[raaaainm]

In studying for the GRE I am presented with this problem:



The answer is E. You are not given a calculator to use on the GRE so it is impossible to calculate the exact value.

I would like a confirmation of my logic here: The ΔG is very negative and is given in units of kJ. That means the magnitude of ΔG is very large, which means that the ln(Kp) expression is positive as Kp > 1.

Therefore the only possible solutions are D and E. At 298K, ΔG cannot be largely negative unless Kp is a very large number. Therefore the only answer is E.

The problem with this logic however is that suppose there is a large large large number of moles being considered in the problem. A ΔG in kJ/mol multiplied by a sufficiently large number of moles could also produce a large negative ΔG. With a small number of moles, E is the only answer. With a large number of moles, D can be an answer as well. Does the temperature at which the problem is evaluated (298K) limit the number of moles possible to a relatively small set?

Can anyone shed light on this?

[Borek]

ΔG is given per "mole of reaction" (in this case - 1 mole of ozone decomposing), not per "large large number of moles".

[raaaainm]

Thank you. For future problems I will assume that they are all dealing with one-mole.

[magician4]

A ΔG in kJ/mol multiplied by a sufficiently large number of moles could also produce a large negative ΔG.

no
multiplying ΔG ( in kJ/mol ) by some value with the dimension "mol" won't gain you another Δ G ( in kJ/mol) thereafter, but something with the dimension of "energy" (i.e. kJ ) instead: the total energy (here: released) if so ans so much moles of reaction would have taken place
... and of course , this value would (here:) increase with more stuff (i.e. ozone) to be destroyed, as the absolute amount of energy released is a clear function of the amount of stuff you treat with the respective process
burn a lot of coal and gain a lot of heat , it's as simple as that

nevertheless, the amount of heat released per a given amount of coal still remains a constant

... and it's the same with ΔG , as it is related to a well-definded amount of things happening (usually one mole of formula reaction)

hence, ΔG is and remains independent from the number of moles processed

regards

Ingo

[Borek]

Thank you. For future problems I will assume that they are all dealing with one-mole.

That's not what "mole of reaction" means. If I tell you that ΔG for the reaction

2A + 3B 4C

is 100 kJ, I doesn't mean 100 kJ per mole - that's not how the reaction is written. It is 100 kJ per 2 moles of A reacting, or 100 kJ per 3 moles of B reacting, or 100 kJ per 4 moles of C produced.

I could also tell you ΔG is 50 kJ per

A + 3/2B 2C

and it would mean exactly the same thing - 50 kJ per "mole of reaction as written". 50 kJ per one mole of A reacting means 100 kJ per two moles of A reacting and so on.

[raaaainm]

Thank you. For future problems I will assume that they are all dealing with one-mole.

That's not what "mole of reaction" means. If I tell you that ΔG for the reaction

2A + 3B 4C

is 100 kJ, I doesn't mean 100 kJ per mole - that's not how the reaction is written. It is 100 kJ per 2 moles of A reacting, or 100 kJ per 3 moles of B reacting, or 100 kJ per 4 moles of C produced.

I could also tell you ΔG is 50 kJ per

A + 3/2B 2C

and it would mean exactly the same thing - 50 kJ per "mole of reaction as written". 50 kJ per one mole of A reacting means 100 kJ per two moles of A reacting and so on.

Thank you Borek, this is incredibly helpful and clarified it much better for me.

Thank you as well magician!