Chemistry question

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17 years ago
#1
OK, so this question in my Chemistry book is driving me batty. I think I must be missing
something obvious.

"At 1000K, the equilibrium constant for the reaction H2O(g) + C(s) <==>
H2(g) + CO(g) is 3.72atm. Find the equilibrium partial pressures of each of the gases in the
equilibrium mixture if the total pressure is 25atm".

OK, so obviously this is a hetrogenous system, so Kp = P(H2).P(CO) / P
(H3) and the carbon is ignored. But where does that get you?

Kp = x(H2).Pt.x(CO).Pt
-----------------
x(H2O).Pt

Kp / Pt = x(H2).x(CO) = some random constant
-----------
x(H2P)

Hmm. Not much help there. But there is no information (that I could see anyway) from which you could
determine the molar amounts of the products or reactants.

I suspect I'm missing something rather blatant. Does anyone care to point out my idiocy?

Tom
0
17 years ago
#2
"Tom Salls" <[email protected]> wrote in message news:[email protected]...
[q1]> "At 1000K, the equilibrium constant for the reaction H2O(g) + C(s) <==>[/q1]
[q1]> H2(g) + CO(g) is 3.72atm. Find the equilibrium partial pressures of each of the gases in the[/q1]
[q1]> equilibrium mixture if the total pressure is 25atm".[/q1]
[q1]>[/q1]
[q1]> OK, so obviously this is a hetrogenous system, so Kp = P(H2).P(CO) / P[/q1]
[q1]> (H20) and the carbon is ignored. But where does that get you?[/q1]
[q1]>[/q1]
[q1]> Kp = x(H2).Pt.x(CO).Pt[/q1]
[q1]> -----------------[/q1]
[q1]> x(H2O).Pt[/q1]
[q1]>[/q1]
[q1]> Kp / Pt = x(H2).x(CO) = some random constant[/q1]
[q1]> -----------[/q1]
[q1]> x(H2O)[/q1]

Oh memories Sorry, I wish i could but have forgotten everything I learnt in A-level chemistry.
Strangely I seem to remember everything from mathematics and physics though. Um the constant can
just be assumed as 1 and scaled accordingly can't it? Oh wait forget about it, i've forgotten.
Programming rulez anyway. Oh wait I think I've forgotten all the stuff I learn this year too. and
year 2's supposed to kill everyone. oh man.

peace,
G.Sharma.
0
17 years ago
#3
On Thu, 11 Jul 2002 1531 +0100s, Tom Salls([email protected]) said...
[q1]>OK, so this question in my Chemistry book is driving me batty. I think I must be missing[/q1]
[q1]>something obvious.[/q1]
[q1]>[/q1]
[q1]>"At 1000K, the equilibrium constant for the reaction H2O(g) + C(s) <==>[/q1]
[q1]>H2(g) + CO(g) is 3.72atm. Find the equilibrium partial pressures of each of the gases in the[/q1]
[q1]> equilibrium mixture if the total pressure is 25atm".[/q1]
[q1]>[/q1]
[q1]>OK, so obviously this is a hetrogenous system, so Kp = P(H2).P(CO) / P[/q1]
[q1]>(H20) and the carbon is ignored. But where does that get you?[/q1]

But total P is P(H2) + P(CO) + P(H2O)

So P(H2) * P(CO) / (25 - P(H2) - P(CO) ) = 3.72 a bit of fiddling here gives you:
P(H2) * P(CO) + 3.72 P(H2) + 3.72 P(CO) = 3.72*25

This is insoluble as is, so assumes that you started with equal amounts of CO and H2, so you will
also end up with equal amounts. Thus it's reduced to a simple quadratic equation.

That's of the top of my head, so could be wrong. Haven't done this since A-level...

chris
0
17 years ago
#4
Tom,

[q1]> "At 1000K, the equilibrium constant for the reaction H2O(g) + C(s) <==>[/q1]
[q1]> H2(g) + CO(g) is 3.72atm. Find the equilibrium partial pressures of each of the gases in the[/q1]
[q1]> equilibrium mixture if the total pressure is 25atm".[/q1]
[q1]>[/q1]
[q1]> OK, so obviously this is a hetrogenous system, so Kp = P(H2).P(CO) / P[/q1]
[q1]> (H20) and the carbon is ignored. But where does that get you?[/q1]

There is no need to bring in the mole fractions.
K(p) = p(H2)*p(CO)/p(H20) = 3.72 atm

The total pressure at equilibrium is 25 atm. It tells you that
K(p1) + p(CO) + p(H20) = 25 atm .

