Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break the water pouch, the ammonium nitrate dissolves in water and the pack gets cold. The heat of solution for ammonium nitrate is 25.4 kJ/mol.

a. Is the dissolution of ammonium nitrate endothermic or exothermic?

b. A cold pack contains 135.0 g of water and 50.0 g of ammonium nitrate. What will be the final temperature of the activated cold pack, if the initial temperature is 25.0 degree C? (Assume that the specific heat of the solution is the same as that for water, 4.184 J/g degree C and no heat is lost).

a) Endothermic

b) T₂ = 53.1 ºC

Explanation:

a) We are told that when the ammonium nitrate dissolves in water the pack gets cold so the system is absorbing heat from the surroundings and by definition it is an endothermic process.

b) Recall that the heat, Q, is given by the formula:

Q = mcΔT where m is the mass of water,

c is the specific heat of water, and

ΔT is the change in temperature

We can determine the value for Q since we are given the heat of solution for the ammonium nitrate. From there we can calculate ΔT and finally answer our question.

Molar mass NH₄NO₃ = 80.04 g/mol

moles NH₄NO₃ = 50.0 g / 80.04 g/mol = 0.62 mol

Q = 25.4 kJ/mol x 0.62 mol = 15.87 kJ = 15.87 kJ x 1000 J = 1.59 x 10⁴ J

Q = mcΔT ⇒ ΔT = Q/mc

ΔT = 1.59 x 10⁴ J / (135 g x 4.184 J/gºC) = 28.1 ºC

T₂ - T₁ = ΔT ⇒ T₂ = ΔT + T₁ = 28.1 ºC + 25.0 ºC = 53.1 ºC