Chemistry HL/SL help

[DropBoite]

1)

1st: MnO₄^- + 8H⁺ + 5e^- --> Mn²⁺ + 4H₂O

2nd: 1/2 Cl₂ + e^- --> Cl^-

Firstly, times the 2nd equation by two to get whole numbers. This gives: Cl₂ + 2e^- --> 2Cl^-

You'll also want to switch that equation around in order to get the electrons on the opposite side. This gives: 2Cl^- --> Cl₂ + 2e^-

Now, you want to equal the number of electrons on each side in order to cancel them out from the redox reaction.

Times the 1st equation by two: 2MnO₄^- + 16H⁺ + 10e^- --> 2Mn²⁺ + 8H₂O

Times the 2nd equation by five: 10Cl^- --> 5Cl₂ + 10e^-

We can add these equations together, to give: 2MnO₄^- + 16H⁺ + 10e^- + 10Cl^- --> 2Mn²⁺ + 8H₂O + 5Cl₂ + 10e^-

Now you can cancel out the common species on both sides (in this case, only the electrons).

Resulting in: 2MnO₄^- + 16H⁺ + 10Cl^- --> 2Mn²⁺ + 8H₂O + 5Cl₂

What is noticed is that you multiplied both equations by something, but if that's not necessary in some case, why do it? I say this because you could have simply multiplied the second equation by 2.5 no? Is it that whole numbers are preferred ? Thanks! Great answers 10/10 for clarity and structure!

[Drake Glau]

You are multiplying the equations to make equal numbers of electrons on both sides of the equations so they will cancel out in the end when you combine the two. You can't have a fraction of an electron so you need to stick to whole numbers. It honestly doesn't matter what you multiply them by, it's the ratio that matters because in the end things just starting canceling out to the numbers you need whether you had 4563 electrons or just 10

[DropBoite]

Guys often I'm told to solved a question of this sort:

" the pH of a mixture of 50.0 cm3 of 0.100 mol dm–3 aqueous ammonia and 50.0 cm3 of 0.0500 mol dm–3 hydrochloric acid solution"

by equating [NH3] to [NH4]. Hell! I just don't get the logic! Ammonia is a weak base, so it's initial concentration in undissociated form is unchanged and stoichiometrically its products are formed in 1:1 ratios, but I haven't ever seen THIS relationship anywhere, can someone please explain this to me quick?We have mock exams in 4 Hours!

Another thing came up:Does ethanoic acid behave the same way?

Because the marking scheme says that ethanoic acid doesn't! Which is peculiar!

[CH3C00H] DOESN'T EQUAL [CH3COO-] apparently!? And how on earth will [CH3COO-] have a initial concentration at all?!

Here are the details: " Determine the pH of a solution formed from adding 50.0 cm3 of 1.00 mol dm–3 ethanoic acid, CH3COOH(aq), to 50.0 cm3 of 0.600 mol dm–3 sodium hydroxide, NaOH(aq)."

(initial)[CH3COOH] = 0.500 mol dm–3 and) eqm [CH3COOH] = 0.200 mol dm–3;

(initial)[CH3COO–] = 0.300 mol dm–3 and) eqm [CH3COO–] = 0.300 mol dm–3;

How does that make sense? Don't get me wrong I can understand the equilibrium concentration being hte difference between acid and base concentration, 0.3 mol dm3 makes sense, but the INITIAL concentration MUST be an error!? *panic*

Exam went well! BUT I still need the help!! Please help!

Edited April 23, 2012 by DropBoite

[Drake Glau]

Anything "weak" doesn't full disocciate. So the concentration of the acid and it's conjugate will NEVER be equal (unless you do stuff to force the equilibrium....which ib won't do to you)

Does your data booklet include values of pKa for some basic weak acids? Or just ka...either work. You would need this value to solve that example with the ethanoic acid.

(initial)[CH3COOH] = 0.500 mol dm–3 and) eqm [CH3COOH] = 0.200 mol dm–3;

(initial)[CH3COO–] = 0.300 mol dm–3 and) eqm [CH3COO–] = 0.300 mol dm–3;

^This makes no sense at all.

CH3COOH<=>CH3COO- + H+

There is no initial concentration of CH3COO-, at equil [H+] and [CH3COO-] are both 0.300moldm3 and [CH3COOH] is 0.200moldm3.Oh well I guess you can get ka from this, don't need the value. Ka=[H+][CH3COO-]/[CH3COOH] (it's just like kc, except for acids)

Is this data included in the example "Determine the pH of a solution formed from adding 50.0 cm3 of 1.00 mol dm–3 ethanoic acid, CH3COOH(aq), to 50.0 cm3 of 0.600 mol dm–3 sodium hydroxide, NaOH(aq)."?