Having said that, you still have two simultaneous equations, but three unknowns. This is just my
guess, but maybe you could assume that p(H2) =
p(CO) . This would be true if you first had a vessel with only water (steam) and carbon in it. The
reaction would then produce equal numbers of moles of hydrogen and carbon monoxide, hence the
partial pressures of hydrogen and CO at equilibrium would be identical, justifying my
assumption.

However, if you started off with a reaction vessel that has differing amounts (numbers of moles)
of hydrogen and CO in it, then the final partial pressures of hydrogen and carbon monoxide would
not be equal.

Or maybe I'm getting something wrong too

q.
0
17 years ago
#5
In article <[email protected] 133.1.4>, [email protected] says...
[q1]> On Thu, 11 Jul 2002 1531 +0100s, Tom Salls([email protected]) said...[/q1]
[q2]> >OK, so this question in my Chemistry book is driving me batty. I think I must be missing[/q2]
[q2]> >something obvious.[/q2]
[q2]> >[/q2]
[q2]> >"At 1000K, the equilibrium constant for the reaction H2O(g) + C(s) <==>[/q2]
[q2]> >H2(g) + CO(g) is 3.72atm. Find the equilibrium partial pressures of each of the gases in the[/q2]
[q2]> > equilibrium mixture if the total pressure is 25atm".[/q2]
[q2]> >[/q2]
[q2]> >OK, so obviously this is a hetrogenous system, so Kp = P(H2).P(CO) / P[/q2]
[q2]> >(H20) and the carbon is ignored. But where does that get you?[/q2]
[q1]>[/q1]
[q1]> But total P is P(H2) + P(CO) + P(H2O)[/q1]
[q1]>[/q1]
[q1]> So P(H2) * P(CO) / (25 - P(H2) - P(CO) ) = 3.72 a bit of fiddling here gives you:[/q1]
[q1]> P(H2) * P(CO) + 3.72 P(H2) + 3.72 P(CO) = 3.72*25[/q1]
[q1]>[/q1]
[q1]> This is insoluble as is, so assumes that you started with equal amounts of CO and H2, so you will[/q1]
[q1]> also end up with equal amounts. Thus it's reduced to a simple quadratic equation.[/q1]
[q1]>[/q1]
[q1]> That's of the top of my head, so could be wrong. Haven't done this since A-level...[/q1]

whooops, sorry, I didn't realise you'd replied. It's comforting you arrived at the same answer
though.

How come you haven't done it since A-level??????????????????????????? ???

M.
0
17 years ago
#6
[email protected] says...
[q1]> But total P is P(H2) + P(CO) + P(H2O)[/q1]
[q1]>[/q1]
[q1]> So P(H2) * P(CO) / (25 - P(H2) - P(CO) ) = 3.72 a bit of fiddling here gives you:[/q1]
[q1]> P(H2) * P(CO) + 3.72 P(H2) + 3.72 P(CO) = 3.72*25[/q1]
[q1]>[/q1]
[q1]> This is insoluble as is, so assumes that you started with equal amounts of CO and H2, so you will[/q1]
[q1]> also end up with equal amounts. Thus it's reduced to a simple quadratic equation.[/q1]

Damn, damn, and damn again. Working through your suggestion does indeed get a plausible answer
(P(H2) == P(CO) == 6.6162atm, P(H20) ==
11.767atm), but I'd hope that a quadratic solution wouldn't be necessary.

This is the official textbook for the syllabus, and the syllabus cheerfully says "Students will be
expected to use the expressions for Kc and Kp to perform calculations but they will /not/ be
expected to perform calculations that require the solution of quadratic equations using the formula
<etc etc>". (It's Edexcel, feel free to spit and hiss now). Ah well, that's what I get for trusting
the textbook!

Thanks for all the help (and thanks to Martina too). It certainly makes sense that the partial
pressure of the products should be the same since they're all present in the same molar amounts
according to the equation.

