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)-->NH4 ^+(aq) + NO3^-(aq). In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 degrees C and the final temperature (after the solid dissolves) is 21.9 degrees C. Calculate the change in enthalpy for the reaction. (Use 1.0g/mL as the density of the solution and 4.18 J/g . degrees C as the specific heat capacity.) Express the answer to TWO significant figures.

The steps would be greatly appreciated!! Thanks so much!

Answer 1

Calculate the change in enthalpy for the reaction?

dH = m C dT

dH = 25 grams water(4.18J/g-C)(-3.9Celsius) = -407.55

dH water = - 408 Joules
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if the water lost 408 Joules, (dH water = -408J)
then the ammonium nitrate gained 408J , (dH reaction = +408J)

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often times questions then ask, what was the dH per gram ammonium nitrate:
408J / 1.25 grams = 326 J /gram

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often they then ask , what was the dH per mole ammonium nitrate:
326 J /gram @ 80.04 g/mol = 26.1 kJ/mole


dH reaction =

Hop this helpS!!