Transcript 2-inorganicchemistryweek1(17)

Inorganic chemistry
 Calculating nuclear binding energy.
 Thermo chemistry.
 Enthalpy of a reaction.
Assistance Lecturer Amjad Ahmed Jumaa
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1
Calculating nuclear binding energy:
energy required to break up a nucleus into its component protons and
neutrons.
masses of nuclei are always less than the sum of the masses of the
nucleons (the protons and neutrons in a nucleus).
mass defect. The difference between the mass of an atom and the sum
of the masses of its protons, neutrons, and electrons.
According to Einstein's mass-energy equivalence relationship:
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E = mc2
Where:
E is energy
m is mass
c is the velocity of light
We can calculate the amount of energy released by writing:
∆E= (∆m) c2
Where:
∆E = energy of products – energy of reactants.
∆m = mass of products – mass of reactants.
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the mass changes in nuclear reactions are approximately a million
Times larger per mole of reactant than those in chemical reactions.
Consider the alpha decay of uranium-238 to thorium-234. The nuclear
equation is
238
92
238.0003
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233.9942
4.00150 amu
The change in mass for above nuclear reaction, starting with molar
amounts, is
∆m = (233.9942 + 4.00150 -238.0003) g = -0.0046 g
•The minus sign indicates a loss of mass. This loss of mass is clearly large
enough to detect.
to calculate the energy change for a nuclear reaction.
This is illustrated in the next example.
1eV = 1.602 x 10-19 J.
1 MeV equals 1.602x10-13 J.
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Example:
Calculating the Energy Change for a Nuclear Reaction
A. Calculate the energy change in joules (four significant
figures) for the
Above nuclear reaction per mole of
:
Atomic and particle masses are given in the above table the speed
of light is (2.998 x 108 m/s)
B. What is the energy change in MeV for one
nucleus?
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solution
Nuclear mass of (
4.001502 amu
) = 4.00260 amu - (2 x 0.00549 amu) =
2.01345 3.01493 4.00150 1.00728 amu
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Hence,
∆m = (4.00150 + 1.00728 - 2.01345 - 3.01493) amu
= - 0.01960 amu
Therefore, the mass change for molar amounts in this
nuclear reaction is (-0.01960 g), or (-1.960 x10-5) kg. The
energy change is
∆E = (∆m) c2 = (-1.960 x 10-5 kg) (2.998 x108 m/s)2
= -1.762 x 1012 kg.m2/s2 or
-1.762 x1012 J.
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B. The mass change for the reaction of one (
) atom is (-0.01960)
amu. First change this to grams. Recall that 1 amu equals 1/12 the
mass of a (C-12) atom, whose mass is 12 g/6.022x1023. Thus, 1 amu =
1 g/6.022x 1023. Hence, the mass change in grams is:
∆m = - 0.01960 amu x
Then,
∆E = (∆m) c2 = (-3.255x 10-29 kg) (2.998 x108 m/s)2
= - 2.926 x 10-12 J.
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Now convert this to MeV:
∆E = - 2.926 x10-12 J x
= - 18.26 MeV.
Example:
Calculate the nuclear binding energy of the light isotopes of
(
), the atomic mass of helium is (3.01603 amu).
Solution :
The binding energy is the energy required for the process:
→
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+
Step (1): calculate the mass difference (∆m) between the products and
reactants:
∆m = [2 (proton mass) + (neutron)] -
atomic mass
= [2 (1.007285 amu) + 1.008665 amu] – 3.01603 amu
= 8.29 x 10-3 amu.
Step (2): use Einstein's equation to calculate the energy change for the process,
∆E.
∆E = (∆m) c2
= 8.29 x 10-3 amu x (3.00 x108 m/s)2
= 7.46 x 1014
Let's convert to useful units (J /He atom)
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=1.24 x 10-12 J / atom.
This is the nuclear binding energy. It's the energy required to break up
one helium-3 nucleus into (2) protons and (1) neutron.
Step (3): when comparing the stability of the two nuclei we must account for
the fact that they have different numbers of nucleous. For this reason, it is
more meaningful to use the nuclear binding energy per nucleon, defined as:
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Nuclear binding energy per nucleon =
For the helium-3 nucleus:
Nuclear binding energy per nucleon =
= 4.13 x 10 -13 J / nucleon.
balancing nuclear transmutation equations:
Nuclear can undergo change as a result of bombardment by neutrons,
protons, or other nuclei. This process is called nuclear transmutation.
consider the synthesis of neptunium (Np) which was the first
transmutation element to be synthesized by scientist:
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First, uranium-238 is bombarded with neutrons to produce
uranium-239.
Second This is a nuclear transmutation; uranium-239 is unstable and
decays spontaneously to neptunium-239 by emitting a (β) particle.