If so, you can use it to find the ka value of the acid and then use it to found out how much in your example would dissociate and then it's just some stoichiometry...

[DropBoite]

OKay I'm going to stop you right there It makes sense to the examiner! Get my flow !?

What do I do now?

Edited April 26, 2012 by DropBoite

[DropBoite]

MODS PLEASE MAKE SURE THE COMMON FOLK CAN DOWNLOAD THE IMAGES-DELETE AFTER READING, THANK YOU!

Hey guys a quick and easy bunch of question to those that know kinetics, no biggie, but I can't get it cause I really wasn't taught well so please help!

Okay so attached are three questions I don't get becasue of the order of the reactions.

Table 1/ question 1) Find the order of [X] . Apparently the answer is 2nd order. No rctn.

2) Find the order of NO. Apparently the order is 2nd again... Rctn: 2NO + 2H2 --> N2+ H2O

3) Find the order of NO, again it's apparently 2nd. Rctn: 2NO + O2 --> 2NO2 (due to stoichiometry?)

Could someone please explain why it's second order because at I just done see how the changes are of that nature!

Edited April 28, 2012 by Summer Glau

[Drake Glau]

1) [x] doubled, rate increased by a factor or 4, 2nd order. It would be first order if doubling the concentration doubled the rate. You can think of it like this for other questions. If the concentration of one changes and the other doesn't you can take the factor of which the one concentration increases (2 in this case, sometimes 3 if tripled, or 4 if quadrupled) and then you compare it to the factor that the rate increased.

Example to make some more sense: [x] quadrupled, [y] stays constant, rate increased by a factor of 16, the order is 2nd because 4²=16 and the order is the power. Remember your rate law (rate=k[x]ⁿ[y]^m) and how to use it

2) (22.8*10^-5)/(10.10*10^-5)=the factor by which the rate increased.

The factor by which the concentration of NO increased (using data that keeps the hydrogen concentration constant) is 1.5

Basically you need to find x for 1.5^x=(22.8*10^-5)/(10.10*10^-5)

3) stoichiometry doesn't affect the order. It's a common misconception so don't fall for it

This one is more intense, ugh...

The order of O2 is 1st order, double the concentration and the rate doubled.

Rate=k[NO]ⁿ[O2] This is your rate law since O2 is first order, we don't know n still.

Now using experiment 2 and 3 you can set up a ratio to get things to cancel out and let you solve for n.

I'll be doing (experiment 3/experiment 2) it doesn't matter which way you divide, just keep it all the same though.

6.00*10*-2=k(7.00*10^-2)ⁿ(7.00*10^-2)

7.50*10^-3=k(3.50*10^-2)ⁿ(3.50*10^-2)

I used an underline to try and make a division bar

Ok, solve for n.

The k will cancel out.

The concentrations of O2 will give you a number that you can move to the other side

Eventually this will give you a Rate=(some number)ⁿ and you can solve for n easily.

As you can see, when they only change one concentration and keep the other constant stuff is easy, you can look at the power of that concentration to get your rate to double/triple/whatever. When everything is changing you have to work out the entire rate law >.>

[DropBoite]

Ksp = [Ni2+]×[OH–]²;

[OH–] = 2[Ni2+] hence Ksp = 4[Ni2+]³;

Thanks Drake! Here's another one, how did they get 4[Ni2+]³ ?? I just don't get it , because the mole ratio is already accounted for? Let me think aloud and tell me if I'm right: I have no idea why there's 4 but I think I get three because it's the same concentratio so simply increase the power to simplify. But why the 4...?

[LMaxwell]

Does the titre count as raw or processed data in an IA?

I know that means go into processed data, but how about the titre?

[Drake Glau]

Ksp = [Ni2+]×[OH–]²;

[OH–] = 2[Ni2+] hence Ksp = 4[Ni2+]³;

Thanks Drake! Here's another one, how did they get 4[Ni2+]³ ?? I just don't get it , because the mole ratio is already accounted for? Let me think aloud and tell me if I'm right: I have no idea why there's 4 but I think I get three because it's the same concentratio so simply increase the power to simplify. But why the 4...?

I'll treat this like a math problem to see if that helps.

Ni2 will be y, OH- will be x

ksp=yx²

x=2y

sub it in

ksp=y(2y)²

simplify and you get 4y³

Hope that made sense All I did was get rid of the chemical symbols and give you just letters lol

Does the titre count as raw or processed data in an IA?

I know that means go into processed data, but how about the titre?

Raw data. You only measured it so it remains raw.

[eelnedross]

Hey, i'm doing Organic Chemistry atm; and i was wondering how does the solubility of organic compounds vary?

