[Waizac]

Hi guys,

Sorry to bother you but I have an assignment to hand in in few weeks. It's 8 questions long, some with sub questions up to 'e'. Bar two sub questions, I've completed it. However, due to a mixture of uncovered topics, mild dyslexia and being an idiot, I'm a little unconfident about it.
I'm not asking for people to go over the whole thing, I would be really grateful for few pointers on even one or two of the questions. If possible I'd rather be told the technique for answering a question rather than the answer, unless it's just a small mistake.

Sorry again, I don't usually ask for help however my last assignment just scraped 67% and I've been having anxiety attacks all day.

I'm a little apprehensive to post it here in case someone from my course googles the questions and finds my answers. Plagiarism and all that.
So if anyone could let me PM my answers I would appreciate it. (Some of the questions use images so it may have to be done by email, if you're comfortable with that, hotmail has a view online option to ensure I'm not sending you a virus or anything).

Sorry for the long post and thanks in advance!

[Waizac]

Okay. Sorry guys, I realise that that was probably a bit too much to ask for.
I've since had some confirmation on my technique and it seems to be going 'okay'. However, I'm still unsure about this one question. Everyone I've spoken to hasn't commented on it but it's still bugging me as the way I answered it just seems to be too simple.
Anyway here's the question and my answer (hoping someone could tell me if I've messed up asap. I'm still incredibly paranoid about it being stolen O_o) Thanks guys.



5.   The experiment described below is to calculate the enthalpy of hydration of sodium carbonate. i.e. the process: Na2CO3(s)  + 10H2O(l)    Na2CO3.10H2O(s)

   An accurately known quantity of anhydrous sodium carbonate was weighed by difference and added to some water contained in a calorimeter. The solution was stirred and the change in temperature was measured. The experiment was repeated using an accurately weighed quantity of hydrated sodium carbonate, Na2CO3.10H2O.  The following results were obtained:

   Expt 1: Using Na2CO3

   Volume of water in calorimeter            50 cm3
   Initial temperature of water               21.3C
   Mass of weighing bottle + Na2CO3         15.262g
   Mass of weighing bottle               10.232g
   Final temperature of water               26.5C

   Expt 2: Using Na2CO3.10H2O

   Volume of water in calorimeter            50 cm3
   Initial temperature of water               20.6C
   Mass of weighing bottle + Na2CO3.10H2O      16.208g
   Mass of weighing bottle               10.976g
   Final temperature of water               15.7C

   In the following calculation you can assume that the density and specific heat capacity of all the solutions are the same as those of water, ie  density = 1 g cm-3 and specific heat capacity = 4.2 J g-1 C -1 .

   (a)   Calculate the heat energy transferred by the water in experiment 1.
                                    
      1g of water = 1cc.  fin temp = 26.5, in temp = 21.3, diff = 26.5 – 21.3 = 5.2

                  5.2 x 4.2 = 21.84 J           

   (b)   Scale up your answer to (a) to calculate the enthalpy change of the process per mole of Na2CO3(s) to three significant figures (take care with the sign)
            Na2CO3(s)  +  excess water    Na2CO3(aq)    
   
           21.84/1000 = -0.02KJMol ('-' added as it is an exothermic reaction)

Once again, thanks!

[rabolisk]

(a) You forgot one part. Check the units that you used. What you multiplied does not cancel out to Joules.

(b) With the answer to (a) being wrong, this is wrong. But even if it had been right, this isn't right. What you did was simply convert joules to kilojoules, but the question is asking to calculate the enthalpy change per one mole of sodium carbonate. The (wrong) answer from (a) is not calculated for one mole of sodium carbonate. You have to use stoichiometry to go from one to the next.

[Waizac]

 Right, I'm not sure if I've followed you correctly. But here are the new answers. :/ Any closer?

 (a)
16.208 - 10.976 =5.232g (mass of Na2CO3 used in exp)

5.2 (difference) x 5.232 x 4.2 (Joules per 1c of temp rise) = 114.27J (5sig fig)

(b)
One mol of NaCO3 = 106g mol ... 106/5.232 = 2.12 ....20.25993 x 114.27 = 2315.1 J mol ...2315.15/1000 = -2.32 KJ mol-1 (to three sig fig, '-' added as it is exothermic).

