Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Frederick95 on March 07, 2008, 08:00:54 AM
Title: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 08:00:54 AM
Calculate the molar enthalpy of combustion of acetone, using the experimental evidence.
Given: Mass (m) of water: 100.0 g Specific heat capacity(c) of aluminum: 0.91 J/ g degrees Celsius Mass (m) of aluminum can: 50.0 g Initial temperature (Ti) of calorimeter: 20.0 degrees Celsius Final temperature(Tf) of calorimeter: 25.0 degrees Celsius Mass (m) of acetone burned: 0.092 g
MY WORK IS AS FOLLOWS
Required: Molar enthalpy of combustion of acetone?
Q = mc DT = (100.0 g) (0.91 J/(goC)) (5.0 oC) = 455 J H = - Q = - 455 J = -0.455 kJ Number of moles (n) = (given mass (g))/(gram-formula mass) =
HOW DO I ACCOMPLISH THIS STEP? THANKS AGAIN
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Borek on March 07, 2008, 08:35:07 AM
Stop starting new threads with the same information, if you have additional questions - reuse your old thread. This is final warning.
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: mo_1984 on March 07, 2008, 03:29:23 PM
i have the same rpoblem as fredrick listed. it is the EXACT same question he asked any1 plz help me how to do it.
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 03:55:05 PM
Dear Frederick95, Dear Mo_1984;
First Part: ΔT is OK!
But Part 2: Q = ….. is wrong! It must be: Q = QWater + QAl !! (Remember: mWater is not equal mAl.)
Frederick95: I think you know, since a short while, how to finish.
I hope it helps for both of you.
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: mo_1984 on March 07, 2008, 04:14:43 PM
yea but if u r putting in the mass of alu then u have to add the heat capacity also???
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 04:20:02 PM
Dear Mo_1984;
You’re right!, – cAl for Al, and cWater for …. !
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: mo_1984 on March 07, 2008, 04:24:03 PM
so when using q= mct formula for m add alu and water masses and for c add heat capacity for both??? and r u sure abt this man because every time i ask some1 they are telling me different answers so i am soo confused.
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 04:42:18 PM
Can you show us an example of this Argos?
Please
Thank you for any further help
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 04:42:43 PM
Dear Mo_1984;
NO!, – I told you: QTotal = QWater + QAl !! You need twice: Q = mcΔT !!!
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 04:48:09 PM
Dear Frederick95, Dear Mo_1984;
QTotal = QWater + QAl = mWcWΔT + mAlcAlΔT.
I hope I have been of help to you.
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 04:48:58 PM
Thank you Argos:)
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 04:57:11 PM
Argos, do you have a link to any website that explains this formula?
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 05:11:40 PM
Dear Frederick95;
Not at the moment, - but think about:
At begin both (Water and Calorimeter, build from Al) have the Temperature Ti. After burning the Acetone both must have the identical Temperature Tf. Conclusion: Both must have taken enough Heat corresponding to their individual Mass m and individual c!
That’s all. You have to make you “A Picture” (see below).
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 05:21:31 PM
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 05:28:06 PM
Argos -----Thanks ;)
But.. My sum is 2317.5 kJ
I must find the molar enthalpy.
Is there a formula for this? I cannot find it:(
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 05:34:05 PM
Dear Frederick95;
I got quite close the same, if , if I divide YOUR result by 1000!, but for how many Moles Acetone, and that’s how many Grams?
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 05:40:58 PM
I got 0.0015 mol C3H6O(acetone) as an answer. But what to do with this?
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 05:46:38 PM
Dear Frederick95;
You got Q, but you need Q per Mole, also Q / Mole! (Use at least 2 figures more!! You can round your end result, - Not before!)
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 05:51:46 PM
AHA!! SO....
2317.5 / 0.0015 mol C3H6O
= 1545000 J?
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 05:59:07 PM
Dear Frederick95;
NO!, – I told you: Use at least 2 figures more!!
