Heat of Reaction question


by nerdmon
Tags: chemistry, enthalpy, heat of formation
nerdmon
nerdmon is offline
#1
Jan25-09, 01:23 AM
P: 5
1. The problem statement, all variables and given/known data
When 50.0 mL of 1.0 M H2SO4 at 26.1 celcius was added to 50.0 mL of .976 M NaOH also at 26.1 celcius, the temperature rose to 32.6 celcius. Calculate the heat of reaction.

2. Relevant equations

H2SO4 (aq) + 2NaOH(aq) --> Na2SO4 (aq) + 2H2O(l)

3. The attempt at a solution
First, I converted both given amount of solutions into moles and found that NaOH was the limiting reactant (made .0488 mol h20 while the sulfuric acid made .1 mol). Then I converted the 50 mL of NaOH into grams (1 g/mL) and plugged that into the q=mcDeltaT equation. I did:
50g*4.184J/gcelcius*6.5 celcius = 1359.8 J
Then I divided that by the mols of NaOH
1359.8 J/.0488 mol = 27864.7541 J/mol

Can anyone confirm if I did this correctly?
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Bacat
Bacat is offline
#2
Jan25-09, 11:16 PM
P: 151
1 g/mL is the density of water, therefore I don't think that you can convert 50 mL of NaOH into 50g of NaOH using it. The density of NaOH should be something like 2.130 g/mL, but I'm not sure if that is the correct value or not.
nerdmon
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#3
Jan25-09, 11:32 PM
P: 5
oh i assumed 4.184 because it is a dilute solution

Bacat
Bacat is offline
#4
Jan25-09, 11:34 PM
P: 151

Heat of Reaction question


I think 4.184 is fine for the spec heat. I'm talking about the density of the material that reacted. But maybe I am misunderstanding.
nerdmon
nerdmon is offline
#5
Jan25-09, 11:45 PM
P: 5
ohh good point.. hmm maybe I don't know... thanks for the heads up on that
ditto_299
ditto_299 is offline
#6
Jan28-09, 08:44 PM
P: 7
so im doing a lab on acid and bases and acid and metal reactions, and i have to calculate the energy change, heat

-what would be a difference in the reactions with the heat??
- which one would be faster??
Bacat
Bacat is offline
#7
Jan28-09, 11:20 PM
P: 151
ditto_299, you should start a new thread and follow the template. We need more information about your problem (and you should show us what you've tried to do so far) before we can help you. Cheers.


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