Question

In need of Calorimetry help as soon as possible.

Answer 1

Because it's late as it is and for a full understanding, it'll just provide the document.

Answer 2

These are what need to be changed with a bit of an explanation

Answer 3

Part 1
#1 Balanced reaction include phase and ions
#3 q surroundings = -q system Use -q for #4 Enthalpy Change.

Part 2
#1 Balanced reaction include phase and ions
#3 Find moles NaOH by coverting NaOH mL to Liters then use Molarity NaOH to convert to mol NaOH
The molarity of NaOH is given in the directions.
#4 Divide enthalpy change from 2 by moles from 3 to calculate total enthalpy change.

Percent error
Recalculate percent error using corrected values in Parts 1 and 2.

The set up is good; however due to prior calculation errors, the answers are incorrect.
Please try this again after going back to correct the answers for Parts 1 and 2.

Show the work applying Hess's Law to calculate the enthalpy change. Add #4 from part 1 to #4 of part 2. (the calculated enthalpy of each reaction)

#5 Classify the reactions in the lab as endothermic or exothermic. Proceed to support this classification and attractive forces between ions and the solute or solvent.

Answer 4

@welshfella @iambatman @Conqueror
would you help out with this at all? c:

Answer 5

Looks like you have a good understanding of the subject.
Do check your calculations, and complete parts #4 and #5.
If no one helps, I'll take another look when I return later in the day.

Answer 6

I appreciate that, thank you mathmate c:

right now my problem is that I don't actually understand it and need help getting there. This was pretty much information gathering with minor adjustments to what I do understand

Answer 7

@mathmate are you available at the moment?

Answer 8

@Luigi0210 would you mind stepping in?

Answer 9

@alanate
We'll go through your work part by part, in the same order as your work.

Answer 10

To summarize, part I deals with determining the enthalpy change for the dissolution of NaOH in water, and part II deals with the enthalpy change for the neutralization of NaOH and HCl.

Answer 11

Alright, thank you c:

Answer 12

For Part I, here are my comments:
using \(\Delta Q = mc\Delta T\)
we have
\(\Delta Q = 4.184\times 205.0\times (27.8-24.2)=3.087~kJ\)
Note the volume is 205 (and not 200) mL.
Divide 3.087 kJ by 0.06338 mol gives enthalpy change = -48.72 kJ/mol
(since energy is dissipated outside of NaOH).

Answer 13

Don't worry, I'm following along (the browser keeps me inactive on the page for some reason)

Answer 14

So the resulting completed enthalpy equation would look like:
\(NaOH_{(s)}->Na^+ + OH^-+48.7~kJ/mol\)
The above equation should be part of conclusion (2).

Answer 15

looks like I had that equation aside from phase symbols and the actual enthalpy c:

Answer 16

For part II, we have very little numerical difference, namely due to 4.184 instead of 4.18.
I get 2.517 kJ disspated heat for 0.5M*0.1L = 0.05 mol gives
2.517/0.05=50.33 kJ/mol of energy dissipated for the neutralization of aqueous NaOH and HCl.
0.05 mol is used because 100 mL of NaOH is the limiting reagent in the neutralization (assuming both reactants are exactly 0.5M).

Answer 17

I will leave it to you to complete the enthalpy equation similar to part I, and which forms conclusion (3).

Answer 18

Thank you for your help mathmate, I appreciate it ^-^

Answer 19

Are you ok with conclusion #1 that relates to Hess's law?

Answer 20

wait, no no, I'll need help with that as well, sorry, I was focused on part I and II

Answer 21

Can you tell me where does Hess's law come into play in this experiment?

Answer 22

hold on, I'm still calculating.
As to my knowledge, Hess's law is a manifestation that enthalpy is a state function.

Answer 23

Pretty much just verifying enthalpy

Answer 24

exactly.
You have performed two experiments, each of which is one stage of the following:
\(NaOH{(s)}+HCl_{(aq)} -> NaCl_{(aq)}+H_2O_{(l)} + (?) kJ \)
The amount of heat can be obtained from standard tables, which should compare with the sum of the energies of each of the two stages of experiments that you have performed.

Answer 25

Sorry, I'm being distracted here. I need help with determining the moles of NaOH on question 3 of part II.
I'm given mL, not grams

Answer 26

mL can be converted to grams of the reactant through molarity.

Answer 27

100 mL of 0.5M NaOH is 0.1L of 0.5M NaOH equals 0.1L*0.5mol/L = 0.05mol

Answer 28

Note how the dimensions (units) cancel out the L to give mol.

Answer 29

Ah, okay, that makes sense, sorry, it's kind of loud and distracting here. Thank you again

Answer 30

no problem. If I am not around, it looks like @hwyl might be interested in helping too!

Answer 31

She seems to have the same assignment c:

Answer 32

But you must have different results (mass, volume, temperatures, etc.)

Answer 33

that's possible but this is only hypothetical being that it's an assignment without any lab work, my masses and temperatures were a given, and not actually measured myself.

Answer 34

oh! I thought they were experimental results! It would have been nice to do the experiment first hand!

Answer 35

in practice they're the same thing

Answer 36

I know it would definitely be, being a virtual student, we have computer visuals at least

Answer 37

you use the same calculation for calorimetry but perhaps a little bit more involved because you need to measure the containers and calculate dissipation

Answer 38

That's why I was surprised to see that the results in part I actually exceeded the expected results, and was looking for possible "negative leak" of energy.

Answer 39

this is just to make sure that the concept is driven home

the same is applied for everything you use in the process but just keep adding

Answer 40

or subtracting

Answer 41

Anyway, are you all set @alanate ?

Answer 42

yes indeed I am c: For like the 14th time, thank you so much xD

Answer 43

You're welcome! :)