Cheers

Tom
0
17 years ago
#7
On Thu, 11 Jul 2002 1815 +0100s, martina([email protected]) said...
[q1]>In article <[email protected] 133.1.4>, [email protected] says...[/q1]
[q2]>> That's of the top of my head, so could be wrong. Haven't done this since A-level...[/q2]
[q1]>[/q1]
[q1]>whooops, sorry, I didn't realise you'd replied. It's comforting you arrived at the same answer[/q1]
[q1]>though. [/q1]
[q1]>[/q1]
[q1]>How come you haven't done it since A-level??????????????????????????? ???[/q1]
[q1]>[/q1]
[q1]>M.[/q1]

You should be careful with that punctuation, people will think you're insane or something. Next
thing you'll be writing Ahaha!!!!! and getting an unhealthy interest in opera....

Haven't done it since A-level since we don't do stuff that easy with numbers - pretty much the only
thing we've used eqm consts for is proving other thermodynamic relations...

<screams at the thought of partition functions>

chris
0
17 years ago
#8
It certainly makes
[q1]> sense that the partial pressure of the products should be the same since they're all present in[/q1]
[q1]> the same molar amounts according to the equation.[/q1]

Yes, but that assumes that the numbers of moles of H2 and CO were the same at the beginning (or that
there was no hydrogen or carbon monoxide present at the beginning). They should have told you that
it's a valid assumption.

M.
0
17 years ago
#9
On Thu, 11 Jul 2002 1802 +0100s, Tom Salls([email protected]) said...
[q1]>[email protected] says...[/q1]
[q2]>> But total P is P(H2) + P(CO) + P(H2O)[/q2]
[q2]>>[/q2]
[q2]>> So P(H2) * P(CO) / (25 - P(H2) - P(CO) ) = 3.72 a bit of fiddling here gives you:[/q2]
[q2]>> P(H2) * P(CO) + 3.72 P(H2) + 3.72 P(CO) = 3.72*25[/q2]
[q2]>>[/q2]
[q2]>> This is insoluble as is, so assumes that you started with equal amounts of CO and H2, so you will[/q2]
[q2]>> also end up with equal amounts. Thus it's reduced to a simple quadratic equation.[/q2]
[q1]>[/q1]
[q1]>Damn, damn, and damn again. Working through your suggestion does indeed get a plausible answer[/q1]
[q1]>(P(H2) == P(CO) == 6.6162atm, P(H20) ==[/q1]
[q1]>11.767atm), but I'd hope that a quadratic solution wouldn't be necessary.[/q1]

Why? It's not like they're difficult, they're barely GCSE standard. Anyway, I can't think of any way
to do these without going via a quadratic...

[q1]>This is the official textbook for the syllabus, and the syllabus cheerfully says "Students will be[/q1]
[q1]>expected to use the expressions for Kc and Kp to perform calculations but they will /not/ be[/q1]
[q1]>expected to perform calculations that require the solution of quadratic equations using the formula[/q1]
[q1]><etc etc>". (It's Edexcel, feel free to spit and hiss now). Ah well, that's what I get for trusting[/q1]
[q1]>the textbook![/q1]

Oh no, I remember now Least it's not as bad as physics, in that you can get away with doing
chemistry like this at A-level without making it 15 times as difficult...

[q1]>Thanks for all the help (and thanks to Martina too). It certainly makes sense that the partial[/q1]
[q1]>pressure of the products should be the same since they're all present in the same molar amounts[/q1]
[q1]>according to the equation.[/q1]
[q1]>[/q1]
[q1]>Cheers[/q1]
[q1]>[/q1]
[q1]>Tom[/q1]

As Martina says, it depends on the initial concentrations. If you start with uneven amounts, it'll
stay like that. Of course it'll try and get close to the eqm conditions you've worked out, but
conservation laws will prevent that.

chris
0
17 years ago
#10
[q1]> You should be careful with that punctuation, people will think you're insane[/q1]

that couldn't be far from the truth, if your previous comments about my sanity, or lack of it, are
to be trusted.

