One elementary laboratory demonstration in introductory chemistry is the mixing of NH4NO3 with water — Physical Chemistry — Thermodynamics Chemistry Question
Ammonium nitrate
One elementary laboratory demonstration in introductory chemistry is the mixing of NH4NO3 with water in a thermally isolated container. In this problem, 80 g NH4NO3 are mixed with 1 kg H2O which are both initially at 0 °C.
Given are: heat capacity of liquid water = 76 J mol–1 K–1, enthalpy of fusion for water = 6.01 kJ mol–1, enthalpy of solution of NH4NO3 in water = 25.69 kJ mol–1, cryoscopic constant for water = 1.86 K kg mol–1. Determine the final state of the system.
Choose the correct answers:
I) The final temperature of the system is equal to:
a) 1.86 K, b) 3.72 K, c) 3.72 °C, d) 1.86 °C, e) –3.72 °C, f) –3.72 K, g) 1.86 K, h) –1.86 °C) i) –3.83 °C.
II) The final state of the system consists of:
a) 1 liquid and 1 solid phase, b) 1 liquid and 2 solid phases, c) 1 liquid phase, d) 1 solid phase, e) 2 liquid phases, f) 2 solid phases, g) 2 liquids and 1 solid phase.
III) The mixing process can be described as (check all that apply):
induced, spontaneous, reversible, irreversible, one where separation of components is impossible by any means, adiabatic, non–adiabatic, isobaric, isothermal, isochoric, isenthalpic, isoenergetic.
IV) The change in entropy (ΔS) of the system is:
a) ΔS > 0, b) ΔS = 0, c) ΔS < 0, d) indeterminate.
Model Answer
Mixing is endothermic and the process is adiabatic, thus heat has to be provided by the solution itself. Since water is at its freezing point, it will tend to freeze, but the solution created will experience a depression of freezing point due to the presence of dissolved ions. The large amount of heat required for solvation will necessitate some freezing of water.
A Hess cycle of 3 steps will be considered.
a) mixing at 0 °C with H1 > 0,
b) lowering of the temperature of the mixture to its final temperature with H2 < 0,
c) freezing of some water ms with H3 < 0.
The final temperature is given by θ2 = – Kf n / (m – ms) where Kf is the cryoscopy constant, 2 is the number of particles per formula weight for NH4NO3, n is the number of moles of NH4NO3, n = 80 g / 80 g mol−1 = 1 mol , m = 1000 g.
ΔH1 = Δhs n = – 25.69 kJ mol–1 × 1 mol = – 25.69 kJ
ΔH2 = m cp θ2 / M where cp is the molar heat capacity of water and M its molar mass (18 g mol–1).
ΔH3 = – ms Δfh / M where Δfh is the molar enthalpy of fusion.
ΔH1 + ΔH2 + ΔH3 = ΔHtotal = 0 because no heat is allowed to be exchanged between the system and its surroundings.
Substituting for θ2 and solving for ms yields the following expression: [Quadratic in ms]. We discount the solution derived from the + sign as unphysical (ms > m) and arrive at the result ms = 28.52 g of ice.
Hence θ2 = –3.83 °C.
If we made the simplification that we expect ms << m, then θ2 is immediately calculated as –3.72 °C, which yields a value for ms = 29.9 g. If this result is used to improve the θ2 value using the exact expression, we get θ2 = –3.83 °C. Then, ms can be further improved to 28.5 g.
The process is spontaneous, irreversible, one where separation of components is possible, adiabatic, isobaric, isenthalpic, nearly isoenergetic.
The equation ΔG = ΔH – T ΔS can be used here because T varies less than 2 %.
ΔG < 0 because the process is spontaneous and ΔH = 0, hence ΔS > 0. This is also to be expected from stability criteria under the constraint ΔH = 0.