Thermodynamics Question
Chemistry

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I am working through an experiment where 3.016g of NH4NO3(s) was dissolved in 50g of H2O in a coffee cup calorimeter. I determined the qrxn = 19.655 kJ/mol (qsoln + qcup). I am now asked to calculate the change in enthalpy for the reaction in moles using deltaHrxn = Sigma deltaHf(products)  Sigma deltaHf(reactants) and the deltaHf NH4NO3 is 495 kJ/mol. I am stuck here, blanking out on how to proceed
Thank you for the opportunity to help you with your question!
The qrxn u calculated is the enthalpy change or the heat lost when the solute dissolved .it is therefore the sigma delta H of the product.divide it by the mol weight 80 to get the delta H in kj/mol unit.
NH_{4}NO_{3} (s) → NH_{4}^{+} (aq) + NO_{3}^{} (aq)
To use the formula deltaHf of each of the ions on the right hand side should be given. Do you have that?
To use the formula deltaHf of each of the ions on the right hand side should be given. Do you have that?
Hi, the equation given is NH4NO3(s) = NH4NO3(aq) I know that it dissociates in ions but they don't give us deltaHf for anything other than the one I outlined. I guess this is why I am confused. All of the data from the experiment that I have been using is: Wgt of NH4NO3 = 3.016g, No of moles calculated to be 0.03768 mols. 50g of H2O. T(initial) 23.5C. T(final) 20.5C. Asked to calculate heat associated with qH2O (which I took to be qsoln) and worked that out to be qsoln = 53.016g x 4.184 J/g.C X (20.523.5) = 665.5J, then asked for heat released/absorbed by deltaHrxn  that is qsoln = 665.6J. Then asked for change in enthalpy of reaction = 0.666kJ. Then they wanted the calorimeter introduced and gave it a C of 25J/C. Now my enthalpy was 19.655kJ/mol. Thats when I arrive at the question I posted, and the only information I have is what I have detailed.
Standard enthalpies for NH4+ (aq) and NO3(aq) are 132.8 kj/mol and 206.6kj/mol respectively.
So Delta Hrxn = 132.8206.6(495)=
http://www.chm.davidson.edu/vce/calorimetry/heatofsolutionofammoniumnitrate.html
Go to this website for clarification.
http://www.chm.davidson.edu/vce/calorimetry/heatofsolutionofammoniumnitrate.html
Go to this website for clarification.
I understand where you are getting the values for the ions and if I do that calculation then the answer is 155.6. However the next question is a %error question between this value and the value I calculated of 19.665kJ/mol. Won't that be a large %error
You can get the values for standard enthalpies of the ions from any standard textbook table.
You can get the values for standard enthalpies of the ions from any standard textbook table.
Go to the website and make sure all calculations are correct. It is shown step by step very nicely over there.
Go to the website and make sure all calculations are correct. It is shown step by step very nicely over there.
Thank you for your help
Qrxn is the heat change for the reaction and it is the sum of the heat changes associated with soln and the cal.
Qcal = CX(20.523.5)= 25x3= 75 J= 0.075 kj
This amount of heat is lost by the cal.
Qsoln= 0.665 kj
This is the heat lost by the solution .
The sum of these two is the heat aborbedaborbed by the solute to dissolve I.e. qrxn.. divide by 80 to get deltaHf of reaction.
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