BORNHABER CYCLES Learning Objectives 1 Define key enthalpy
BORN-HABER CYCLES Learning Objectives: 1. Define key enthalpy change definitions. 2. Describe how to construct Born. Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds. 3. Explain how Born-Haber cycles can be used to calculate enthalpies of solution. Specification Reference: 3. 1. 8. 1
Lattice Enthalpy Definition(s) THERE ARE TWO DEFINITIONS OF LATTICE ENTHALPY 1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions. ’ Example Na+(g) + Cl¯(g) Na+ Cl¯(s) 2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions. ’ Example Na+ Cl¯(s) Na+(g) + Cl¯(g) MAKE SURE YOU CHECK WHICH IS BEING USED LO 1: Define key enthalpy change definitions.
Lattice Enthalpy Definition(s) 1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions. ’ Values Example highly EXOTHERMIC strong electrostatic attraction between oppositely charged ions a lot of energy is released as the bond is formed relative values are governed by the charge density of the ions. Na+(g) + Cl¯(g) Na+(g) + Cl–(g) Na. Cl(s) LO 1: Define key enthalpy change definitions. Na+ Cl¯(s)
Lattice Enthalpy Definition(s) 2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions. ’ Values Example highly ENDOTHERMIC strong electrostatic attraction between oppositely charged ions a lot of energy must be put in to overcome the attraction relative values are governed by the charge density of the ions. Na+ Cl¯(s) Na+(g) + Cl–(g) Na. Cl(s) LO 1: Define key enthalpy change definitions. + Cl¯(g)
Calculating Lattice Enthalpy SPECIAL POINTS you CANNOT MEASURE LATTICE ENTHALPY DIRECTLY it is CALCULATED USING A BORN-HABER CYCLE greater charge densities of ions = greater attraction = larger lattice enthalpy Effects Melting point the higher the lattice enthalpy, the higher the melting point of an ionic compound Solubility solubility of ionic compounds is affected by the relative values of Lattice and Hydration Enthalpies LO 1: Define key enthalpy change definitions.
Lattice Enthalpy Values Cl¯ Br¯ F¯ O 2 - Na+ -780 -742 -918 -2478 K+ -711 -679 -817 -2232 Rb+ -685 -656 -783 Mg 2+ -2256 Ca 2+ -2259 -3791 Units: k. J mol-1 Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions. Na+ Cl¯ K+ Cl¯ The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together. LO 1: Define key enthalpy change definitions.
Born-Haber Cycle For Sodium Chloride k. J mol-1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> Na. Cl(s) – 411 Enthalpy of sublimation of sodium Na(s) + 108 Enthalpy of atomisation of chlorine ½Cl 2(g) ——> Cl(g) + 121 Ist Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ + 500 Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g) – 364 Na+(g) + Cl¯(g) ——> Na. Cl(s) ? Lattice Enthalpy of Na. Cl LO 1: Define key enthalpy change definitions. ——> Na(g)
Born-Haber Cycle - Na. Cl 1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> Na. Cl(s) This is an exothermic process so energy is released. Sodium chloride has a lower enthalpy than the elements which made it. VALUE = - 411 k. J mol-1 Na(s) + ½Cl 2(g) 1 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Na. Cl 1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> 2 Na. Cl(s) Enthalpy of sublimation of sodium Na(s) ——> Na(g) + ½Cl 2(g) 2 This is an endothermic process. Energy is needed to separate the atoms. Sublimation involves going directly from solid to gas. VALUE = + 108 k. J mol-1 Na(s) + ½Cl 2(g) 1 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Na. Cl 1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> 2 Enthalpy of sublimation of sodium Na(s) 3 Na. Cl(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl 2(g) ——> Cl(g) Na(g) + Cl(g) 3 Na(g) + ½Cl 2(g) 2 Breaking covalent bonds is an endothermic process. Energy is needed to overcome the attraction the atomic nuclei have for the shared pair of electrons. VALUE = + 121 k. J mol-1 Na(s) + ½Cl 2(g) 1 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Na. Cl 1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> 2 Na+(g) + Cl(g) Enthalpy of sublimation of sodium Na(s) 3 Na. Cl(s) ——> Na(g) 4 Enthalpy of atomisation of chlorine ½Cl 2(g) ——> Cl(g) Na(g) + Cl(g) 4 Ist Ionisation Energy of sodium 3 Na(g) ——> Na+(g) + e¯ Na(g) + ½Cl 2(g) 2 All Ionisation Energies are endothermic. Energy is needed to overcome the attraction the protons in the nucleus have for the electron being removed. VALUE = + 500 k. J mol-1 Na(s) + ½Cl 2(g) 1 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Na. Cl 