nstant cold packs sometimes used to treat athletic injuries contain solid ammoni

Question

nstant cold packs sometimes used to treat athletic injuries contain solid ammonium nitrate and liquid water separatedbya thin divider. When the divider is broken, the ammonium nitrate dissolves endothermically:

NH4NO3(s) NH4+(aq) + NO3–(aq)

To measure the enthalpy of this reaction, you dissolve 1.25 g of ammonium nitrate in enough water to make 25.0 mL of solution. The initial temperature is 25.8 °C, and the final temperature is 21.9 °C. (Assume that the density of the solution is 1.00 g /mL and that the heat capacity of the solution is 4.184 J/g°C

Explanation / Answer

Answer – Given, mass of NH4NO3 = 1.25 g , water volume = 25.0 mL

ti = 25.8oC, tf = 21.9oC , density = 1.009 g/mL

Specific heat capacity , C = 4.184J/goC

Mass of water = 25.0 mL = 25.0 g

Total mass of solution = 1.25 +25 = 26.25 g

We know formula for calculating the heat

q = m*C*t

   = 26.25 g * 4.184 J/goC * (21.9 – 25.8)oC

   = -428.3 J

So water release the heat

Heat loos = heat gain

So NH4NO3 gets heat for dissociation,

So, Hrxn = -q

                   = -(-428.3 J)

                    = 428.3 J

                     = 0.4283 kJ

Moles of NH4NO3 = 1.25 g / 80.04 g.mol-1

                                = 0.0156 moles

So enthalpy of this reaction per mole is

Hrxn = 0.4283 kJ/ 0.0156 mole

             = 27.4 kJ/mol

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