OCR Chemistry A F325 Equilibria, Energetics and Elements Wed 13 June 2012

Another general question:

2A + 1B ---> 2C

Say you react 0.6 mol of A, so you need 0.3 mol of B. How many mol of C would you form? 0.6 as well?
Kind of getting confused in equilibrium questions with initial concentrations and they don't tell you the amount of product formed.

yes 0.6 mol of C

depends on what you say by "react". Are you losing 0.6 mol of A due to the reaction you mean?

Reply 1381

VQG

11

January 2011 titration question 7ai)

You work out the number of moles of S203- reacted, but why do you divide that answer by 4 to work out the amount of O2? I've looked over that question, but I still can't understand it...

QP: http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61491_pp_11_jan_gce_f325.pdf?
MS: http://pdf.ocr.org.uk/download/ms_11/ocr_61999_ms_11_gce_f325.pdf?

Reply 1382

VQG

11

Original post by DavidMRoper

yes 0.6 mol of C

depends on what you say by "react". Are you losing 0.6 mol of A due to the reaction you mean?

Yeah as in initial concentration being 1.0 mol dm-3, and you're left with 0.4 mol dm-3 at equilibrium.

Reply 1383

VQG

11

Sorry for the barrage of questions it's just that I've had to teach myself all of the last module, and electrochemical cells/half cells to myself over half term with no guidance :frown:

Reply 1384

Killjoy-

8

Original post by VQG

January 2011 titration question 7ai)

You work out the number of moles of S203- reacted, but why do you divide that answer by 4 to work out the amount of O2? I've looked over that question, but I still can't understand it...

QP: http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61491_pp_11_jan_gce_f325.pdf?
MS: http://pdf.ocr.org.uk/download/ms_11/ocr_61999_ms_11_gce_f325.pdf?

I'd explained this one earlier:

Original post by Killjoy-

You'd have half the moles of I2 reacting...
But from the middle equation moles of Mn(OH)3 is twice moles of I2.
And moles of O2 is a quarter of the moles of Mn(OH)3.

Thus the moles of O2 you started with is a quarter of the moles of S2O32–(aq).

You may want to follow the quote to see what was originally asked.

(edited 11 years ago)

Reply 1385

VQG

11

Original post by Killjoy-

I'd explained this one earlier:

You may want to follow the quote to see what was originally asked.

Right thanks a lot :smile:

Reply 1386

VQG

11

But even then in the first and second step:
In the first step you have 4Mn(OH)3, then all of a sudden you only react 2Mn(OH)3 in the second step :s-smilie:

Do you just multiply the moles in step 2 and 3 x2 to make it all equal? Then it makes sense.

Reply 1387

walkbesideme

Original post by arvin_infinity

Oh man..someone else clarify this but textbook says...octahedral complexes containing bindentate ligands can show cis/trans iso..

am almost checking if I've gone crazy and cannot read :colondollar:

Octahedral complexes can only show cis/trans isomerism when they contain 2 of the same bidentate ligand and 2 of another unidentate ligand molecule
for example [Fe(C2O4)2(H2O)2]2+ would show cis/trans isomerism however a molecule such as [Fe(C2O4)2(NH2CH2CH2NH2)]2+ would not

Reply 1388

walkbesideme

Original post by VQG

But even then in the first and second step:
In the first step you have 4Mn(OH)3, then all of a sudden you only react 2Mn(OH)3 in the second step :s-smilie:

Do you just multiply the moles in step 2 and 3 x2 to make it all equal? Then it makes sense.

Nope remember the Stoichiometry of the equation are just relative mole relationships therefore even though it says 4 in one setp and 2 in the next you still have the same number of moles

Reply 1389

VQG

11

Original post by walkbesideme

Nope remember the Stoichiometry of the equation are just relative mole relationships therefore even though it says 4 in one setp and 2 in the next you still have the same number of moles

Ah okay makes sense now.
Does anyone know whether we need to learn the structure of EDTA? Ethanedionate ion came up in one of the papers.

Reply 1390

walkbesideme

Original post by VQG

Ah okay makes sense now.
Does anyone know whether we need to learn the structure of EDTA? Ethanedionate ion came up in one of the papers.

I'm sure we don't; can you remember which paper this was?

Reply 1391

sumsum123

2

Original post by VQG

Ah okay makes sense now.
Does anyone know whether we need to learn the structure of EDTA? Ethanedionate ion came up in one of the papers.

