Multiple choice question - equilibrium

[Kaatie]

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Consider the following reaction:
FeO(s) + CO(g) (double arrow) Fe(s) + CO2(g) ΔH = –283kJ
Which of the following changes to equilibrium conditions
would favour the formation of iron?
A the addition of more finely powdered FeO
B an increase in temperature
C a decrease in pressure
D the removal of carbon dioxide

can't decide between a and d (im guna feel really dumb if its neither lol)

[singh.with.me]

u sure the questions typed right??
as in, maybe option D in meant to be about carbon MONoxide??

if not, noooo idea =[

i feel pretty dumb myself =]

[gurmies]

I can't see any real distinction between A and D for this reason:

If you add more FeO, equilibrium will shift to the right, thus favouring the production of Fe. Similarly, if you get rid of CO2, the equilibrium will have to shift to that same right side to make more CO2 + Fe I assume. I like A slightly better than D though.

[study-freak]

Originally Posted by Kaatie

Consider the following reaction:

FeO(s) + CO(g) (double arrow) Fe(s) + CO2(g) ΔH = –283kJ

Which of the following changes to equilibrium conditions

would favour the formation of iron?

A the addition of more finely powdered FeO

B an increase in temperature

C a decrease in pressure

D the removal of carbon dioxide



can't decide between a and d (im guna feel really dumb if its neither lol)

It's D. See below for brief explanation

Originally Posted by gurmies

I can't see any real distinction between A and D for this reason:

If you add more FeO, equilibrium will shift to the right, thus favouring the production of Fe. Similarly, if you get rid of CO2, the equilibrium will have to shift to that same right side to make more CO2 + Fe I assume. I like A slightly better than D though.

No, there is a distinction. Adding FeO does not affect the equilibrium since there is no change in concentration of any substance. FeO is a solid.
Thus, my answer is D.

[Pwnage101]

its definitely D. in A they are testing whther you know increasing the surface area by making FeO a powder increases the reaction rate, but this happens for both the forwarda and reverse reaction, and does not affect they equilibrium/yield, just how fast it achieves this equilibrium....

[Fortify]

Is it D?

[gurmies]

Originally Posted by Pwnage101

its definitely D. in A they are testing whther you know increasing the surface area by making FeO a powder increases the reaction rate, but this happens for both the forwarda and reverse reaction, and does not affect they equilibrium/yield, just how fast it achieves this equilibrium....

True that.

[kaz1]

Originally Posted by Pwnage101

its definitely D. in A they are testing whther you know increasing the surface area by making FeO a powder increases the reaction rate, but this happens for both the forwarda and reverse reaction, and does not affect they equilibrium/yield, just how fast it achieves this equilibrium....

Sorry but this really confuses me. Wouldn't adding powdered FeO increase the concentration which would then in turn force the reaction to the products side?

Can someone also explain adding FeO would increase the reaction rate as well?

[Pwnage101]

Originally Posted by kaz1

Sorry but this really confuses me. Wouldn't adding powdered FeO increase the concentration which would then in turn force the reaction to the products side?

Can someone also explain adding FeO would increase the reaction rate as well?

The question, as presented by the OP, states:

"Consider the following reaction:
FeO(s) + CO(g) (double arrow) Fe(s) + CO2(g) ΔH = –283kJ
Which of the following changes to equilibrium conditions
would favour the formation of iron?
A the addition of more finely powdered FeO
B an increase in temperature
C a decrease in pressure
D the removal of carbon dioxide "


now i take 'the addition of more finely podered FeO' to mean 'instead of adding FeO in blobs, illa dd FeO as a powder'. I can see how u might take it as the addition of MORE FeO, which (you are right) would be correct. However, like i said in my previous post, i believe they added a) to test wheher you know about reaction rates compared to equilibrium.

You would agree, though, D is definitely correct? they can get away with this Q in a HSC because u have to choose the MOST CORRECT answer, and to me D is most definitely correct, while A is incorrect if you take it as i took it (more finely powdered FeO instead of a blob of FeO, not as in add more FeO and let it be finely podered).

Sometimes you have to analyse the Q and teh options - why would they mention 'more finely podered FeO'? Then it clicks - reaction rate increases. But by now you would have eliminated B and C as correct options, so youa re left with A and D. And you have doubt over A because it could mean what i took it to mean, so your safest option is D, which is definitely correct.

PS. Nail the MC in exams, trust me it helps =)

[study-freak]

What I thought was:
Add more FeO (ignoring the "finely powdered" part).
This increases the amount of FeO in the reaction vessel.
However, FeO is just a solid. It is not going to vaporise or dissolve to a liquid for reaction.
Hence FeO concentration remains the same whereas its amount increases.

For example, if I add more HCl to 0.1M HCl solution, I'll be increasing its concentration.
But if I add more 0.1M HCl solution to 0.1M HCl solution, the amount of HCl present increases but its concentration remains the same.

Correct me if I'm wrong.

[Pwnage101]

yeh, but u ignored the 'finely powdered' part...LOL...you cant do that, your changing the question.

Although your conclusion is correct, i believe your interpretationis wrong.

Also, when u say "if I add more 0.1M HCl solution to 0.1M HCl solution, the amount of HCl present increases but its concentration remains the same." - TRUE, but in an equilibrium, we assume the volume of the system tremains constant, since it is in a closed system. Here you increased teh amoun t of HCL as well as the volume of water (and thus teh volume of the system), which is not consistent with the question, which deals witha closed systm...

as for 'For example, if I add more HCl to 0.1M HCl solution, I'll be increasing its concentration.' - yes if u add 10mol/L HCL, no if u add 0.001mol/L HCl.

Just learn from this in the future, every word is placed there for a reason in an exam, learn to think critically....

[study-freak]

hmm i c..
thanks for reply

[Kaatie]

ohh not more iron. thanks guys. very badly worded

[FFC]

Originally Posted by study-freak

It's D. See below for brief explanation


No, there is a distinction. Adding FeO does not affect the equilibrium since there is no change in concentration of any substance. FeO is a solid.
Thus, my answer is D.

correct