That will give you: H = - 1462.6 kJoules / mole for Acetone.
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 06:01:08 PM
1462.6 kJoules?
But why not 2317.5 kJ?
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 06:03:28 PM
Dear Frederick95;
Why should it?
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 06:05:16 PM
Because 2317.5 kJ is the value we got for q
1462.5 does not = q
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 06:17:32 PM
Dear Frederick95;
Please read correct: H = - 1462.6 kJoules / mole for Acetone. (That's a little more precise = 2.320 kJoules / 0.001586 moles.)
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 06:30:25 PM
OHHHH! :D
So Enthalpy = 2.320 kJoules / 0.001586 moles
Therefore Enthalpy(H) is 1462 kJ/mol?
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 06:34:40 PM
Dear Frederick95;
Don’t miss the Sign, as you told in your Question!
Quote
H = - Q = - 455 J = -0.455 kJ Number of moles (n) = (given mass (g))/(gram-formula mass)
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 06:36:10 PM
Enthalpy -1462 kJ/mol (because it is exothermic)?
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 06:39:45 PM
Dear Frederick95;
Precisely!
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: Frederick95 on March 07, 2008, 06:41:00 PM
THANK YOU ARGO YOU ARE THE BEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :) :) :) :) :) :) ;D ;D ;D ;D :D :D :D :D :D
I will not forget you!!!:):)
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 07, 2008, 06:46:32 PM
Dear Frederick95;
You’re welcome! ─ Soon (but not toooo soon!) again.
Good Luck! ARGOS++
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: mo_1984 on March 07, 2008, 10:16:01 PM
i am so lost plz check if this is right and if not help me fix it. PLZ PLZ PLZ
1) Using the table of standard enthalpies of formation, I calculated the theoretical value for the molar enthalpy of combustion of acetone by: Writing a balanced equation: C3H6O (l) + 4O2 (g) --> 3CO2 (g) + 3H2O (l)
Then each compound, I wrote out the enthalpy of formation.
After I put these enthalpy values into the equation and now I used the number of moles to find the theoretical value:
∆H = ∑n ∆H°f(prod.) - ∑n ∆H°f(react.)
= [3(-393.5) + 3(-285.8] – [(-248.1) + 4(0.0)] kJ/mol = [-2037.9 – (-248.1)] kJ/mol = -1789.8 kJ/mol 2) Calculate the molar enthalpy of combustion of acetone, using the experimental evidence. ∆T = t2 – t1 = 25.0oC – 20.0oC = 5.0oC
QTotal = QWater + QAl = mWcWΔT + mAlcAlΔT = (100.0g)(4.18J/(g*oC)(5 oC) + (100.0g)(0.91 /(g*oC)(5 oC) = 2090J + 455J = 2545 J Molar Mass of Acetone: C3H6O C3 = 3 X 12 = 36 H6 = 6 X 1 = 6 O = 1X 16 = 16 = 58 g/mol Moles = Mass / molar mass = 0.092g / 58g/mol = 1.59 X 10-3 mol H = - Q = -2545J (it’s an exothermic reaction therefore it needs to have a negative value to get the correct percent error/enthalpy change) n ∆H°f = ∆H ∆H°f = ∆H/ n ∆H°f = -2545J / 1.59 X 10-3 mol = -1600.63 kJ/ mol 3) Calculate the percentage error for this experiment. Error % = [the accepted answer - your observed answer] / the accepted answer (then multiply by 100) = [(-1789.8kJ/mol) - (-1600.63kJ/mol)] / (-1789.8kJ/mol) X 100% = 10.6%
Title: Re: molar enthalpy of combustion of acetone next step?
Post by: ARGOS++ on March 08, 2008, 11:00:12 AM
Dear Mo_1984;
Question 1: Well done!
Question 2: Frederick95 and I got a different value as you may see above! You did Reply # 9 not correct! ( mAl ). You did also same Mistake as Frederick95 first: Tooo less significant figures!
Question 3: Needs to be recalculated, because Question 2!