Lanthanum pentanickel, LaNi5(s), is under consideration for solid-state hydrogen storage. LaNi5(s) i — Thermodynamics Chemistry Question
Problem Context
Lanthanum pentanickel, LaNi5(s), is under consideration for solid-state hydrogen storage. LaNi5(s) is a conductive metallic crystal, and it forms hydrides in two phases:
• an α phase α-LaNi5Hx(s) observed at lower H2 pressure, characterized as a solid-state solution
• a β phase β-LaNi5H6.39(s) observed at higher H2 pressure, characterized by metal-hydrogen bonding
Species | ∆H°f, kJ mol–1 | S°, J mol–1 K–1 | Species | ∆H°f, kJ mol–1 | S°, J mol–1 K–1
H2(g) | 0 | 130.7 | LaNi5(s) | –162 | 217
Ni(s) | 0 | 29.9 | α-LaNi5Hx(s) | –186 | 223
La(s) | 0 | 56.9 | β-LaNi5H6.39(s) | ? | ?
LaNi5(s) is placed in vacuum chambers, one at 30.0 °C and one at 50.0 °C. Pure H2(g) is added to each chamber, and the weight-percent hydrogenation of LaNi5(s) is recorded as a function of pressure.
Calculate ∆G°f of LaNi5(s) at 298 K.
Model Answer
For the reaction
La(s) + 5 Ni(s) ® LaNi5(s)
∆H°rxn = (–162) – (0 + 5) kJ mol–1 = –162 kJ mol–1
∆S°rxn = (217) – (56.9 + 5[29.9]) J mol–1 K–1 = 10.6 J mol–1 K–1
∆G°rxn, 298K = ∆G°f, 298K of LaNi5(s) = ∆H°rxn – (298 K)∆S°rxn
∆G°f, 298K of LaNi5(s) = –165 kJ mol–1
Show that the maximum degree of hydrogenation x for α-LaNi5Hx(s) is approximately 0.43.
Model Answer
The maximum mass% of H in α-LaNi5Hx(s) is 0.10%. So in 100. g of α-LaNi5Hx(s), there are 0.10 g/(1.008 g mol–1) = 0.099 mol H and 99.9 g/(432.35 g mol–1) = 0.231 mol LaNi5. So x = 0.099/0.231 = 0.43.
Calculate ∆G°rxn at 30 °C and at 50 °C for the hydrogenation of the α phase to the β phase.
Model Answer
The balanced reaction is:
a-LaNi5H0.43 + 2.98 H2(g) ® b-LaNi5H6.39
If ∆Grxn (nonstandard; not the standard ∆G°) is positive, then the a phase is stable, while if it is negative the b phase is stable. The two phases can coexist if ∆Grxn = 0. The hydrogen pressure at which the two phases coexist can be read from the steep part of the graphs, i.e., 3.1 bar at 30 °C and 5.4 bar at 50 °C. At these pressures,
∆G = 0 = ∆G° + RTln(Q)
∆G° = –RTln(Q) = –RTln(PH2–2.98)
∆G°(30 °C) = –R(303.2 K)ln[(3.1)–2.98] = 8.50 kJ mol–1
∆G°(50 °C) = –R(323.2 K)ln[(5.4)–2.98] = 13.5 kJ mol–1
Calculate ∆H°f and S° for β-LaNi5H6.39(s).
Model Answer
One can get ∆H°rxn and ∆S°rxn for the hydrogenation reaction in (c) by solving the two simultaneous equations:
8.50 kJ mol–1 = ∆H°rxn – (303.2 K)∆S°rxn
13.5 kJ mol–1 = ∆H°rxn – (323.2 K)∆S°rxn
∆H°rxn = –67.3 kJ mol–1, ∆S°rxn = –250. J mol–1 K–1
From the tabulated values:
∆H°rxn = ∆H°f(b) – (–186 kJ mol–1) – 2.98(0) = –67.3 kJ mol–1
∆H°f(β-LaNi5H6.39(s)) = –253 kJ mol–1
∆S°rxn = S°(b) – (223 J mol–1 K–1) – 2.98(130.7 J mol–1 K–1) = –250. J mol–1 K–1
S°(β-LaNi5H6.39(s)) = 362 J mol–1 K–1