[q1]> Haven't done it since A-level since we don't do stuff that easy with numbers - pretty much the[/q1]
[q1]> only thing we've used eqm consts for is proving other thermodynamic relations...[/q1]
[q1]>[/q1]
[q1]> <screams at the thought of partition functions>[/q1]

[q1][/q1]

M.
0
17 years ago
#11
[email protected] says...
[q1]> It certainly makes[/q1]
[q2]> > sense that the partial pressure of the products should be the same since they're all present in[/q2]
[q2]> > the same molar amounts according to the equation.[/q2]
[q1]>[/q1]
[q1]> Yes, but that assumes that the numbers of moles of H2 and CO were the same at the beginning (or[/q1]
[q1]> that there was no hydrogen or carbon monoxide present at the beginning). They should have told you[/q1]
[q1]> that it's a valid assumption.[/q1]

The question was exactly as I gave it, unfortunately.
0
17 years ago
#12
[email protected] says...
[q1]> On Thu, 11 Jul 2002 1802 +0100s, Tom Salls([email protected])[/q1]
[q2]> >Damn, damn, and damn again. Working through your suggestion does indeed get a plausible answer[/q2]
[q2]> >(P(H2) == P(CO) == 6.6162atm, P(H20) ==[/q2]
[q2]> >11.767atm), but I'd hope that a quadratic solution wouldn't be necessary.[/q2]
[q1]>[/q1]
[q1]> Why? It's not like they're difficult, they're barely GCSE standard. Anyway, I can't think of any[/q1]
[q1]> way to do these without going via a quadratic...[/q1]

I don't remember any quadratics in GCSE Chemistry! I think the question is actually wrong for the
syllabus, since I can't find any other way of solving it that doesn't come down to a quadratic.

Still, all good practice I s'pose...

Tom
0
17 years ago
#13
On Thu, 11 Jul 2002 1919 +0100s, Tom Salls([email protected]) said...
[q1]>[email protected] says...[/q1]
[q2]>> On Thu, 11 Jul 2002 1802 +0100s, Tom Salls([email protected])[/q2]
[q2]>> >Damn, damn, and damn again. Working through your suggestion does indeed get a plausible answer[/q2]
[q2]>> >(P(H2) == P(CO) == 6.6162atm, P(H20) ==[/q2]
[q2]>> >11.767atm), but I'd hope that a quadratic solution wouldn't be necessary.[/q2]
[q2]>>[/q2]
[q2]>> Why? It's not like they're difficult, they're barely GCSE standard. Anyway, I can't think of any[/q2]
[q2]>> way to do these without going via a quadratic...[/q2]
[q1]>[/q1]
[q1]>I don't remember any quadratics in GCSE Chemistry![/q1]

Me neither, I meant GCSE maths, if not earlier.

[q1]>I think the question is actually wrong for the syllabus, since I can't find any other way of[/q1]
[q1]>solving it that doesn't come down to a quadratic.[/q1]

I tried too, couldn't find one without a lot more info. And I'm not sure even then it'd be doable.

[q1]>Still, all good practice I s'pose...[/q1]
[q1]>[/q1]
[q1]>Tom[/q1]

Yeah, it's not like it'll hurt you to do a quadratic. They're hardly difficult...

chris
0
17 years ago
#14
[email protected] says...
[q1]> Tom Salls([email protected]) said...[/q1]
[q2]> >I don't remember any quadratics in GCSE Chemistry![/q2]
[q1]> Me neither, I meant GCSE maths, if not earlier.[/q1]

Is GCSE maths a requirement for A-level Chem these days? I don't remember my GCSE Maths (or my
A-level Maths come to that), but do remember being very glad when it was over.

[q1]> Yeah, it's not like it'll hurt you to do a quadratic. They're hardly difficult...[/q1]

It's an irrelevance to my course -- re-remembering how to do a quadratic when it's not needed just
wastes brain cycles.

Tom
0
17 years ago
#15
[q2]> >> Why? It's not like they're difficult, they're barely GCSE standard. Anyway, I can't think of[/q2]
[q2]> >> any way to do these without going via a quadratic...[/q2]
[q2]> >[/q2]
[q2]> >I don't remember any quadratics in GCSE Chemistry![/q2]
[q1]>[/q1]
[q1]> Me neither, I meant GCSE maths, if not earlier.[/q1]
[q1]>[/q1]
[q2]> >I think the question is actually wrong for the syllabus, since I can't find any other way of[/q2]
[q2]> >solving it that doesn't come down to a quadratic.[/q2]
[q1]>[/q1]
[q1]> I tried too, couldn't find one without a lot more info. And I'm not sure even then it'd be doable.[/q1]
[q1]>[/q1]
[q2]> >Still, all good practice I s'pose...[/q2]
[q2]> >[/q2]
[q2]> >Tom[/q2]
[q1]>[/q1]
[q1]> Yeah, it's not like it'll hurt you to do a quadratic. They're hardly difficult...[/q1]

I think I never got a question requiring the knowledge of quadratics in chemistry. The people who
wrote the textbook probably wanted you to try something different.

