how to solve this?
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Since densities of gases are proportional to their molecular masses
then the mean molecular mass of iodine vapors is 184. Molecular mass concerns 1 mole hence
254x+(1-x)127=184
where x is a number of moles of molecular iodine present in 1 mole of iodine vapors.
I hope from this you can easily calculate % of thermal dissociation of iodine in this temperature which is ~ 60%.
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Got 44.89%, but the right answer is 37.93%.
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x - moles of undissociated I2
(1-x)/2 - moles of dissociated iodine
x+(1-x)/2 = (1+x)/2 - moles of iodine in account
((1-x)/2)/((1+x)/2)=(1-x)/(1+x) - thermal dissociation degree (times 100 if in %)
And result should be done with 2 significant digits (previously I gave an aproximate percent of undissociated part of iodine)
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Still, can't figure out this:
I2 ::equil:: 2I
x-y 2y
x-y=184/254
x-y=0.724? 2 unkowns.
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Pure molecular iodine I2 has a molar mass of 253.8 g/mol, so its vapor density relative to H2 is 126.9.
100% dissociated iodine is I and has a molar mass of 126.9 g/mol, so its vapor density relative to H2 is 63.45.
You are told density of the mixture (relative to H2) is 92. This is enough to calculate exact composition of the mixture.
This is a simple application of the Avogadro's law.
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126.9x+63.45y=92
x+y=1
From here x=0.45, y=0.55 so it is 55% but it isn't the right answer. Where I've mistaken?
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126.9x+63.45y=92
x+y=1
From here x=0.45, y=0.55 so it is 55%
No, it is not 55% from these numbers. You are asked about what fraction of iodine dissociated, not what is molar fraction of I.
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I never done equilibrium+vapor density, don't understand how to calculate without Keq.
Still, can't figure out this:
I2 ::equil:: 2I
x-y 2y
x-y=184/254
x-y=0.724? 2 unkowns.
Is it: 126.9(x-y)+63.45*2y=92 where I have to calculate y?
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Stop whining what you don't know, start to think how to apply what you do know. Keq is not necessary to solve the problem.
If you have 1 mole of a I2/I mixture, and you know there is 0.45 mole of I2 and 0.55 mole of I, how much I2 was there initially?
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0.45+0.55/2=0.725mol of I2
0.725-0.45=0.275mol that dissociated
0.275/0.725=37.93%
Got it now, thanks for the help.