TrueTears question thread

[dummy]

Due to the change in pressure, the extent of the reaction may be altered.
However, reactions would still occur with the consumption of reactants and production of products in the ratio accordant to the reaction given.

If you are talking about the ratio of the amount / concentration of substances present in the solution, no, it will differ from the one in the initial condition.

If you are talking about the ratio of substances in the reactions that happen (despite its yield of producing the product), yes, it will stay the same.

[TrueTears]

Due to the change in pressure, the extent of the reaction may be altered.
However, reactions would still occur with the consumption of reactants and production of products in the ratio accordant to the reaction given.

If you are talking about the ratio of the amount / concentration of substances present in the solution, no, it will differ from the one in the initial condition.

If you are talking about the ratio of substances in the reactions that happen (despite its yield of producing the product), yes, it will stay the same.

reason? proof?

[dummy]

The reaction reaches a point where the number of forward reactions = the number of backward reactions.
Therefore, reactions still occur according to their mole ratio.
ie, (mole) ratio (of particles in reactions) is constant
________________________________________________________________________________

Since pressure changed, more forward reactions occur.
Number of moles of reactants and products therefore changed.
Concentration therefore altered.
ie,ratio (of each particles present in solution) is not constant

[TrueTears]

So the ratio of the amount of substances (in mol) at the new equilibrium (after pressure change) is the same as the old equilibrium?

[dummy]

So the ratio of the amount of substances (in mol) at the new equilibrium (after pressure change) is the same as the old equilibrium?

Since pressure changed, more forward reactions occur.
Number of moles of reactants and products therefore changed.
Concentration therefore altered.
ie,ratio (of each particles present in solution) is not constant

I hope this could help, mate.

[kamil9876]

Lol, by some fluke, in TT's example the X:Y ratio is constant, though this almost never happens so it's very bad practice. You know that the ratios of how much X was consumed to how much Y is produced is constant, and you know this because of conservation of matter etc.

In TT's example, say A moles of X were consumed, that means that 3A moles of Y were consumed. THe ammount of X and Y left is:





But the X:Y:Z ratio is not conserved. And if instead you started of with 4 moles of Y rather than 3 then the ratio would not be a constant (dependant on A).


I don't know why you are worrying so much about the ratio of the actual ammounts. Just use the fact that the ratio of the consumption is is constant and you can work out everything from there. As you can see you have no reason for believing that the ratio of the ammounts stay constant or change unless you do some completely unnecesary mathematics.

[TrueTears]

So the ratio of the amount of substances (in mol) at the new equilibrium (after pressure change) is the same as the old equilibrium?

Since pressure changed, more forward reactions occur.
Number of moles of reactants and products therefore changed.
Concentration therefore altered.
ie,ratio (of each particles present in solution) is not constant

I hope this could help, mate.

No it's not, its the same.

[dummy]

as in n(x):n(y):n(z) present in the solution = the same?
After change in condition, forward reaction increased, more n(Z) produced, less n(x) and n(y) present.
I can't see how does the ratio remain the same, TrueTears.

Or are you talking about the mole ratio between n(x):n(y) ??

[TrueTears]

Mol ratio between n(x) : n(y)

So the ratio of the amount of substances (in mol) at the new equilibrium (after pressure change) is the same as the old equilibrium?

Maybe this could help, mate.

[dummy]

but i said its not constant?

So the ratio of the amount of substances (in mol) at the new equilibrium (after pressure change) is the same as the old equilibrium?

Since pressure changed, more forward reactions occur.
Number of moles of reactants and products therefore changed.
Concentration therefore altered.
ie,ratio (of each particles present in solution) is not constant

I hope this could help, mate.

No it's not, its the same.

I suggested two ways of intepreting this question, and this is one of them.

[TrueTears]

But it is constant for the question I provided or did you think there was another invisible question that magically appeared out of nowhere?

[dummy]

reached, obviously the K value stays the same, but does the ratio of X:Y:Z still stay the same?

[TrueTears]

So the ratio of the amount of substances (in mol) at the new equilibrium (after pressure change) is the same as the old equilibrium?

Since pressure changed, more forward reactions occur.
Number of moles of reactants and products therefore changed.
Concentration therefore altered.
ie,ratio (of each particles present in solution) is not constant

I hope this could help, mate.

clearly incorrect.

[lacoste]

X + 3Y -> 2Z
 you begin with 1 mol of X 3 mol of Y and no Z
 now the first equilibrium it gets to is
 0.7 mol of X 2.1 mol of Y but only 0.6 mol of Z
 temperature is constant. the K value is clearly (0.6^2)/(0.7 * 2.1^3)
 if the pressure is increased [meaning forward reaction is favoured] at the new equilibrium reached, obviously the K value stays the same, but does the ratio of X:Y:Z still stay the same?
 or how about just X:Y?

Hey TT, is this from checkpts? i havnt done checkpts yet so idk.

[TrueTears]

No, it was something that poped up in my mind.