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November 01, 2025, 11:39:49 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 448755 times)  Share 

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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #975 on: June 19, 2013, 08:53:47 pm »
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Ah...yeah ok I wasn't particularly clear with that in my working.
Although if you looked at the original question which had VO2NO3, it's slightly less ambiguous there.
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Re: Chemistry 3/4 2013 Thread
« Reply #976 on: June 19, 2013, 08:59:09 pm »
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Ah...yeah ok I wasn't particularly clear with that in my working.
Although if you looked at the original question which had VO2NO3, it's slightly less ambiguous there.

Yeah you're right, I should've noticed that

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Re: Chemistry 3/4 2013 Thread
« Reply #977 on: June 23, 2013, 12:40:09 pm »
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I'm having trouble with this equilibrium question. Specifically, calculating the mols present once the equation reaches equilibrium. Can someone help me out?
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #978 on: June 23, 2013, 06:04:05 pm »
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It's a stoichiometry problem.

2 mol NO2 initially, 1.3 mol NO2 left. n(NO2) used up = n(SO3) formed = 0.7 mol
Also n(SO2) used up = n(NO2) used up = 0.7 mol
so 1.3 mol SO2 left as well

We have 1.3 mol SO2 and NO2 with 0.7 mol NO and SO3. As volume = 1 L, number of moles and the concentration are numerically identical. Also number of moles in the reactions stays constant so no need to worry about units.
Equilibrium constant = [SO3]*[NO]/([SO2]*[NO2] = 0.7^2/1.3^2=49/169 no units
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Re: Chemistry 3/4 2013 Thread
« Reply #979 on: June 29, 2013, 04:04:48 pm »
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Haven't got a clue as to how to do part c...

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psyxwar

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Re: Chemistry 3/4 2013 Thread
« Reply #980 on: June 29, 2013, 04:22:44 pm »
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Haven't got a clue as to how to do part c...

(Image removed from quote.)
Okay so your balanced equation is:

2SO3 (g) + CO2 (g) -> CS2 (g) + 4O2 (g)

From then on its just stoichiometry. You have 0.028 mol CS2 in your equilibrium mixture, therefore the number of moles of O2 is 4 times this. You can then work out the molarity of all of the reactants and products.

Putting all of the values into K = ([CS2][O2]^4)/([SO3]^2[CO2]) should give you your answer. Units as appropriate.

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Re: Chemistry 3/4 2013 Thread
« Reply #981 on: June 29, 2013, 04:26:13 pm »
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That's not right, because reactants and products are not necessarily in stoichiometric ratios at equilibrium. The solutions also make no sense.
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psyxwar

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Re: Chemistry 3/4 2013 Thread
« Reply #982 on: June 29, 2013, 04:27:26 pm »
+2
That's not right, because reactants and products are not necessarily in stoichiometric ratios at equilibrium. The solutions also make no sense.
It is right though. You only start off by mixing the reactants and therefore your products are definitely in stoichiometric ratios.

Think of it this way -- to make four moles O2 you also have to make one mole CS2. Even if we were to consider the back reaction, it still uses the molecules in these ratios.
« Last Edit: June 29, 2013, 04:29:10 pm by psyxwar »
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Re: Chemistry 3/4 2013 Thread
« Reply #983 on: June 29, 2013, 04:32:10 pm »
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The solutions suggest otherwise. :P
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psyxwar

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Re: Chemistry 3/4 2013 Thread
« Reply #984 on: June 29, 2013, 04:41:19 pm »
+2
The solutions suggest otherwise. :P
Okay so I just did it on paper just then:

n(CS2) = 0.028, therefore c(CS2) = 0.0014M
n(SO3) = 0.022, therefore c(SO3) = 0.0011M
n(CO2) = 0.014, therefore c(CO2) = 0.0007M
n(O2) = 0.112 therefore c(O2) = 0.0056M

Inputting them into the equilibrium equation and we get K = 1.626 x 10^-3 M^2, correct sig figs and we get 1.6 x 10^-3 M^2. What does the answer say?
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Re: Chemistry 3/4 2013 Thread
« Reply #985 on: June 29, 2013, 04:44:00 pm »
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Nevermind, a friend told me that there was a typo in the school's solutions because he got the same answer as you. Your way made sense (it was what I did), but I couldn't understand why it wasn't in line with the answer I was provided with.

Thanks for the help. :) I have to say, you must be going pretty well in Chemistry if you're already understanding Unit 4 content!
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Re: Chemistry 3/4 2013 Thread
« Reply #986 on: June 29, 2013, 04:50:48 pm »
+1
no psyxwar is correct. although 2 mol of SO3 might not necessary give you 1 mol of CS2, the products will still be in a 1:4 ratio. and unless you have a dodgy set of solutions, psyxwar's answer does coincide with that provided by the solution key.

edit: lol, too slow. xD
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psyxwar

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Re: Chemistry 3/4 2013 Thread
« Reply #987 on: June 29, 2013, 04:53:50 pm »
+1
Nevermind, a friend told me that there was a typo in the school's solutions because he got the same answer as you. Your way made sense (it was what I did), but I couldn't understand why it wasn't in line with the answer I was provided with.

Thanks for the help. :) I have to say, you must be going pretty well in Chemistry if you're already understanding Unit 4 content!
No worries! :) Haha yeah I'm doing pretty well in chem.
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Edward21

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Re: Chemistry 3/4 2013 Thread
« Reply #988 on: July 02, 2013, 09:25:35 pm »
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1st post of July ;)

Do we need to know, actually rephrase, be able to write out the electron configurations in subshell form eg. 1s2, 2s2, 2p6 ?
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Re: Chemistry 3/4 2013 Thread
« Reply #989 on: July 02, 2013, 09:29:22 pm »
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1st post of July ;)

Do we need to know, actually rephrase, be able to write out the electron configurations in subshell form eg. 1s2, 2s2, 2p6 ?

I haven't seen that asked for in a year 12 examination before. However, it could be tested as it is a prerequisite skill. I would imagine that you could see a question relating to that in a year 11 topic test or examination. Nonetheless, I'd still recommend knowing how to write it out, just incase.
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