Edit: second attempt ^

[rabolisk]

(a) No. What does 4.2 signify? Also, there's no reason why you would have 5 sig figs for the answer. You should bring an extra digit over for calculation in (b), but when you report the answer for this part, 5 sig figs isn't right.

(b) This would be correct if your value from (a) was correct. Check sig figs though.

[Waizac]

4.2 is the amount of Joules required to heat 1g of water by one degree. Do I not use that?
I have no knowledge on any rule systems with sig figs. The examiners so far haven't commented on them, but it would be good know. Are there any official systems set out?


5.2 (difference) x 5.232 x 4.2 (Joules per 1c of temp rise) = 114.26688J

(b)
One mol of NaCO3 = 106g mol ... 106/5.232 = 2.12 ....20.25993 x 114.26688 = 2315.0389901184 J mol ...2315.0389901184/1000 = -2.315 KJ mol-1 (to four sig fig, '-' added as it is exothermic).

[rabolisk]

Yes, 4.2 is the amount of energy required to heat 1 g of water by 1 degree celsius. But what you did in your calculation is not reflective of that.

[Waizac]

Right, is this more on track? And btw, thanks for all this help, I seriously appreciate it!

5.2 (difference) x 5.232 x (4.2 x 50) (Joules per 1c of temp rise, and '50' the amount of grams of water present in calorimeter) = 5678.4J

[rabolisk]

That's correct. You should technically have 2 sig figs in your answer.

[Waizac]

Wahey! Thanks a lot, that must've taken you a lot of patience.


Just to confirm (b) with the new (a) results.

(106/5.232) x 5678.4/1000 = 115.0440367 . = -115KJmol-1

[Waizac]

I think I might have hit a snag here.
Going over the questions again more carefully, I came up with these answers:
Here's from the previously worked out technique

(a) 5492.76 J (b) -115.75 kJ

(Looking back at (a), I wasn't sure why the weight of the reactant was included so I tried it without using that value (I'm sure this is probably wrong but I'll include them anyway). I got (a) -1092J (b) -23.01KJ)

Questions (c) and (d) are exactly the same except for experiment two which is on the second post.
I got (a) -5383.73 J (b) 109.1KJ

(again without using weight value in (a)
(a) -1029 (b) 20.9KJ)

This seemed to be right until reaching (e) and (f)

(e)   Use the results from (b) and (d), the energy cycle below, together with Hess's law, to calculate the enthalpy change of the reaction, to three significant figures:
Na2CO3(s)  +  10H2O(l)  =  Na2CO3.10H2O(s)
                                            H1  
            Na2CO3(s)  +  10H2O(l)  =  Na2CO3.10H2O(s)

         H3    +excess water          +excess waterH2

                                 Na2CO3(aq)

By my understanding, H1 should be equal to H1 = H3 - H2   

So it would either be

-115.6 -(109.1) = -224.8kJ

Or,

-23.01 - ( 20.9) = -43.91kJ   

Both of these seemed acceptable until here:   

   (f)   The accepted value for this enthalpy change is ΔH  = -91.1 kJ mol-1. What is the % error in the experimental result, and what are the most likely causes of error?         

Both of my answers to (e) seem too far off from -91.1 to be correct. This could be intentional but I don't think the % of error would be that high?

[rabolisk]

You are correct for (a) and (b). My apologies, the answer should be 1092 rather than 5383.7. What did you get for c and d? Your approach is correct in (e), I believe. It seems like you messed up in (c) or (d). I got an answer that is close enough to -91.

[Waizac]

Yeah, I used the 1gmol value for Na2CO3 instead of Na2CO3.10H2O for question (d). I get -79.2kJ in the end, was that similar to yours?

And once again I really can't thank you enough. If I could give you all the snacks in the world, I would!

[rabolisk]

-79.2 kJ is what I got, and I'm pretty sure that's correct.