Chemistry 3/4 2013 Thread

[zvezda]

Having trouble with this question:

Consider the following thermochemical equations. 

C(s) + O2(g)  -->  CO2(g)                          ΔH = -393.5 kJ/mol
S(s) + O2(g)  -->  SO2(g)                          ΔH = -296.1 kJ/mol
CS2(l) + 3O2(g) -->  CO2(g) + 2SO2(g)      ΔH = -1072kJ/mol

The enthalpy of reaction (in kJ/mol) for C(s) + 2S(s) --> CS2(l) is
A. -1762
B. -873.3
C. +86.3
D. +382.4

Help is appreciated 

Are you familiar with Hess' Law?

[Limista]

Are you familiar with Hess' Law?

No. What is it and how will it help me with this question?

[Edward21]

Having trouble with this question:

Consider the following thermochemical equations. 

C(s) + O2(g)  -->  CO2(g)                          ΔH = -393.5 kJ/mol
S(s) + O2(g)  -->  SO2(g)                          ΔH = -296.1 kJ/mol
CS2(l) + 3O2(g) -->  CO2(g) + 2SO2(g)      ΔH = -1072kJ/mol

The enthalpy of reaction (in kJ/mol) for C(s) + 2S(s) --> CS2(l) is
A. -1762
B. -873.3
C. +86.3
D. +382.4

Help is appreciated 

Basically, this a really fun type of question if you have a little OCD, because it all should work out perfectly, nice and neat 
Work out what you need from the equation. You need 1 mol of C, 2 mol of S going to 1 mol of CS2.

Numbers refer to the order of the 3 reactions present 
1) Do nothing

2) Multiply all by 2 (INCLUDING THE DELTA H)

3) Reverse the whole thing (THUS DELTA H VALUE IS NOW POSITIVE) now an endothermic reaction

Rewrite all 3 equations with the changes one above the other. 1) -393.5kJ/mol 2) -592.2kJ/mol 3) +1072kJ/mol should be the Delta H values obtained
You'll notice the 1 mol O2 and the 2 mol O2 cancel out on the left side with the 3 mol O2 on the right of reaction 3!!
Then cancel out the CO2, the 2SO2 and you're left with the equation you wanted by adding them together. THEN add the delta H values, to follow suit.
It seems a little complex, but once you do a couple you'll be flying 

[Scooby]

Just wondering... I know we need K to be large and for the analysis to be carried out at a low temperature... but why do we need an excess of SCN^- as opposed to Fe³⁺?

[Limista]

Basically, this a really fun type of question if you have a little OCD, because it all should work out perfectly, nice and neat 
Work out what you need from the equation. You need 1 mol of C, 2 mol of S going to 1 mol of CS2.

Numbers refer to the order of the 3 reactions present 
1) Do nothing

2) Multiply all by 2 (INCLUDING THE DELTA H)

3) Reverse the whole thing (THUS DELTA H VALUE IS NOW POSITIVE) now an endothermic reaction

Rewrite all 3 equations with the changes one above the other. 1) -393.5kJ/mol 2) -592.2kJ/mol 3) +1072kJ/mol should be the Delta H values obtained
You'll notice the 1 mol O2 and the 2 mol O2 cancel out on the left side with the 3 mol O2 on the right of reaction 3!!
Then cancel out the CO2, the 2SO2 and you're left with the equation you wanted by adding them together. THEN add the delta H values, to follow suit.
It seems a little complex, but once you do a couple you'll be flying 

Thanks a billion 

[scribble]

Just wondering... I know we need K to be large and for the analysis to be carried out at a low temperature... but why do we need an excess of SCN^- as opposed to Fe³⁺?

youre trying to find the concentration of Fe3+ so you want to try turn all (or as much as possible) of Fe3+ to FeSCN2+ which is red and can detected. this can be done by adding SCN-.
if the Fe3+ is in excess then theres some thats still colourless and some thats red and youre not really going to be able to determine the concentration.

[SocialRhubarb]

why do we need an excess of SCN^- as opposed to Fe³⁺?

The aim of the whole experiment is to determine the concentration of the Fe³⁺ in a solution, which we want to do by driving the forwards reaction and converting as much of the Fe in solution into FeSCN²⁺ ions which we pick up using UV-visible spectroscopy.

If we're trying to find the concentration of the iron (III) ions in solution, while adding excess iron (III) ions will drive the forwards reaction, it will also kind of void the measurement we wanted to make in the first place because we've just changed the concentration of iron (III) in solution. We can't really just subtract the concentration we add from the end result because if the iron (III) really is in excess, some will remain in solution. So we drive the forwards reaction by adding SCN^- instead of iron (III).

Additionally, D is right because A, B and C are wrong.

[Edward21]

Just wondering... I know we need K to be large and for the analysis to be carried out at a low temperature... but why do we need an excess of SCN^- as opposed to Fe³⁺?

I think it's alluding to the pure fact that if you're investigating the conc. of Fe3+ you wouldn't add more of it in your investigation!! Adding the SCN- will ensure the position of equilibrium moves to the right, red coloured complex for the UV-Vis investigation, as there is a fixed amount of Fe3+ there may be a maximum red colour reached at which an absorbance would be measured to ascertain the Fe3+ concentration ie. adding excess SCN- at low temperature in identical conditions to the sample, to known standard solutions! This way you could plot a calibration curve, oh how very neat all the spectroscopy is tying in with the Unit 4 stuff 

[lzxnl]

youre trying to find the concentration of Fe3+ so you want to try turn all (or as much as possible) of Fe3+ to FeSCN2+ which is red and can detected. this can be done by adding SCN-.
if the Fe3+ is in excess then theres some thats still colourless and some thats red and youre not really going to be able to determine the concentration.

