Question #f536b

1 Answer
Nov 22, 2015

Answer:

Here's what I got.

Explanation:

The idea here is that you need to use mass of the nucleus and the combined mass of the nucleons it contains to determine the mass defect.

Once you know the mass defect per nucleus, you can use Einstein's equation to determine the binding energy per nucleus, then finally use Avogadro's number to find the binding energy per mole of nuclei.

Now, a nuclues' mass defect tells you the difference between the actual mass of the nucleus and the combined mass of the nucleons it contains.

A nucleon is a proton or a neutron. In your case, you know that the iron-56 nucleus has a mass of #"55.92068 u"#, but that the combined mass of its protons and neutrons is equal to #"56.44914 u"#.

This means that the mass defect for a single iron-56 nucleus will be

#color(blue)(M_"defect" = m_"nucleons" - m_"nucleus")#

#M_"defect" = "56.44914 u" - "55.92068 u" = "0.52846 u"#

Here's where Einstein's equation comes into play. The idea here is that the mass defect of a nucleus can be attributed to the binding energy of the nucleus, which tells you how much energy is required to split the nucleus into its constituent nucleons.

You can therefore, using Einstein's equation, convert the mass defect into binding energy

#color(blue)(E = M_"defect" * c^2)" "#, where

#E# - the binding energy
#M_"defect"# - the mass defect of the nucleus
#c# - the speed of light in a vacuum, equal to #"299,792,458 m s"^(-1)#

The trick now is to realize that you must convert the mass defect from unified atomic mass units, or #u#, to kilograms - you will see why a little later on.

By definition, one unified atomic mass unit is approximately equal to

#"1 u" = 1.660539 * 10^(-27)"kg"#

This means that the mass defect per nucleus for iron-56 will be equal to

#0.52846color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-27)"kg")/(1color(red)(cancel(color(black)("u")))) = 8.7773 * 10^(-28)"kg"#

This means that the binding energy per nucleus for iron-56 will be

#E = 8.7753 * 10^(-28) * (299792458)^2 color(white)(x) overbrace("kg m"^2 "s"^(-2))^(color(red)("Joules"))#

#E = 7.8868 * 10^(-11)"J"#

Now, nuclear binding energy is usually given per mole of nuclei. This means that you're going to have to use Avogadro's number to get

#7.8868 * 10^(-11)"J"/color(red)(cancel(color(black)("nucleus"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("nuclei"))))/"1 mole" = 4.749 * 10^(13)"J/mol"#

Expressed in kilojoules per mole, the answer will be

#4.749 * 10^(13)color(red)(cancel(color(black)("J")))/"mol" * "1 kJ"/(1000color(red)(cancel(color(black)("J")))) = color(green)(4.749 * 10^(10)"kJ/mol")#