When 5.00 g of ammonium nitrate (NH_4NO_3) was dissolved in 50.0 mL of water in
ID: 925462 • Letter: W
Question
When 5.00 g of ammonium nitrate (NH_4NO_3) was dissolved in 50.0 mL of water in a coffee-cup calorimeter, the temperature dropped to 18.0 degree C from 22.0 degree C. If the heat capacity of the calorimeter is 13.7 J/ degree C, and specific heat capacity of the solution is 4.18 J/g degree C, Calculate: DeltaH(in kJ per mol of NH_4NO_3) for the solution process (AW;N=14.0, H=1.0, O=16). Is the solution process exothermic or endothermic? Would NH_4NO_3 solution serve as an effective hot pack or cold pack?Explanation / Answer
we know that from our knowledge of thermodynamics, as the pressure is kept constant(atmospheric pressure)
For NH4NO3, we know that:
m(mass) =80.05 g,
For water:
m= 50 g,
csp = 4.184 J/g °C
q=mcspT
qcal=(4.18J/g°C)(55g)(18°C22°C)= -919.6 J
and we already know that q of calorimeter = -q of reaction so
q of reaction = 0.919 kJ
moles of NH4NO3= 5/80.05 = 0.063
delta H = q/ moles
delta H = 0.919/0.063 = 14.58 kJ/mole
b)
as the delta H of reaction is positive, the reaction is endothermic
c)
as the reaction is endothermic, it will serve as an effective cold pack