Question

A 3.52 g sample of ammonium nitrate (NH_4NO_3) is added to 80.00 mL of H_2O in a constant-pressure calorimeter of negligible heat capacity. The water is stirred until all ammonium nitrate has dissolved. The temperature of the solution drops from 21.6 to 18.1 degree C (NH_4NO_3) (s) + H_2O (l) rightarrow NH_4(aq) + NO_3(aq) delta H = ?

Assume that the specific heat and density of the resulting solution are equal to those of water, 4.18 JW/g and 100g/ml, respectively and assume that no heat is lost to the calorimeter itself, nor to the surroundings.

a. ls this process endothermic or exothermic (circle the correct answer?
b. calculate the heat of solution(delta H)NH_4NO_3 expressed in kilojoules per mole of NH_4NO_3? Heat of solution (delta H) =______ kl/mol

Answer 1

(a) The process is an endothermic reaction. This is because there is a fall in temperature of water.

(b) Heat of solution = +26.6 KJ/mol

Explanation:

Endothermic reaction is a reaction that absorbs heat from the surroundings. It occurs when bond breaking is greater than bond forming.

The dissolution of NH4NO3 in water is an endothermic reaction. it absorbs heat from the water it dissolves in it. That is the reason for the fall in temperature of water.

The heat loss by water = - M × c × ΔT

M = mass of water. Assume 1 ml of water to be 1 g

c = specific heat of water (4.18J)

ΔT = change in temperature (final temperature - initial temperature)

= 18.1 - 21.6

= - 3.5

The heat loss by water = - 80 × 4.18 × (- 3.5)

= 1170.4 joules

= 1.17 KJ

Step 2: calculate the number of mole of NH4NO3 present.

no of mole = mass / molar mass

Molar mass of NH4NO3 = 80 g/mol

no of mole = 3.52 / 80

= 0.044 mol

Step 3: Calculate the Heat of solution (ΔHsol)

(ΔHsol) = Heat lost by water / no of mole of NH4NO3

= 1.17 / 0.044

= + 26.6 KJ/mol

(Please check the attached document for more explanation)