Answer to Question #17063 in Organic Chemistry for Kenton
18.4 oc. Calculate H (in kJ/ mol NH4N03) for the solution process
NH4NO3(S) ----> NH4 +(aq) + N03 - (aq)
Assume that the specific heat of the solution is the same as that of pure water.
(b) Is this process endothermic or exothermic?
Molar mass of NH4NO3 = 14+4+14+48 = 80 g/mol
So, 3.88 g is equal to
3.88/80 = 0.0485 mol
Now, mass of soln = 60+3.88 g = 63.88 g
= 63.88x4.18x(23-18.4) = 1228 J
So, 0.0485 mol absorbs 1228 J
absorbs 25319 J
So, heat of dissolution of NH4NO3 is 25.319
This process is endothermic.
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Calculate the minimum value of ∆S (J mol-1 K1) associated with dissolving NaNO3 in water at 23oC. For this process ∆H=25.41 kJ mol-1.
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It should be exothermic