# Heat of Reaction question

by nerdmon
Tags: chemistry, enthalpy, heat of formation
 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?
 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.
 P: 5 oh i assumed 4.184 because it is a dilute solution
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.
 P: 5 ohh good point.. hmm maybe I don't know... thanks for the heads up on that
 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??