Wording of Answers - Please help if possible

[stephanieteddy]

One of the issues I have with this unit of Chem is that I struggle with the wording of my answers and explaining what is happening.

What is the best way to say that the forward reaction is favoured? Equilibrium is pushed to the right?

I still don't understand how it can be 'pushed' to the right. For example, if I remove a product from the equilibrium mixture, does that mean the system with favour the forward reaction? Also, does this then mean that the rate of the forward rxn is increased? If not, what is actually happening when the 'forward rxn is favoured'?

Thank you so much.

[stephanieteddy]

Pwetty pwease

[daniel034]

I normally just equilibrium will shift in favour of the forward reaction. And it means the rate of the forward reaction has increased relative to the rate of the reverse reaction [either forward reaction increasing (e.g addition of reactants or reverse decreasing (e.g removal of products)]. The rate of the forward reaction will then decrease until dynamic equilibrium is reestablished!

[stephanieteddy]

Okay cool! My teacher said that the rates of reaction aren't altered, but I was a bit unsure. So you're saying when there is a shift to favour the forward rxn, the rate of forward rxn has increased? 100% sure? I'll blame you if I get it wrong on Tuesday!

[thushan]

Woah woah woah hold it there - daniel means RELATIVE to the back reaction. Just so there are no misunderstandings - in THIS case, IMMEDIATELY after the removal of product, the rate of the forward reaction does not change (concentration of reactants initially don't change), but the rate of the back reaction decreases as concentration of products have decreased (remember reaction rate is dependent on concentration of reacting species!). This means that the rate of the forward reaction is higher than that of the back, and hence the forward reaction is favoured.

[daniel034]

Nah i'm saying that either the forward reaction increases or the reverse reaction decreases or both.
Haha, i'm scared now 

[thushan]

Yeah, yeah I know - but I think steph misunderstood you, hence my post

[stephanieteddy]

You guys are the best! Thank you! I feel a lot better now! I always struggle to answer/understand what it actually means by 'forward reaction is favoured'!

Silly me, DUH! How could the forward reaction increase if the reactant concentration hasn't changed! RELATIVE to the back reaction, the forward reaction is now higher, which means forward reaction is 'favoured'.

One more quick question, sorry, when I get a question that says explain why the rate of reaction has increased with perhaps an increase in temperature, can I just say: Higher temperature results in higher kinetic energy, therefore there will be a higher proportion of collisions that are fruitful, resulting in a higher rxn rate?

And if a question asks why the increase in temperature changes the K value (for an endothermic), can I say: Since the reaction is endothermic, the system shifts to absorb the extra energy and hence the forward reaction is favoured increasing the K value? (that last bit sounds a bit dodgey to me)

[stephanieteddy]

Sorry Daniel!

[daniel034]

No worries! This is helping me as much as it's helping you. Good luck for Tuesday

[thushan]

You guys are the best! Thank you! I feel a lot better now! I always struggle to answer/understand what it actually means by 'forward reaction is favoured'!

Silly me, DUH! How could the forward reaction increase if the reactant concentration hasn't changed! RELATIVE to the back reaction, the forward reaction is now higher, which means forward reaction is 'favoured'.

One more quick question, sorry, when I get a question that says explain why the rate of reaction has increased with perhaps an increase in temperature, can I just say: Higher temperature results in higher kinetic energy, therefore there will be a higher proportion of collisions that are fruitful, resulting in a higher rxn rate?

And if a question asks why the increase in temperature changes the K value (for an endothermic), can I say: Since the reaction is endothermic, the system shifts to absorb the extra energy and hence the forward reaction is favoured increasing the K value? (that last bit sounds a bit dodgey to me)

First question - yup, and also mention that there are also more collisions (just collisions) per unit time as well.

As for second question - Hmm. This is a toughie. I would accept that, but I'd prefer it if you said "since the reaction is endothermic, the system shifts to absorb the extra energy and hence the forward reaction is favoured; since there is no initial change in concentration of any of the species and hence no initial change in Q, K itself must have increased initially and Q followed it subsequently."

[daniel034]

As for second question - Hmm. This is a toughie. I would accept that, but I'd prefer it if you said "since the reaction is endothermic, the system shifts to absorb the extra energy and hence the forward reaction is favoured; since there is no initial change in concentration of any of the species and hence no initial change in Q, K itself must have increased initially and Q followed it subsequently."

What does Q stand for sorry?

[stephanieteddy]

Ahhh! I see, because with all the other changes made to a system that cause a shift, (pressure, adding, removing), the concentration of at least one of the species has been altered, causing that net shift in whichever direction.

However, in the case of changing temperature, there is no change in concentration, meaning that the K value itself must have changed and hence Q( do I have to specify that this is the reaction quotient or is it okay just to put Q?) eventually changes as it approaches equilibrium.

I feel like I should be paying you for this. Thank you so much, I will sleep a lot easier on Monday night now!

[thushan]

Brilliant - you've got it!

Dan - Q is the reaction quotient (concentration fraction).

[daniel034]

Cheers! Makes sense now