Problem 5.54 Part A When a 3 8- sample of salid ammonium nitrate dissoves in 60
ID: 555527 • Letter: P
Question
Problem 5.54 Part A When a 3 8- sample of salid ammonium nitrate dissoves in 60 0 g af water in a coffee-cup calorimeter, the temperature drops from 23.0 "Cto 184 C Calculate water (in kJ/mol N NO3) for the solution process NH NO3 s) ) NH NO (aq Assume that the specific heat of the solution is the same as that of pure a Express your answer using two signilicant ligures. kJ/mol Submit My Answers SeU Part B Is this process 9 endothermic or exothermic? O exothermic Submit My Answers Give Up ContinueExplanation / Answer
Ans. Part A: Moles of NH4NO3 = Mass /Molar mass
= 3.88 g / (80.04344 g/ mol)
= 0.048473679 mol
# Total mass of solution = Mass of NH4NO3 + Mass of H2O
= 3.88 g + 60.0 g
= 63.88 g
# Amount of heat lost by the solution is given by-
q = m s dT - equation 1
Where,
q = heat lost
m = mass of solution
s = specific heat of solution
dT = Final temperature – Initial temperature
Or, dT = 18.40C – 23.00C = -4.60C
Putting the values in equation 1-
q = 63.88 g x (4.184 J g-10C-1) x (-4.60C)
Hence, q = -1229.460032 J
The –ve sign of q indicates that heat is being released by solution.
# The amount of heat lost by solution must be equal to the amount of heat gained by NH4NO3 during its solvation.
So,
Amount of heat gained by NH4NO3 during solvation = 1229.460032 J
Now,
Molar enthalpy of solvation = Amount of heat gained / Moles of NH4NO3
= 1229.460032 J / 0.048473679 mol
= 25363.456 J/ mol
= 25.36 kJ mol
Hence, dH for NH4NO3 solvation = 25.36 kJ/mol or +25.36 kJ/mol
# Part B. The reaction is endothermic because solvation of NH4NO3 absorbs heat.