Aren't quadratics covered by GCSE maths syllabi? I think my A-level Physics syllabus assumed we can
solve such equations.

M.
0
17 years ago
#16
"> It's an irrelevance to my course -- re-remembering how to do a quadratic
[q1]> when it's not needed just wastes brain cycles.[/q1]

Oi, Sir! Where has sheer love of learning gone? Now, when I was young... )

M.
0
17 years ago
#17
[email protected] says...
[q1]> I think I never got a question requiring the knowledge of quadratics in chemistry. The people who[/q1]
[q1]> wrote the textbook probably wanted you to try something different.[/q1]

Think you're right, Martina. And I wouldn't have minded if they'd given any example that used a
quadratic, or given any indication that this question was in some way special. But no, it was just
tacked on the end of a chapter as one of a set of four assessment questions.

Given that I'm currently learning Chem and Biol solely from textbooks, I am appalled by how badly
written and organised they seem to be. I think I'm going to have to join a proper course to minimise
wasting time like this -- I now have the knowledge to solve a problem which is not in the syllabus
and so will not come up in the exam. Bugger.

Tom
0
17 years ago
#18
[email protected] says...
[q1]> "> It's an irrelevance to my course -- re-remembering how to do a quadratic[/q1]
[q2]> > when it's not needed just wastes brain cycles.[/q2]
[q1]>[/q1]
[q1]> Oi, Sir! Where has sheer love of learning gone? Now, when I was young... )[/q1]

I'm an old fart.

Sheer love of learning has been painfully burnt out of me. ;>

Tom
0
17 years ago
#19
[q2]> > I think I never got a question requiring the knowledge of quadratics in chemistry. The people[/q2]
[q2]> > who wrote the textbook probably wanted you to try something different.[/q2]
[q1]>[/q1]
[q1]> Think you're right, Martina. And I wouldn't have minded if they'd given any example that used a[/q1]
[q1]> quadratic, or given any indication that this question was in some way special. But no, it was just[/q1]
[q1]> tacked on the end of a chapter as one of a set of four assessment questions.[/q1]
[q1]>[/q1]
[q1]> Given that I'm currently learning Chem and Biol solely from textbooks, I am appalled by how badly[/q1]
[q1]> written and organised they seem to be. I think I'm going to have to join a proper course to[/q1]
[q1]> minimise wasting time like this -- I now have the knowledge to solve a problem which is not in the[/q1]
[q1]> syllabus and so will not come up in the exam. Bugger.[/q1]

wooo hooo, studying on your own? So did I, and wasn't it a pain in the neck at times!

It's always good to have additional textbooks to help you out.

Biology: Green, Stout & Taylor, Biological Science is excellent for many aspects of biology -
especially cell biology, genetics, anatomy and physiology.

Chemistry: Chemistry in Context (5th edition iirc)

HTH

M.
0
17 years ago
#20
On Thu, 11 Jul 2002 2004 +0100s, Martina([email protected]) said...
[q1]>[/q1]
[q1]>"> It's an irrelevance to my course -- re-remembering how to do a quadratic[/q1]
[q2]>> when it's not needed just wastes brain cycles.[/q2]
[q1]>[/q1]
[q1]>Oi, Sir! Where has sheer love of learning gone? Now, when I was young... )[/q1]
[q1]>[/q1]
[q1]>M.[/q1]

No no no, it's "when I were a {lad|lass}..." in an generically regional accent.

Anyway, I can't think of anyone I know who hasn't had the love of learning burnt out of them by
now... though the compscis are still in the "why do this? because I can!" phase.

Although if anyone knows some way to learn how to make supervisors nicer then please let me know,
that's one thing I'd love to learn

chris
0
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