Yerh I remember it coming up! I did learn how to draw it and the function of EDTA just incase. You never know. (random way I remember its structure is if you look at it its the 'en' bidentate NH2CH2CH2NH2 but with the hydrogens of the nitrogens replaced with ethanoate ions, CH3COO-)

Reply 1392

mathsclown

7

My quick entropy notes if anyone needs :smile:

Some reactions occur at room temperature even though they are endothermic, so an alternative factor other than energy
must be present.

Entropy is the measure of disorder or randomness present in a substance. Solids have low entropy while liquids and gases have high levels of entropy. Temperature is also a factor as it gives rise to an increase in kinetic energy and so increases entropy.

Enthalpy and entropy are linked together, the balance between them will determine whether or not a reaction will occur spontaneously.

Second law of thermodynamics is linked to entropy, in our everyday lives, our activties increase entropy ie.
- heat leaving a hot room to a cold one
- dropping of a box of matches that were in order in a box
- melting of ice above 0 degrees
- melting of a salt in liquid water
So entropy, for any system, processes occur such that the entropy of the total universe increases.

By looking at a chemical equation we can see whether or not entropy will increase or not, this is done by looking at the number of particles, if the number of particles increases from reactants to products then entropy increases.

(edited 11 years ago)

Reply 1393

mathsclown

7

And the entropy changes:

Standard entropy changes
Entropy is affected by temperature and so standard entropy of substances have found under standard conditions. The units are JKmol^-1 and to find deltaS: deltaS = deltaSproducts - deltaSreactants
(If negative then the reaction is not feasible)

Gibbs free change
Whether or not a reaction will occur is dependant on enthalpy (exo means more likely) and entropy (high positive values at higher temps) => as temperature effects entropy this is also a factor.

These factors can be combined to give gibbs free energy equation which measures the feasibility of a reaction
deltaG = deltaH - TdeltaS, T in kelvin and deltaS in KJmol^-1, if negative value is found then reaction is feasible.
NOTE: like electrode potentials, it only shows reaction feasibility, not that it will occur as it doesn't take into account other factors ie. activation energy.

To find out at what temperature a reaction is feasible the following steps can be taken:

make deltaG => 0
so 0 = deltaH - TdeltaS
deltaH = TdeltaS
t = deltaH/deltaS and then just sub-in values

Reply 1394

otrivine

14

any one wants to do question session

Reply 1395

pinkcherrytart

13

construct a redox equation for the following:

aqueous thiosulfate ions S2032- are oxidised to sulfate VII ions SO42- bu chlorine gas which is reduced to cloride ions.

The correct equation is:

S2O32- + 4Cl2 +5H2O-> 2so42- +8Cl + 10H+

I just dont seem to get how :s

Reply 1396

Oromis263

2

Original post by pinkcherrytart

construct a redox equation for the following:

aqueous thiosulfate ions S2032- are oxidised to sulfate VII ions SO42- bu chlorine gas which is reduced to cloride ions.

The correct equation is:

S2O32- + 4Cl2 +5H2O-> 2so42- +8Cl + 10H+

I just dont seem to get how :s

Ok, have you learnt the method of finding the redox equation using oxidation numbers? I'll demonstrate with your example, just give me a minute. :smile:

Step 1. Find the change in oxidation number of the sulphur and chlorine.

Step 2. Calculate overall change in oxidation numbers.

The rest of the method is shown in this picture. :smile:

(I apologise for combining the multiplying component step, and the balancing step into one equation. :smile:

)

(edited 11 years ago)

Reply 1397

Dreamweaver

10

Original post by pinkcherrytart

construct a redox equation for the following:

aqueous thiosulfate ions S2032- are oxidised to sulfate VII ions SO42- bu chlorine gas which is reduced to cloride ions.

The correct equation is:

S2O32- + 4Cl2 +5H2O-> 2so42- +8Cl + 10H+

I just dont seem to get how :s

First write the half equations. Oxygen always comes out as water so add water to the side with less oxygen. Add H+ to balance the hydrogens. Then add electrons to balance the charges. Then add the half equations together but make sure you multiply them out to make sure the electrons cancel out.

Reply 1398

Oromis263

2

Original post by otrivine

any one wants to do question session

Sure, I'm up for it, I should probably make sure I remember this module. :smile:

Reply 1399

Smiley Face :)

What colour is [Cu(NH3)4(h20)6] ?
What is the definition of a transition metal?
Name a multidente ligand?
What is an optical isomer?
What colour is the [Fe(OH)3] preciptitate?
Tell me what pKa is?
Why do we need a salt bridge?
Explain the use of tranistion metal ions if haemoglobin?
What doe cis platin do?

OCR Chemistry A F325 Equilibria, Energetics and Elements Wed 13 June 2012

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