# Www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating nuclear binding energy.  Thermo chemistry.  Enthalpy of a reaction.

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www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating nuclear binding energy.  Thermo chemistry.  Enthalpy of a reaction. 1

www.soran.edu.iq 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:

www.soran.edu.iq E = mc 2 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) c 2 Where: ∆E = energy of products – energy of reactants. ∆m = mass of products – mass of reactants.

www.soran.edu.iq  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 233.9942 4.00150 amu

www.soran.edu.iq  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.

www.soran.edu.iq 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 10 8 m/s) B. What is the energy change in MeV for one nucleus?

www.soran.edu.iq Nuclear mass of ( ) = 4.00260 amu - (2 x 0.00549 amu) = 4.001502 amu 2.01345 3.01493 4.00150 1.00728 amu solution

www.soran.edu.iq 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) c 2 = (-1.960 x 10 -5 kg) (2.998 x10 8 m/s) 2 = -1.762 x 10 12 kg.m 2 /s 2 or -1.762 x10 12 J.

www.soran.edu.iq 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/6.022x10 23. Thus, 1 amu = 1 g/6.022x 10 23. Hence, the mass change in grams is: ∆m = - 0.01960 amu x Then, ∆ E = (∆m) c 2 = (-3.255x 10 -29 kg) (2.998 x10 8 m/s) 2 = - 2.926 x 10 -12 J.

www.soran.edu.iq 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: → +

www.soran.edu.iq Step (1): calculate the mass difference (∆m) between the products and reactants: ∆m = [2 (proton mass) + (neutron)] - = [2 (1.007285 amu) + 1.008665 amu] – 3.01603 amu = 8.29 x 10 -3 amu. atomic mass Step (2): use Einstein's equation to calculate the energy change for the process, ∆E. ∆E = (∆m) c 2 = 8.29 x 10 -3 amu x (3.00 x10 8 m/s) 2 = 7.46 x 10 14 Let's convert to useful units (J /He atom)

www.soran.edu.iq =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:

www.soran.edu.iq 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:

www.soran.edu.iq 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).

www.soran.edu.iq 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.

www.soran.edu.iq 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.

www.soran.edu.iq 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.

www.soran.edu.iq  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.

www.soran.edu.iq 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. = H p -H r

www.soran.edu.iq  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.

www.soran.edu.iq  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) + 2H 2 O (l) 2NaOH (aq) + H 2 (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.

www.soran.edu.iq 2 mol Na(s) + 2 mol H 2 O (l) ΔH = −368.6 k (368.6 kJ of heat is released) 2 mol NaOH (aq) + 1 mol H 2 (g)

www.soran.edu.iq  From the above diagram (an enthalpy diagram) we can see when 2 mol Na(s) and 2 mol H 2 O(l ) react to give 2 mol NaOH(aq) and 1 mol H 2 (g), 368.6 kJ of heat is released, and the enthalpy of the system decreases by 368.6 kJ.  Consider the reaction of methane, CH 4 (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:

www.soran.edu.iq CH 4 (g) +2O 2 (g) CO 2 (g) + 2H 2 O (l); ∆H = - 890.3 kJ  The calculation involves the following conversions: Grams of CH 4 moles of CH 4 kilojoules of heat 10.0 g CH 4 x Grams of A (reactant or Product) Grams of A (reactant or Product) x Conversion factor: g A to mol A (using molar mass) Conversion factor: g A to mol A (using molar mass) x Conversion factor: mol A to KJ (using enthalpy of reaction) Conversion factor: mol A to KJ (using enthalpy of reaction) Kilojoules of heat

www.soran.edu.iq Example : How much heat is evolved when 9.07 x10 5 g of ammonia is produced according to the following equation?(Assume that the reaction occurs at constant pressure.) N 2 (g) + 3H 2 (g) 2NH 3 (g); ∆H = - 91.8 kJ  Problem Strategy The calculation involves converting grams of NH 3 to moles of NH 3 and then to kilojoules of heat.

www.soran.edu.iq Grams of NH 3 moles of NH 3 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 NH 3 is accompanied by qp = -91.8 kJ. Solution: 9.07 x 10 5 g NH 3 x

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