Question

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).

Answer 1

Answer:

$T_f=-7.81^0C$

Explanation:

Hello,

a) In this case, since the heat associated with the dissolution of ammonium nitrate is positive, such reaction is endothermic as it absorbs heat.

b) Now, for computing the temperature once the dissolution is done, we apply (considering that it is a cooling process):

$T_f=T_0-\frac{\Delta H}{mCp}$

Nonetheless, we should first compute the moles of the mixture as:

$n_{mix}=135.0gH_2O*\frac{1molH_2O}{18gH_2O}+50.0gNH_4NO_3*\frac{1molNH_4NO_3}{80gNH_4NO_3}=8.125mol$

Thus, the total absorbed heat is:

$\Delta H=25.4kJ/mol*8.125mol=206.375kJ$

Now, the temperature is:

$T_f=25.0^0C-\frac{25.4kJ}{(135.0+50.0)g*4.184x10^{-3}kJ/g^0C} \\\\T_f=-7.81^0C$

Best regards.