Instant cold packs 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 endothermic reaction: NH4NO3 (s) → NH+4 (aq) + NO-3 (aq) In order to measure the enthalpy change for this reaction, 2.25 g of NH4NO3 is dissolved in enough water to make 15.0 mL of solution. The initial temperature is 25.8 oC and the final temperature (after the solid dissolves) is 21.9 oC. Calculate the change in enthalpy for the reaction in kJ. (Use 1.0 g/mL as the density of the solution and 4.18 J/g. oC as the specific heat capacity.)

Question

Chemistry

Charlee Mccall

26 June, 12:35

Instant cold packs 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 endothermic reaction: NH4NO3 (s) → NH+4 (aq) + NO-3 (aq) In order to measure the enthalpy change for this reaction, 2.25 g of NH4NO3 is dissolved in enough water to make 15.0 mL of solution. The initial temperature is 25.8 oC and the final temperature (after the solid dissolves) is 21.9 oC. Calculate the change in enthalpy for the reaction in kJ. (Use 1.0 g/mL as the density of the solution and 4.18 J/g. oC as the specific heat capacity.)

+4

Answers (1)

Know the Answer?

Harold Harrington · Answer 1

8.693kJ

Explanation:

The enthalpy change of a reaction is the heat produced per mole of reaction.

For the reaction:

NH₄NO₃ (s) → NH₄⁺ (aq) + NO₃⁻ (aq)

The heat produced (Using coffe cup calorimeter formula) is:

Q = - m*C*ΔT

Where Q is heat, m is mass of solution (15.0mL ≡ 15.0g - Density of 1g/mL), C is specific heat capacity of solution (4.18J/g°C) and ΔT is change in temperature (21.9°C - 25.8°C = - 3.9°C).

Replacing:

Q = - 15.0g*4.18J/g°C*-3.9°C

Q = 244.5J is heat of reaction per 2.25g of NH₄NO₃.

Moles in 2.25g of NH₄NO₃ - 80.0g/mol-:

2.25g * (1mol / 80.0g) = 0.0281 moles of NH₄NO₃

Thus, enthalpy change of reaction is:

244.5J / 0.0281mol = 8693J = 8.693kJ