Chemistry 3/4 2013 Thread

[scribble]

hahaha, we'll they can't put TOO much stuff onto the course, not everyone in victoria are knowledge hungry as you guys

[lzxnl]

You clearly haven't seen the physics course. They've taken stuff out of the course when we have more time to learn everything due to the cutting of the mid years.
Also, in magnetism, the only two cases we EVER consider are right angles and being parallel. We don't, for instance, consider the magnetic force on a wire in a magnetic field which is NOT perpendicular to the wire. Would it kill VCAA to put the extra trig in?

Look at IB physics. I have friends that tell me what is done in year 11 isn't repeated in year 12. That means they get through more content in around 70% of the time; the IB physics course is over according to my friend. That's freakishly fast.

[scribble]

ahahahaaaaaaa don't talk to me about physics, i got a 29 in that subject, not even kidding.
As far as im concerned, everything should not be in the physics course because it's all yucky and doesn't make sense. :')

but aren't they cutting content so there's not as much to remember come end of year?

[Homer]

An equilibrium mixture of gases was made by mixing sulfur trioxide and carbon dioxide. It consisted of 0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2 in a 20 L vessel. Calculate the value of the equilibrium constant at that temperature.

2SO3 (g) + CO2 (g)  CS2 (g) + 4O2 (g)

[brightsky]

You clearly haven't seen the physics course. They've taken stuff out of the course when we have more time to learn everything due to the cutting of the mid years.
Also, in magnetism, the only two cases we EVER consider are right angles and being parallel. We don't, for instance, consider the magnetic force on a wire in a magnetic field which is NOT perpendicular to the wire. Would it kill VCAA to put the extra trig in?

Look at IB physics. I have friends that tell me what is done in year 11 isn't repeated in year 12. That means they get through more content in around 70% of the time; the IB physics course is over according to my friend. That's freakishly fast.

yeah my teacher was complaining the other day about the fact that they cut weeks worth of material from the vce physics course, but only cut around a lesson's worth of material from the vce chem course. i actually regret my decision not to do physics.

An equilibrium mixture of gases was made by mixing sulfur trioxide and carbon dioxide. It consisted of 0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2 in a 20 L vessel. Calculate the value of the equilibrium constant at that temperature.

2SO3 (g) + CO2 (g)  CS2 (g) + 4O2 (g)

so Kc = [O2]^2[CS2]/[SO3]^2[CO2]
you've given the mole amount of each species. to calculate the molarity, just divided each by 20. then, plug it into the formula above.

[Homer]

what about the O2 molarity?

[scribble]

it says that the mixture is made by mixing so3 and co2. this means initially theres only reactants/no products. it tells you the mol of CS2 at equilibrium. and the ratio of O2 and CS2 is 4:1, so n(O2)eq=4n(CS2)eq

[Alwin]

An equilibrium mixture of gases was made by mixing sulfur trioxide and carbon dioxide. It consisted of 0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2 in a 20 L vessel. Calculate the value of the equilibrium constant at that temperature.

2SO3 (g) + CO2 (g) ⇄ CS2 (g) + 4O2 (g)

Me personally, I always find using tables makes things look pretty I'll go step by step for you, so you can see what's happening.
NOTE THAT I ASSUME WE MUST FIND THE AMOUNT OF O₂ FIRST otherwise the question makes no sense

	2SO3 (g) +	CO2 (g)	⇄	CS2 (g) +	4O2 (g)
Initially:	?	?		?	?
Reacted:	?	?		?	?
Equilibrium:	0.022	0.014		0.028	?

Now, since in the question it states that only sulfur trioxide and carbon dioxide are present initially, and CS₂ present in equilibrium, clearly the forward reaction dominated. Hence we know that 0.028mol of CS₂ formed and can calculate the amount of 2SO₃ and CO₂ reacted using stoichiometry:

	2SO3 (g) +	CO2 (g)	⇄	CS2 (g) +	4O2 (g)
Initially:	?	?		?	?
Reacted:	0.056	0.028		0.028	0.112
Equilibrium:	0.022	0.014		0.028	?

We can now complete the table, as follows:

	2SO3 (g) +	CO2 (g)	⇄	CS2 (g) +	4O2 (g)
Initially:	0.078	0.042		0	0
Reacted:	0.056	0.028		0.028	0.112
Equilibrium:	0.022	0.014		0.028	0.112

Since we now have our amounts of products and reactants, we can go about calculating the equilibrium constant at that temperature





really hoping I didnt screw up somewhere, oh and ofc hoping it helps
EDIT: and... out ninja'd XD i knew i was taking too long on the tables

[brightsky]

that will be 4 times the molarity of SO3. think about it this way. you start with a certain amount of SO3 and CO2 (the question doesn't tell you what the initial amount is but it doesn't matter). you chuck them in a vessel and they react to form CS2 and O2. for every CS2 molecule that is formed, 4 O2 molecules are formed. so in the end, the amount of CS2 and the amount of O2 will be in a 1:4 ratio, even though the reaction does not go to completion (i.e. even though 2 moles of SO3 and 1 mole of CO2 doesn't give you 1 mole of CS2 and 4 moles of O2). note that the amount of SO3 LOST and the amount of CO2 LOST will be in a 2:1 ratio. the final amounts of SO3 and CO2 just depend on how much of each you have in the first place.

[scribble]

^If ever in doubt, Alwins method is spot on and foolproof. the nInitial nReacted and nEquilibrium table thing usually makes what you're doing crystal clear. especially when you're doing more annoying questions.

[psyxwar]

Since we now have our amounts of products and reactants, we can go about calculating the equilibrium constant at that temperature

How do you type it like that? :O

[Homer]

i see thanks guys and psyxwar i usually use http://www.codecogs.com/latex/eqneditor.php

[9_7]

Hey can somebody please help me with this question? D:

The following mixtures are heated at constant volume. Describe the change in the amount of hydrogen gas for each case.
a) N2 + 3H2 <-> 2NH3            (exothermic)
I thought if you changed the hydrogen gas, in this case as a reactant, it would be a forward equilibrium so there would be a decrease in hydrogen gas. But the answer says that it is an increase? Where am I wrong :s thanks a lot!!

[psyxwar]

Hey can somebody please help me with this question? D:

The following mixtures are heated at constant volume. Describe the change in the amount of hydrogen gas for each case.
a) N2 + 3H2 <-> 2NH3            (exothermic)
I thought if you changed the hydrogen gas, in this case as a reactant, it would be a forward equilibrium so there would be a decrease in hydrogen gas. But the answer says that it is an increase? Where am I wrong :s thanks a lot!!

I think you're misinterpreting the question. It says what is the change on H gas amount when you heat the mixture, not when you change the amount of H gas. Increasing the temp of an exothermic reaction causes the equilibrium position to be shifted left, hence more reactants (and thus more H gas).

[9_7]

I think you're misinterpreting the question. It says what is the change on H gas amount when you heat the mixture, not when you change the amount of H gas. Increasing the temp of an exothermic reaction causes the equilibrium position to be shifted left, hence more reactants (and thus more H gas).

Oh crap right.. Thanks mate