Instant cold packs, often used to ice athletic injuries on the field, contain am

Question

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction:

NH4NO3(s)NH+4(aq)+NO3(aq)

In order to measure the enthalpy change for this reaction, 2.00 g of ammonium nitrate is dissolved in enough water to make 40.0 mL of solution. The initial temperature is 39.9 C and the final temperature (after the solid dissolves) is 36.2 C.

Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.00g/mLas the density of the solution and 4.18J/gCas the specific heat capacity.)

Thanks All!!

Explanation / Answer

Calculation of total mass of solution

Total mass of solution = 40.0 mL x 1.0 g / 1 mL = 40.0 g

Sp. Heat capacity is 4.184 J /g 0 C

Initial Temperature = 39.9 deg C

Final temperature = 36.2 deg C

Les use following equation

q rxn + q solution = 0

so q solution = -q rxn

here q rxn is heat given by reaction

lets calculated q rxn

-q rxn = (qsol)

We know q sol = m C Delta T

Here m is mass in g , C is sp heat capacity and delta T is change in T

Lets plug the given values.

-q rxn = [40.0 x 4.184 x ( 36.2 – 39.9 ) ] J

= -117.152 J

So qrxn = 117.152 J

We know enthalpy of reaction delta H = qrxn/ n

n is number of moles

n = mass / molar mass

n of ammonium nitrate

= 2.00 g / molar mass of ammonium nitrate = 2.00 g / 80.052 g per mol

= 0.02498 mol

Delta H = 117.152 J / 0.05498 mol

= 4689.13 J/mol

Delta H in kJ = 4.689 kJ

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