Estimate the surface energy of (110) face of K crystals given just the information in the periodic table. Express your answer in the unit of J/cm2.

Answer 1

Someone MIGHT be able to help you IF you indicate what YOU THINK about each of these problems you've posted.

Answer 2

First Consider the Lattice Type of Potassium , and that is BBC

Then determine where the 110 "plane" is

Now, if you can imagine the (110) position you can see that cutting the unit cell at that position will rapture two K-K bonds. Hence to know that surface energy you need to know what is the value of breaking a K-K bond!!! then divide it by the area of that (110) surface.

To know K-K bond energy, you can start with (Ha) Atomisation energy of K given in the periodic table:

89.54 kJ/mol.

Ok you can convert that per atom value

So divide it by Avogadros number

89.54 KJ/mol x (1 mol / 6.02E23)

Now look at the BCC structure again and you see that each Na atom has 8 atomic neighbors, hence for one K atom to atomize it need to break 8 bonds, so now K-K bond energy = 89.54 KJ/mol x (1 mol / 6.02E23) / 8 !!!

Now that you have that take note that the unit is KJ/bond now. You must now calculate the area of 110 plan. To do that you have to solve for a (Lattice Constant)

a = (2 * Molar Volume / Avogadros) ^ 1/3

Now the area of the plane is just a*a^(1/2)

Answer 3

You will then notice that your final units is KJ/(bond cm^2)

You need to remove that bond unit so you must divide your final answer with the number of bonds present at the surface you wish to break / slip, and again that number of bonds is 2 !!!