1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> 2 Na+(g) + Cl(g) Enthalpy of sublimation of sodium Na(s) 3 Na. Cl(s) ——> 5 Na(g) 4 Enthalpy of atomisation of chlorine ½Cl 2(g) ——> Na+(g) + Cl–(g) Cl(g) Na(g) + Cl(g) 4 Ist Ionisation Energy of sodium 3 Na(g) ——> Na+(g) + e¯ 5 Electron Affinity of chlorine Cl(g) + e¯ ——> Na(g) + ½Cl 2(g) 2 Cl¯(g) Electron affinity is exothermic. Energy is released as the nucleus attracts an electron to the outer shell of a chlorine atom. VALUE = - 364 k. J mol-1 Na(s) + ½Cl 2(g) 1 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Na. Cl 1 Enthalpy of formation of Na. Cl Na(s) + ½Cl 2(g) ——> 2 Na+(g) + Cl(g) Enthalpy of sublimation of sodium Na(s) 3 Na. Cl(s) ——> 5 Na(g) 4 Enthalpy of atomisation of chlorine ½Cl 2(g) ——> Na+(g) + Cl–(g) Cl(g) Na(g) + Cl(g) 4 Ist Ionisation Energy of sodium 3 Na(g) ——> Na+(g) + e¯ 5 Electron Affinity of chlorine Cl(g) + e¯ 6 ——> Na(g) + ½Cl 2(g) 2 Cl¯(g) Lattice Enthalpy of Na. Cl Na(s) + ½Cl 2(g) Na+(g) + Cl¯(g) ——> Na. Cl(s) Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other. 6 1 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Na. Cl CALCULATING THE LATTICE ENTHALPY Na+(g) + Cl(g) Apply Hess’s Law 6 = - - 5 - 4 3 - 5 + 1 2 4 The minus shows you are going in the opposite direction to the definition = - (-364) - (+500) - (+121) - (+108) + (-411) = - 776 k. J mol-1 Na+(g) + Cl–(g) Na(g) + Cl(g) 3 Na(g) + ½Cl 2(g) OR… 2 Ignore the signs and just use the values; If you go up you add, if you come down you subtract the value 6 = 5 - 4 - 3 - 2 - Na(s) + ½Cl 2(g) 1 1 = (364) - (500) - (121) - (108) - (411) = - 776 k. J mol-1 6 Na. Cl(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Born-Haber Cycle - Mg. Cl 2 1 Enthalpy of formation of Mg. Cl 2 Mg(s) + Cl 2(g) ——> 2 Mg. Cl 2(s) Mg 2+(g) + 2 Cl(g) Enthalpy of sublimation of magnesium Mg(s) ——> 5 Mg(g) 6 3 Enthalpy of atomisation of chlorine ½Cl 2(g) ——> 4 Mg 2+(g) + 2 Cl–(g) Mg(g) + 2 Cl(g) 3 2 nd Ionisation Energy of magnesium Mg(g) + Cl 2(g) 2 Electron Affinity of chlorine Cl(g) + e¯ 7 4 Ist Ionisation Energy of magnesium Mg+(g) ——> Mg 2+(g) + e¯ 6 + 2 Cl(g) x 2 Cl(g) Mg(g) ——> Mg+(g) + e¯ 5 Mg+(g) ——> Cl¯(g) x 2 Mg(s) + Cl 2(g) 1 Lattice Enthalpy of Mg. Cl 2 Mg 2+(g) + 2 Cl¯(g) ——> Mg. Cl 2(s) 7 Mg. Cl 2(s) LO 2: Describe how to construct Born-Haber cycles to calculate enthalpy changes and compare these values to the perfect ionic model to provide evidence for the covalent character of ionic compounds.
Enthalpy Changes in Solution - Ionic Compounds Enthalpy of Solution: – The enthalpy change that takes place when 1 mole of a solute dissolves in a solvent to form an ‘infinitely’ dilute solution. Na. Cl(s) + aq Na. Cl(aq) Hsol = +5 k. J mol-1 • Hsol - Enthalpy of Solution. • Hsol can be exothermic or endothermic. • Hsol can be measured experimentally. LO 3: Explain how Born-Haber cycles can be used to calculate enthalpies of solution.
Enthalpy of Solution • The sum of two imaginary steps: Reverse of the Lattice Enthalpy ( HLE) + The Hydration Enthalpy ( Hhyd) (of the cations and anions). Lattice Enthalpy: The enthalpy change that takes place when 1 mole of a solid ionic lattice forms from its gaseous ions. Na+(g) + Cl-(g) Na+Cl-(s) HLE = -776 k. J mol-1 • HLE is defined exothermically. • The more closely ions pack together in the solid lattice the more exothermic is HLE. The smaller and more highly charged the ions (the greater their charge density) the closer they pack. Enthalpy of Hydration The enthalpy change that takes place when 1 mole of gaseous ions become hydrated (surrounded by water molecules). Na+(g) + aq Na+(aq) Hhyd = - 406 k. J mol-1 • The smaller and more highly charged the ions (the greater their charge density), the more exothermic is Hhyd. LO 3: Explain how Born-Haber cycles can be used to calculate enthalpies of solution.
Enthalpy of Solution Calculation: H sol = - H LE + ( H hyd(cation) + H hyd(anion)) • Standard conditions of temperature and pressure (298 K, 1 atm. ) are implied above. – Thermochemical definitions often refer to Standard Molar Enthalpy Changes (changes measured under standard conditions). Na+(g) + Cl-(g) Hhyd = -771 k. J mol-1 - HLE = +776 k. J mol-1 Hsol = - -776 + -771 = + 5 k. J mol-1 Na+(aq) + Cl-(aq) Na. Cl(s) + aq Hsol = +5 k. J mol-1 LO 3: Explain how Born-Haber cycles can be used to calculate enthalpies of solution.
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