Chemistry 3/4 2013 Thread

[lzxnl]

What is the best way to explain why a certain spontaneous reaction occurs?
1) the strongest oxidant reacts with the strongest reductant
2)...

Do we mention the E0 values of the half equations?

Because the electrolysis of molten KCl involves the strongest oxidant with the strongest reductant but its still a forced reaction.
I don't know how to explain why molten KCl to K and Cl2 isn't spontaneous.

You're right; molten KCl only has K+ and Cl-

In electrolysis, the strongest reductant is more readily oxidised, and vice-versa.

[achre]

2012 exam 2 question 9 MC.
An 11th hour question, but: how do we now that the value of Q isn't still increasing? It's 400 at 30 seconds, why does the fact that it's greater than K mean the system is moving to consume the products? Maybe it gets up to 410 at 35 seconds before tending towards K?

[barydos]

2012 exam 2 question 9 MC.
An 11th hour question, but: how do we now that the value of Q isn't still increasing? It's 400 at 30 seconds, why does the fact that it's greater than K mean the system is moving to consume the products? Maybe it gets up to 410 at 35 seconds before tending towards K?

Equilibrium is reached is when the concentration fraction equals the equilibrium constant.
So if CF (concentration fraction) is greater than K, it hasn't achieved equilibrium yet. More reactants will be produced to lower the CF as it approaches equilibrium (since reactants are in the denominator, so more reactants = lower CF). Until it CF = K.
Also.. yeah generally when things react, they tend to like to be in equilibrium.

[Patches]

Question which is more my misunderstanding of the equilibrium quotient than anything else:

when we, say, add something to a system and the equilibrium position changes (overall reaction rate increases), does this change the value of Q? How then does it get back to K if K doesn't change at constant temperature?

[barydos]

Question which is more my misunderstanding of the equilibrium quotient than anything else:

when we, say, add something to a system and the equilibrium position changes (overall reaction rate increases), does this change the value of Q? How then does it get back to K if K doesn't change at constant temperature?

By adding something that moves the equilibrium position to the right, the yield of products is increased, which thus increases Q. But after a period of time, it'll come back to K. As in, mixtures will always return to K.
Q is simply defined as the concentration fraction AT a certain time. So yes Q does change at that time. But eventually it return to K that's what mixtures do.

[ahat]

Equilibrium is reached is when the concentration fraction equals the equilibrium constant.
So if CF (concentration fraction) is greater than K, it hasn't achieved equilibrium yet. More reactants will be produced to lower the CF as it approaches equilibrium (since reactants are in the denominator, so more reactants = lower CF). Until it CF = K.
Also.. yeah generally when things react, they tend to like to be in equilibrium.

Going on what Anonymiza said, it may also help to understand that in an industrial setting, changes will be made so that the forward reaction will be favoured. Despite with time, the fact products would go back to their equilibrium concentrations, this is prevented by removing the products as they are formed. Thus, you get more product and the forward reaction will continue to occur.

[Daenerys Targaryen]

For zwitter ion stuff.

If the amino acid has say another carboxylic group branching from the R group, would that lost its H (COOH) too?

[Lejn]

For zwitter ion stuff.

If the amino acid has say another carboxylic group branching from the R group, would that lost its H (COOH) too?

In a condition that allows the other carboxyl to donate, yes, it will donate as well.

[ineedhelp2]

Anyone able to help?

8.g of an unsaturated hydrocarbon, C6H10, was treated with 60g of Br₂ dissolved in a suitable solvent. On completion of the reaction, 16.2g of HBr had been evolved and 12g of Br₂ remained in excess. Calculate the number of bromine atoms in each molecule of C6H10 which are involved in:
a) substitution
b) addition

I just found all the moles for each compound and didn't know where to go from there. The answer for both of them are 2 Br atoms.

Thanks in advance

[neonperson]

I think I better go to sleep before I regret it in the morning.

Good luck everyone! Even though we are "competing" with each other, I genuinely wish you all the best!!!!!!! May this year's cohort astound those examiners!!!!!! YAAAAAY!!!!!!!

[SocialRhubarb]

Haha, this is from a STAV paper, right?

Is the mass supposed to be 8.2? I'm just guessing it is because there's a dot after the eight and it would give us nicer numbers.

So we have 0.1 mol of this hydrocarbon. Lets look at the products of the reaction:
16.2g of HBr, or approximately 0.2 mol. How do we form HBr? Through a substitution reaction between bromine and a hydrogen atom on the hydrocarbon. Each substitution reaction adds one bromine to the hydrocarbon and forms one molecule of HBr. There is 0.2 mol of HBr and 0.1 mol of the hydrocarbon, so there have been two substitution reactions on each hydrocarbon, each adding one bromine atom. Hence, 2 atoms of bromine per hydrocarbon molecule.

We also have 12g of Br₂ in excess, which means that 48g of bromine has reacted, which is 0.3 mol of bromine. Now, 0.2 mol of that bromine went to producing HBr, which means that we still have 0.1 mol of bromine unreacted. Where does this go? Addition reaction! Chemistry is exciting. So we have 0.1 mol of bromine undergoing an addition reaction with 0.1 mol of the hydrocarbon, which means each hydrocarbon is undergoing one addition reaction. Each addition reaction adds to bromine atoms to the hydrocarbon, so the answer is again, 2 bromine atoms per hydrocarbon molecule.

[lolipopper]

good luck bitches. Thanks for helping me out for my numerous questions and indirectly giving me hope in chemistry. Love you all for that!

[Edward21]

Ok, I feel a little like I'm waking up a sleeping beast here by finding this thread, but I just really wanted to notify everyone that has contributed here since the start of Year 12. I want to thank each and every single one of you for all of your help, I cannot believe I got a 50 in chemistry and I really believe coming to this thread and refining my skills got me that edge I really needed. I hope you all got what you want, and sorry if this random post is... somewhat annoying to the moderators or people who want to forget VCE Chemistry ever happened    but I really wasn't expecting to do so well in this subject and decided to let all you guys know how thankful I am that I became part of this thread, I think I went here almost every day and just trying to answer each question that was brought up here was fantastic revision. I think we all helped each other in our own way, and I really do hope you got what you wanted in this subject.

And the other subjects you did for that matter!

Love you guys <3