January 25, 2021, 03:24:11 PM
Forum Rules: Read This Before Posting

Topic: Equilibrium issue  (Read 9047 times)

0 Members and 1 Guest are viewing this topic.

AWK

• Retired Staff
• Sr. Member
• Posts: 7872
• Mole Snacks: +546/-92
• Gender:
Re: Equilibrium issue
« Reply #15 on: April 20, 2012, 12:08:49 PM »
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%.
AWK

Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
Re: Equilibrium issue
« Reply #16 on: April 20, 2012, 12:29:10 PM »
Got 44.89%, but the right answer is 37.93%.

AWK

• Retired Staff
• Sr. Member
• Posts: 7872
• Mole Snacks: +546/-92
• Gender:
Re: Equilibrium issue
« Reply #17 on: April 21, 2012, 04:59:07 PM »
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)
AWK

Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
Re: Equilibrium issue
« Reply #18 on: June 10, 2012, 04:09:24 AM »
Still, can't figure out this:
I2 2I
x-y     2y

x-y=184/254
x-y=0.724? 2 unkowns.

Borek

• Mr. pH
• Deity Member
• Posts: 26252
• Mole Snacks: +1706/-402
• Gender:
• I am known to be occasionally wrong.
Re: Equilibrium issue
« Reply #19 on: June 10, 2012, 04:45:20 AM »
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
Re: Equilibrium issue
« Reply #20 on: June 10, 2012, 08:41:48 AM »
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?

Borek

• Mr. pH
• Deity Member
• Posts: 26252
• Mole Snacks: +1706/-402
• Gender:
• I am known to be occasionally wrong.
Re: Equilibrium issue
« Reply #21 on: June 10, 2012, 12:51:48 PM »
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
Re: Equilibrium issue
« Reply #22 on: June 10, 2012, 01:24:19 PM »
I never done equilibrium+vapor density, don't understand how to calculate without Keq.
Still, can't figure out this:
I2 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?

Borek

• Mr. pH
• Deity Member
• Posts: 26252
• Mole Snacks: +1706/-402
• Gender:
• I am known to be occasionally wrong.
Re: Equilibrium issue
« Reply #23 on: June 10, 2012, 02:43:03 PM »
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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
Re: Equilibrium issue
« Reply #24 on: June 10, 2012, 02:52:57 PM »
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.