When 5.00 g of ammonium nitrate (NH_4NO_3) was dissolved in 50.0 mL of water in

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

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