home > q&a board > chemistry > topic: calorimetry

Q BgQuestion:

Scholar
Karma Points: 206
Respect (90%):
posted by  Vintagechick on 10/16/2008 8:34:34 PM  |  status: Live  

Calorimetry

Course Textbook Chapter Problem
General Chemistry Chemistry: A molecular Approach 6 N/A
Question Details:
A calorimeter contains 17.0 mL of water at 13.5\, ^{\circ}C. When 2.20 g of \rm X (a substance with a molar mass of 55.0 g/mol) is added, it dissolves via the reaction

\rm X{(s)}+H_2O{(l)} \rightarrow X{(aq)}

and the temperature of the solution increases to 26.0\, ^{\circ}C.

Calculate the enthalpy change, Delta H, for this reaction per mole of \rm X.

Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 \rm J/(g \cdot {^\circ C}) and 1.00 \rm g/mL] and that no heat is lost to the calorimeter itself, nor to the surroundings.

Express your answer numerically in kilojoules per mole.
Bonus Point Alert! Earn +4 additional karma points for helping this annual member.

AAnswers:

Answer Question
Mentor
Karma Points: 642
posted by Arcanum on 10/16/2008 10:57:27 PM  |  status: Live
Asker's Rating: Lifesaver   
Vintagechick's comment:
"great! Thanks for being so detailed. I can really use this to solve other similar problems"
Response Details:
Ok, I haven't done problems like this in awhile, but I think I will take an educated crack at this :+)
 
What needs to be done is that we need to account how many grams are in the solutions which we have 2.20grams of X, ok good start, and then we have 17mL of water, which this is easy conversion since 1 mL water = 1 gram of water Horray, so now all we have to do is add them together (17g + 2.20g) = 19.2 grams in our solution.
 
Ok, now we need to determine the heat change, which is taking the final temperature which is 26.0 C = 299k and our inital which is 13.5C = 286.5K So sweet there we go...and we know the Specfic heat capacity so we are all set...
 
qSolution = (19.2g)(4.18J/mol)(299k - 286.5k) = 1003.2 J
qrxn = - qsolutions = -1003.2 J
 
Now that we have that we can solve for the ΔH
 
ΔH = (-1003.2J / 2.20grams) * (55.0g/mol)/ (1 mol) = -25.08KJ/mol
 
Which that should be your answer, looks reasonable, and should work :+)
 
If you need any more help with this problem just PM and ill try to help you out some more :+)
 
Pffhh, if I knew everything, I would spend my entire time on here answering questions, but until then ill spend my entire time on here asking questions... :+)
Answer Question
Ask New Question

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different from tutoring »

Need Help In

Useful Links