To balance a nuclear transmutation reaction, follow the same rules
used to balance nuclear equation.
1-The total number of protons plus neutrons in the products and the
reactants must be the same (conservation of mass number).
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2-The total number of nuclear charges in the products and in the reactants
must be the same (conservation of atomic number).
Example:
Write and balance the following reactions. When aluminum-27 is
bombarded with (α-particles), phosphorus-30 and one other particle are
produced. Phosphorus-30 has a low (n /p) ratio and decays spontaneously by
positron emission.
Solution:
The first reaction is a nuclear transmutation. You are given both reactants and
one of these two products in the problem. To balance the equation, remember
that both mass number and atomic number must be conserved.
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To balance the mass number, the missing particle (X) must have a mass
number of (1). To balance the atomic number, (X) must have an atomic
number of (0). (X) must be a neutron.
The second reaction is spontaneous so there is only one reactant,
phosphorus-30. The problem indicates that phosphorus -30 decays
by positron emission. Let's write down what we know so far.
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To balance the mass number, the missing element (X) must
have a mass of (30). To balance the atomic number, (X) must
have an atomic number of (14). (X) must be silicon-30.
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Thermo chemistry:
Is the branch of physical chemistry which deals with the thermal or
heat changes caused by chemical reactions.
Enthalpy of a reaction:
The enthalpy of a system is defined as the sum of the internal
energy and the product of its pressure and volume.
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H=E+PV.
Where:
(E) is the internal energy
(P) Is the pressure
(V) Is the volume of the system.
Enthalpy is also called heat content. Enthalpy is also a function
of the state and it is not possible to measure its absolute volume.
However a change in enthalpy (∆H), accompanying a process
can be measured accurately and is given by the expression:
∆H = H products –H reactants.
= Hp-Hr
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Thus if (∆V).
 the thermal effect observed will be the sum of the change in
internal energy(∆E), and the work done in expression, that is:
∆H =∆E+P∆V.
Therefore, while the heat change in a process is equal to its
change in internal energy (∆E) at constant volume, it gives at
constant pressure the enthalpy change (∆H). That is:
∆E= heat change in a reaction at constant volume.
∆H= heat change in a reaction at constant pressure.
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For reactions involving solids and liquids only the change in volume (∆V) is
very small and the term:
P×∆V.
For such reactions ∆H is equal to ∆E
25°C of sodium metal and water, carried out in a beaker open to the
atmosphere at 1.00 atm pressure.
2Na(s) + 2H2O (l)
2NaOH (aq) + H2(g)
The metal and water react vigorously and heat evolves. Experiment shows
that 2 mol of sodium metal reacts with 2 mol of water to evolve 368.6 kJ of
heat. Because heat evolves, the reaction is exothermic, and you write qp = 368.6 kJ. Therefore, the enthalpy of reaction, or change of enthalpy for the
reaction, is ∆H = -368.6 kJ.
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2 mol Na(s) + 2 mol H2O (l)
ΔH = −368.6 k
(368.6 kJ of heat
is released)
2 mol NaOH (aq) + 1 mol H2(g)
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From the above diagram (an enthalpy diagram) we can see when 2 mol Na(s)
and 2 mol H2O(l ) react to give 2 mol NaOH(aq) and 1 mol H2(g), 368.6 kJ of
heat is released, and the enthalpy of the system decreases by 368.6 kJ.
Consider the reaction of methane, CH4 (the principal constituent of
natural gas),
burning in oxygen at constant pressure. How much heat could you obtain
from 10.0 g of methane, assuming you had an excess of oxygen?
The answer for 1 mol of methane. The thermo chemical equation is:
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CH4 (g) +2O2 (g)
CO2 (g) + 2H2O (l); ∆H = - 890.3 kJ
The calculation involves the following conversions:
Grams of CH4
moles of CH4
kilojoules of heat
10.0 g CH4 x
Grams of A
(reactant or
Product)
x
Conversion
factor:
g A to mol A
(using molar
mass)
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x
Conversion factor:
mol A to KJ (using
enthalpy of
reaction)
Kilojoules
of heat
Example : How much heat is evolved when 9.07 x105 g of
ammonia is produced according to the following
equation?(Assume that the reaction occurs at constant
pressure.)
N2 (g) + 3H2 (g)
2NH3 (g); ∆H = - 91.8 kJ
Problem Strategy The calculation involves converting grams of
NH3 to moles of NH3 and then to kilojoules of heat.
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Grams of NH3
moles of NH3
kilojoules of heat
You obtain the conversion factor for the second step from
the thermo chemical equation, which says that the production
of 2 mol NH3 is accompanied by
qp = -91.8 kJ.
Solution:
9.07 x 105 g NH3 x
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