Posted by a Canadian on Wednesday, February 13, 2013 at 5:41pm.
I did a lab today in class and part of the lab is to experimentally find the enthalpy for the dissolution of ammonium chloride by dissolving 4.0 g of NH4Cl(s) in 50.0 mL of H2O(l).
I got a -5°C temperature change and calculated delta H to be 15 kJ/mol. According to the Internet, the actual value is 14.7 kJ/mol? So I'm a little over. My initial worry was that For sources of error, we only ever discussed things that could cause heat loss - not anything that would actually result in a higher heat gain. And "maybe we measured wrong" definitely wouldn't work as a legitimate source of error. But I thought about it a bit more and the positive 15 kJ/mol indicates that the reactants absorbed that much energy right? Which would result in a lower temperature. And heat loss through the calorimeter, etc would make the temperature lower and make it seem like the reactants absorbed more energy than they really did, and so the enthalpy change was calculated to be a higher value. I was wondering if this is right?
I hope I made sense.
It's also very possible that the Internet/my calculations wrong.
Also, the other part of my lab was to find the value of delta H for the reaction between 25 mL of 1.0M HCl(aq) and 25 mL of 1.0 M NH3(aq). I tried looking for the correct value on the Internet to check if my value, -41.84 kJ/mol, was close and to make sure I didn't completely mess up my lab, but I couldn't find it. I was wondering if anyone just happened to know..?
- Chemistry - DrBob222, Wednesday, February 13, 2013 at 6:00pm
I found four quotes for NH4Cl. Three of them gave 14.7 kJ/mol and the fourth gave 333533 cal/mol (or 14.78 kJ/mol).
For the enthalpy of neutralization, here is a reference that gives 57.3 kJ/mol for a strong acid and a strong base.
- Chemistry - DrBob222, Wednesday, February 13, 2013 at 6:10pm
I don't think your theory about the loss of heat holds water (no pun intended). If your dT was -5C, I would assume the T outside the calorimeter was greater than that so heat would have flowed from outside to inside making the temperature go up in the calorimeter.
A possible source of error could be that you used 4.184 J/g for the heat capacity of the solution as you assume the Cp for NH4Cl soln and pure water were the same. Perhaps they are not.My best guess is that Cp for NH4Cl solution is lower than 4.184 and that would make the experimental results low. Another possible source. How accurately could you read the thermometer. 0.1 degree out of 5 makes a 2.0% error. Your 14.7 is about 2% low.
- Chemistry - a Canadian, Wednesday, February 13, 2013 at 6:22pm
So for the neutralisation one I'm way off, right? Is it unacceptably off? Because my teacher said we'd have tomorrow to continue if we didn't finish, so I might just do that part again.
That makes more sense, thanks so much for answering (for all the other times too--never had a chance to really thank you because I always saw the final post too late)!
- oops--typo--Chemistry - DrBob222, Wednesday, February 13, 2013 at 8:48pm
I think that should read 3533 cal/mol and that x 4.184 = 14.78 kJ/mol
For the neutralization the value I found is for STRONG acid/STRONG base and NH3 is a weak base. That changes the value but I don't know by how much. But I know it will be lower for weak vs strong.
I found HCN + NaOH = 45.3 kJ/mol so your value may not be that far off. Ask you prof what it should be.
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