I saw an excercise with an ester and ketone; CH₃COCH₃ and CH₃COOCH_3. The answer was that the ester is less soluble in water. Why is that?

My theory is that the ester has a higher M_r, and that makes the compound less soluble in water.

Thank you!

[Keel]

Hey, i'm doing Organic Chemistry atm; and i was wondering how does the solubility of organic compounds vary?

I saw an excercise with an ester and ketone; CH₃COCH₃ and CH₃COOCH_3. The answer was that the ester is less soluble in water. Why is that?

My theory is that the ester has a higher M_r, and that makes the compound less soluble in water.

Thank you!

The solubility of esters vs ketones have a long history. Have a read:

http://www.thestudentroom.co.uk/showthread.php?t=710253

[DropBoite]

Ksp = [Ni2+]×[OH–]²;

[OH–] = 2[Ni2+] hence Ksp = 4[Ni2+]³;

Thanks Drake! Here's another one, how did they get 4[Ni2+]³ ?? I just don't get it , because the mole ratio is already accounted for? Let me think aloud and tell me if I'm right: I have no idea why there's 4 but I think I get three because it's the same concentratio so simply increase the power to simplify. But why the 4...?

I'll treat this like a math problem to see if that helps.

Ni2 will be y, OH- will be x

ksp=yx²

x=2y

sub it in

ksp=y(2y)²

simplify and you get 4y³

Hope that made sense All I did was get rid of the chemical symbols and give you just letters lol

Does the titre count as raw or processed data in an IA?

I know that means go into processed data, but how about the titre?

Raw data. You only measured it so it remains raw.

Thanks Drake, you the man! Sorry for seriously late reply haha! ....

[SerUmbras]

Hey, i'm doing Organic Chemistry atm; and i was wondering how does the solubility of organic compounds vary?

I saw an excercise with an ester and ketone; CH₃COCH₃ and CH₃COOCH_3. The answer was that the ester is less soluble in water. Why is that?

My theory is that the ester has a higher M_r, and that makes the compound less soluble in water.

Thank you!

Well, the keytone has more places where water can hydrogen bond than the ester.

Basically, the solubility relies on two things;

1. The number of Hydrogen bonding sites (dipoles, lone electron pairs, OH bonds etc) on the molecule.

2. The length of the molecule.

The first is fairly self explanatory. The more places the molecule can H-bond, the more water it can attach to, the higher its solubility. The second is a bit funny.

If your molecule is CH₃COOCH₃, then it has 2 n.e.p (non-bonding electron pairs) and a rather polar section in the middle. This gives it a good opportunity to bond with water. However, if the molecule is really, REALLY, long, like CH₃(CH₂)₂₇COOCH₃, then the huge long non-polar chain of carbons will begin to overwhelm the effect of the polarity on the end. The bonds that it can make don't help about 27/28's of the molecule to dissolve, and so it becomes less soluble.

Hope that helped.

Edited May 12, 2012 by illidan101

[eelnedross]

Hey guys, i've got another question. I'm not too sure how the reactions of the alkanes with halogens works. I've seen something related to free-radicals, but i don't have any idea how to solve a problem related to that. Anyone care to explain me how it works?

Thank you in advance.

[Drake Glau]

The molecular halogen turns into atomic halogens through homolytic cleavage.

The alkane loses an H creating a carbocation.

One of the radicals from the halogen attaches to the carbocation.

The H attaches to the other radical made from the homolytic cleavage and gives you a hydrohalogen thingy.

That's it in a nutshell.

[eelnedross]

The molecular halogen turns into atomic halogens through homolytic cleavage.

The alkane loses an H creating a carbocation.

One of the radicals from the halogen attaches to the carbocation.

The H attaches to the other radical made from the homolytic cleavage and gives you a hydrohalogen thingy.

That's it in a nutshell.

Lol. Alright, i've understood that one of the products is, for example: HCl, HBr and HI.

Nonetheless, what does a carbocation refer to?

[Drake Glau]

The carbon that has a partial positive charge when the H leaves it.

[eelnedross]

The carbon that has a partial positive charge when the H leaves it.

Something like this: C₂H₅⁺? Then it joins a chlorine ion, Cl^-? To form C₂H₅Cl.

[Drake Glau]

The molecule itself does not have a positive charge. But I guess you CAN think about it like that. ONLY the C that had an H removed has the positive charge and it is VERY small and VERY unstable, it is not a formal charge like the way you are representing it. That small unstably charged positive carbon will bond to the chlorine RADICAL, it's not an "ion" because it, again, does not have a formal charge. The chlorine atom has an extra electron which is what makes it a radical.