To phrase it slightly more clearly, if you have an excess of iron(III) ions, you will still have some unreacted iron(III) ions as well as your thiocyanate iron(III) ions, but the latter is what you're analysing with UV-Vis. Therefore, your analysis will only tell you how much thiocyanate iron(III) ions there are, not how much iron(III) there was initially.

In contrast, if you add an excess of thiocyanate, most of the iron(III) ions will react and form thiocyanate iron(III) ions which you can then detect with UV-Vis. As now, the concentration of thiocyanate iron(III) is now really close to the initial concentration of iron(III), your analysis will be a whole lot more accurate.

[09Ti08]

I'm not sure whether I'm the only one who finds this awkward or not, help would be greatly appreciated!

Now, my problem is the unit of enthalpy change, which is kJ/mol. kJ makes sense to me, but the latter bit doesn't. For example, if we have a reaction like this:
2Mg(s)+O2(g)->2MgO(s), , the question is to find how much energy is released when 7.29g of Mg is burned. 
I know this is the way to do it: E=(7.29/24.3)*(1200/2)=180kJ. But when I examine this closely, I find a problem: If the unit of enthalpy change is kJ/mol, then why do we need to divide by 2 (the coefficient of Mg in the balanced equation)? Does the unit refer to the equation as a whole ?

Thank you!

[lzxnl]

You mean the mol bit?
What's a mole? A mole is a number that is equal to the number of carbon-12 atoms that make up 12 grams. In short, it is roughly equal to 6.02*10^23 units.
So a mole of reaction means 6.02*10^23 copies of this reaction.
So with your reaction, if 1.204*10^24 magnesium atoms and 6.02*10^23 oxygen atoms reacted together to form 1.204*10^24 magnesium oxide units (better term for fundamental particles of ionic compounds?), you would have one mole of reaction and hence an enthalpy decrease of 1200 kJ

So your last statement is correct; unit refers to the equation as a whole. That's why if you double the coefficients, the enthalpy change doubles too.

[9_7]

The pH in a swimming pool needs to be maintained within a narrow range, preferably between 7 and 8. This is usually achieved with a combination of hydrochloric acid and calcium hypochlorite. The hypochlorite ion reacts with H3O+ ions (either from water or the added acid) to produce hypochlorous acid. Hypochlorous acid acts as an antibacterial agent.
OCI-(aq) + H3O+(aq) -- HOCI(aq) + H2O(l)
a)   Given that hypochlorous acid has an acidity constant of 2.88x10^-8 M, determine:
i)   The concentration of HOCI in an HOCI solution of pH 6.4
ii)   The percentage ionisation of HOCI at this concentration
b)   What is the effect on the pH of a swimming pool water of a decrease in concentration of HOCI over a period of time. Use Le Chatelier’s principle to explain your answer.

Can someone please show me what they got? I don't know if my answers are correct ((

[Edward21]

The pH in a swimming pool needs to be maintained within a narrow range, preferably between 7 and 8. This is usually achieved with a combination of hydrochloric acid and calcium hypochlorite. The hypochlorite ion reacts with H3O+ ions (either from water or the added acid) to produce hypochlorous acid. Hypochlorous acid acts as an antibacterial agent.
OCI-(aq) + H3O+(aq) -- HOCI(aq) + H2O(l)
a)   Given that hypochlorous acid has an acidity constant of 2.88x10^-8 M, determine:
i)   The concentration of HOCI in an HOCI solution of pH 6.4
ii)   The percentage ionisation of HOCI at this concentration
b)   What is the effect on the pH of a swimming pool water of a decrease in concentration of HOCI over a period of time. Use Le Chatelier’s principle to explain your answer.

Can someone please show me what they got? I don't know if my answers are correct ((

i) [HOCl] = 5.50X10^-6M
ii) 7.24%
b) Well the system will partially oppose the decrease in HOCl by increasing the rate of the forward reaction, the position of equilibrium then moves to the right, this means that less H3O+ ions will be produced as a result, thus the pH rises as the molarity of H+ ions in the solution decreases/

[9_7]

o.o I think I got mine incorrect, could I see your solution??

[09Ti08]

The pH in a swimming pool needs to be maintained within a narrow range, preferably between 7 and 8. This is usually achieved with a combination of hydrochloric acid and calcium hypochlorite. The hypochlorite ion reacts with H3O+ ions (either from water or the added acid) to produce hypochlorous acid. Hypochlorous acid acts as an antibacterial agent.
OCI-(aq) + H3O+(aq) -- HOCI(aq) + H2O(l)
a)   Given that hypochlorous acid has an acidity constant of 2.88x10^-8 M, determine:
i)   The concentration of HOCI in an HOCI solution of pH 6.4
ii)   The percentage ionisation of HOCI at this concentration
b)   What is the effect on the pH of a swimming pool water of a decrease in concentration of HOCI over a period of time. Use Le Chatelier’s principle to explain your answer.

Can someone please show me what they got? I don't know if my answers are correct ((

A/
i. I don't really care about the molecular formula of hypochlorous acid, you can just write HA<=>H+ + A-
Now
ii. = =7.23%
B/
Concentration of HOCl decreases => equilibrium position shifts to the right => H3O+ is consumed => concentration of H 3O+ decreases => pH increases