Inorganic help please! 6. Not long ago there was a flurry of interest in the pos
ID: 958854 • Letter: I
Question
Inorganic help please!
6. Not long ago there was a flurry of interest in the possibility of "cold fusion" of hydrogen deuterium isotope, 2H) in metallic palladium. The original idea came from the enormous solubility of hydrogen gas in palladium. Palladium metal has an foc The hydrogen atoms occupy the octahedral holes. If hydrogen atoms fill 70% of the octahedral holes and the lattice does not expand upon hydrogenation, how many grams of hydrogen (1H) will be contained in one cubic centimeter of the palladium hydride? Compare this to the density of liquid hydrogen gas H H). Are you surprised?Explanation / Answer
Density is mass per unit volume. Hence, if we calculate the mass of hydrogen in grams in one unit cell and then divide that by the volume of the unit cell in cubic centimeters, we will get the density of hydrogen in that structure.
There are 4 Pd atoms in the fcc unit cell and for every Pd atom, there is one possible octahedral hole. Hydrogen only fills 70% of these. Therfore, the mass of hydrogen within one unit cell can be calculated as follows,
(4 H atoms) x (0.7) x (1.008 g mol-1) x [1 mol/ 6.022 x1023 atoms] = 4.687 x 10-24
Now we will calculate the volume of one unit cell in cm3:
as we know, Palladium (Pd) adopts the fcc structure with a unit cell length of 3.8908 Å.
3.8908 Å = 3.8908 x 10-10 m = 3.8908 x 10-8 cm
volume of unit cell = (3.8908 x 10-8 cm)3 = 5.8900 x 10-23 cm3
Now we know that Density = mass/Volume
Density = (4.687 x 10-24)/(5.8900 x 10-23 cm3) = 0.07957 g cm-3
How does this compare to liquid hydrogen? After doing some unit conversion as follows,
Density of liquid hydrogen = 70.8 Kg/m3
= (70.8 Kg m-3)[1000 g/ 1 kg] [ 1 m3 / 1 x 106 cm3] = 0.078 g cm-3
The density of hydrogen in palladium hydride is slightly greater than the density of liquid hydrogen!
So it is advantageous to carry around some palladium hydride than a tank of liquid hydrogen