Calculating Heat of Formation for NH4Cl Using Calorimetry

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SUMMARY

This discussion focuses on calculating the heat of formation for ammonium chloride (NH4Cl) using calorimetry. The experiment involved mixing 50ml of 1.51M ammonia (NH3) with 50ml of 1.59M hydrochloric acid (HCl), resulting in a temperature change (deltaT) of 8.6°C. A second part of the experiment involved dissolving 4.082g of NH4Cl in 100ml of water, yielding a deltaT of -2.8°C. Participants are guided to write out the chemical reactions and determine the heat capacity of the ammonium chloride solution to calculate deltaH per mole for each reaction.

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Homework Statement


i did a lab calorimeter lab with 50ml 1.51M nh3 and 50ml 1.59M hcl yeilding nh4cl (aq)
the deltaT was 8.6 degC
second part is 100ml of h2o and 4.082g nh4cl
deltaT was -2.8 degC

Homework Equations


how do i find deltaH per mole for each reaction?

The Attempt at a Solution

 
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Hint: Write out the reactions for the first part and the second part. What form was the ammonia? Was it aqueous? How should that be written?
 
its nh3(aq) + hclz(aq) = nh4cl(aq)
and nh4cl(s) = nh4cl(aq)

but where do i go from there?
 
NH3 in water is NH4+ OH-
HCl in water is H3O+ Cl-

one of the two reactions is:

NH4+(aq) + Cl-(aq) ----> NH4Cl(aq)

Can you determine the other reaction?

You will need the heat capacity of the ammonium chloride solution. Can you determine what the concentration will be?
Alternatively, you could assume the heat capacity of pure water, but it won't be an exact solution. Can you think of why you need to obtain the heat capacity?
 

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