Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr
(www.ionicviper.org) on March 17, 2009. Copyright Margret J. Geselbracht 2009. This work is licensed
under the Creative Commons Attribution Non-commercial Share Alike License. To view a copy of this
license visit http://creativecommons.org/about/license/.
H2 STORAGE IN LANTHANUM NICKEL HYDRIDE
1. Hydrogen is a cleaner, greener alternative to fossil fuels. However, not many
of us would care to drive a car with a big tank of hydrogen on the back (ever
heard of the Hindenburg?). Metal hydrides are a safe alternative for storage
of hydrogen. Certain metals act like a hydrogen sponge; hydrogen is able to
neatly fit into the empty spaces in the metallic structure. Image downloaded
from http://www.flickr.com/photos/ulrichp/1362599/sizes/m/ under a Creative
Commons license
a) An intermetallic phase containing lanthanum and nickel readily and
reversibly adsorbs hydrogen at room temperature. The unit cell of the host material is shown
below on the left with layer-by-layer drawings at fractional heights of the unit cell along the z
direction on the right. The lanthanum atoms are the large gray balls and the nickel atoms are the
smaller green balls. The lattice parameters that describe the hexagonal unit cell are a = b =
5.023 Å; c = 3.983 Å. Determine the stoichiometry of the unit cell.
z
z = 0 layer
z = ½ layer
z = 1 layer
There are 8 lanthanum atoms at all 8 corners of the unit cell. On average, each one will
contribute 1/8 to this unit cell, so these will total up to one lanthanum per unit cell.
There are 8 nickel atoms on the unit cell faces, 2 on the bottom layer, 2 on the top layer
and 4 more on the faces at z = 1/2. Each of these contribute ½ to this unit cell. There is
also a single nickel atom in the center of the unit cell which belongs entirely to this unit
cell. So 8 (1/2) + 1 = 5 total nickel atoms.
The stoichiometry of the unit cell is LaNi5.
b) Under typical conditions, this intermetallic phase will take up 6 H atoms per unit cell, fitting in the
empty spaces between the metal atoms. The volume of the lattice expands by about 25% when
the hydride is formed. Given this information, calculate how many grams of hydrogen will be
contained in one cubic centimeter of the lanthanum nickel hydride. How does this number
compare to the density of liquid hydrogen (70.8 kg/m 3)?
Useful info: Volume of hexagonal unit cell =
3 (a 2 )(c)
= 0.866 (a2)(c)
2
11010 m 100 cm 
Unit cell of LaNi5: a = 5.023 Å 
 = 5.023  10-8 cm
 

 1 Å  1 m 

Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr
(www.ionicviper.org) on March 17, 2009. Copyright Margret J. Geselbracht 2009. This work is licensed
under the Creative Commons Attribution Non-commercial Share Alike License. To view a copy of this
license visit http://creativecommons.org/about/license/.
11010 m 100 cm 
c = 3.983 Å 
 = 3.983  10-8 cm

 1 Å  1 m 
Volume of unit cell = (0.866) (5.023  10-8 cm)2 (3.983  10-8 cm) = 8.703  10-23 cm3
Account for 25% expansion
of unit cell in hydride:

Vhydride unit cell = (8.703  10-23 cm3) (1.25) = 1.088  10-22 cm3
Mass of hydrogen per unit cell
1.008 g 

1 mol
= (6 atoms) 

 = 1.004  10-23 g
 1 mol 6.022 1023 atoms 
 1.004 1023 g 
So, density of hydrogen per unit cell = 
= 0.0923 g/cm3
-22
3 
1.088
10
cm



To compare to the density of liquid hydrogen, we need to do a bit of unit conversion:
(100 cm)3 = (1 m)3 so
1 106 cm3 = 1 m3
1000 g = 1 kg
 1 kg 110 6 cm3 
3
(0.0923 g/cm3) 
 = 92.3 g/cm

3
1000 g  1 m

Wow! Look at that! The lanthanum nickel hydride stores hydrogen with a greater
density than liquid hydrogen itself!

c) Aside from the volumetric storage capacity of hydrogen, another important consideration is the
added mass of the hydrogen “sponge” material. Calculate the weight % of hydrogen in the
lanthanum nickel hydride.
 6(MW H)  6(1.008 g mol -1 ) 
Weight % H in LaNi5H6 = 
= 1.379 % H
 
-1 
MW LaNi 5H 6   438.42 g mol 
d) One kilogram of hydrogen provides about the same amount of energy as 1 gallon of gasoline.
Calculate how many kg of lanthanum nickel hydride you would have to carry in your “tank” to
provide the same amount
 of energy as 10.0 gallons of gasoline.
1 kg H 2 1000 g  1 mol H 2 2 mol H 1.008 g H 438.42 g LaNi 5H 6  1 kg 
(10.0 gal gas) 







6.048 g H
1 gal gas  1 kg 2.016 g H 2 1 mol H 2  1 mol H 
1000 g 
= 725 kg LaNi5H6
