± Coulomb\'s Law and Lattice Energy Atoms involved in bonds are usually more sta
ID: 540275 • Letter: #
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± Coulomb's Law and Lattice Energy Atoms involved in bonds are usually more stable than they would be otherwise. This means that energy is released in the formation of the bond. For ionic bonds, Coulomb's law is used to calculate this energy: where E is energy in joules (J), k is a constant equal to 2.31 x 102 J m. i and q2 are the charges of the two ions, and d is the distance between the nuclei of the two ions in meters (m). Whereas the bond energy calculated by Cou lomb's law pertains to a single ionic bond, lattice energy pertains to al the ionic bonding among a group of atoms arranged in a crystal lattice. Lattice energy is calculated by using the Coulomb's law equation, but with a different constant (unique to each substance) that takes into account the crystalline structure of the substance. By convention, lattice energy is defined as the amount required to either break an ionic solid into individual gaseous ions, or to form an ionic solid from gaseous ions. Part A Calculate the amount of energy released in the formation of one mole of MgSe bonds (not lattice energy). The radius of the magnesium ion is 0.65 A , and the radius of the selenide ion is 1 98 A Note that 1 A 10-10 m Express your answer with the appropriate units to three significant figures. You did not open hints for this part ANSWERExplanation / Answer
Energy released in the formation of one molecule of MgSe is given by
E = (k qMg* qSe)/d = (2.31 x 10-28 J. m * 2 * 2) / (0.65+1.98) *10-10 m = 3.513 x 10-18J
To calculate the enrgy released for one mole of MgSe formation,we need to mutiply above number with
Avagrdo's Number i.e. 6.023 X 1023 .
= 3.513 x 10-18 * 6.023 X 1023 = 21.159 X 105 J
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