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

Answer 1

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

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

Specific heat capacity , C = 4.184J/g^oC

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/g^oC * (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