TOPIC 15 ENERGETICS/THERMOCHEMISTRY 15.1 ENERGY CYCLES

ESSENTIAL IDEA The concept of the energy change in a single step reaction being equivalent to the summation of smaller steps can be applied to changes involving ionic compounds. NATURE OF SCIENCE (3.2) Making quantitative measurements with replicates to ensure reliability – energy cycles allow for the calculation of values that cannot be determined directly.

UNDERSTANDING/KEY IDEA 15.1.A Representative equations (e.g. M + (g) → M + (aq) ) can be used for enthalpy/energy of hydration, ionization, atomization, electron affinity, lattice, covalent bond and solution.

GUIDANCE The following enthalpy/energy terms should be covered: ionization, atomization, electron affinity, lattice, covalent bond, hydration and solution.

The lattice enthalpy (ΔH lat º ) is defined as the change in enthalpy that occurs when one mole of a solid ionic compound is separated into its gaseous ions under standard conditions. Electron affinity (ΔH ea º ) is the enthalpy change when one mole of gaseous atoms attracts one mole of electrons. Enthalpy change of atomization (ΔH atom º ) is the heat change when one mole of gaseous atoms are formed from the element in its standard state.

The bond enthalpy for a covalent bond (E) or (ΔH º ) is defined as the energy needed to break one mole of bonds in gaseous molecules under standard conditions. Ionization energy (ΔH ie º ) is the energy required to remove a mole of electrons from a mole of gaseous atoms to form a mole of cations in the gaseous state.

ΔH atom º = M (s) → M (g) E = bond enthalpy (covalent bond) ΔH ie º = M (g) → M + (g) + e - ΔH ea º= X (g) + e - → M - (g) ΔH lat º= MX (S) → M + (g) + X - (g) ΔH sol º = MX (S) → M + (aq) + X - (aq) ΔH hyd º = M + (g) → M + (aq)

APPLICATION/SKILLS Be able to construct Born- Haber cycles for group 1 and 2 oxides and chlorides.

APPLICATION/SKILLS Be able to calculate enthalpy changes from Born-Haber or dissolution energy cycles.

Born-Haber Cycle This is an energy cycle based on Hess’s Law. This is used because lattice energies cannot be determined directly. The formation of an ionic compound from its elements takes place in a number of steps.

Formation of NaCl Na (s) + ½ Cl 2(g) → NaCl (s) ΔH f ˚ = -411 kJ/mol Step 1: Atomize sodium ΔH atom ˚ = +107 Na (s) →Na (g) Step 2: Form one chlorine gas ½E(Cl-Cl) = ½(+243) break the Cl 2 bond ½ Cl 2(g) → Cl (g) E is bond enthalpy given in table Step 3: Remove one e- from the ΔH i ˚= +496 gaseous sodium atom Na (g) → Na + (g) + e- Step 4: One e- is added to the ΔH e ˚= -349 gaseous chlorine atom Cl (g) + e - → Cl - (g) Step 5: The gaseous ions form ΔH lat ˚ = ? one mole of solid NaCl. Na + (g) + Cl - (g) → NaCl (s)

Na (s) + ½Cl 2(g) ΔH f ˚ (NaCl)= ΔH atom ˚+½E(Cl-Cl) + ΔH i ˚(Na) + ΔH e ˚(Cl) - ΔH lat ˚(NaCl) -411 = ½ (243) (-349) - ΔH lat ˚(NaCl) ΔH lat ˚(NaCl) = kJ/mol ΔH atom ˚ = +107 Na (g) +½Cl 2(g) ½E(Cl-Cl) = ½(+243) Na (g) + Cl (g) ΔH i ˚(Na)= +496 ΔH f ˚ (NaCl)= -411 kJ/mol From Table 10 NaCl (s) Na + (g) + e- + Cl - (g) ΔH e ˚(Cl)= -349 Na + (g) + Cl - (g) ΔH lat ˚(NaCl) = ?

OXYGEN EXCEPTION Note that when oxygen is used in an ionic compound that there is a different treatment for (ΔH ea º ). If your cation is a 2+ ion, you will be using 2 ionization energies. The 2 nd ionization energy will have to be given to you in the problem. The 1 st can be found in the data booklet. Oxygen will have an exothermic 1 st (ΔH ea º ) and an endothermic 2 nd (ΔH ea º ). The 2 nd is endothermic because you are trying to add an electron to a negatively charged species and you have to overcome the neg-neg repulsions.

APPLICATION/SKILLS Be able to relate size and charge of ions to lattice and hydration enthalpies.

Two things affect lattice enthalpy: The higher the charge – the stronger the lattice enthalpy. The smaller the ion – the stronger the lattice enthalpy.

UNDERSTANDING/KEY IDEA 15.1.B Enthalpy of solution, hydration enthalpy and lattice enthalpy are related in an energy cycle.

The enthalpy change of solution (ΔH sol º ) is defined as the change in enthalpy that occurs when one mole of a solute is dissolved in a solvent to form aqueous ions in an infinitely dilute solution under standard conditions. Dissolving an ionic cmpd to make aq ions. The enthalpy change of hydration (ΔH hyd º ) is defined as the change in enthalpy that occurs when one mole of gaseous ions is dissolved to form an infinitely dilute solution of one mole of aqueous ions under standard conditions. Refers to the individual ions

INFINITELY DILUTE SOLUTIONS The interaction between the solute and the solvent water molecules depends upon the concentration of the solution. The enthalpy of solution strictly refers to the ideal situation of infinite dilution. To obtain this value, measure the enthalpy changes for solutions with increasing volumes of water until a limit is reached.

SOLUTIONS Ionic compounds are crystal lattices. They readily dissolve in water. The ions are strongly attracted to the polar water molecule. The partial positive charge on the hydrogen attracts the negative ions.

The partial negative charge on the oxygen in the water molecule attracts the positive ions. Ions separated from a crystal lattice in this manner become surrounded by water molecules and are said to be hydrated. The strength of interaction between the polar water molecules and the separated ions is given by the hydration enthalpy.

The enthalpy of hydration of individual ions cannot be measured directly because both types of ions are present and the contribution of each ion cannot be disentangled. The problem is resolved by measuring the enthalpy of hydration of the H + ion and then combining this value with the hydration enthalpy of different compounds to obtain values for individual ions.

Remember the enthalpy of hydration of an ion is the enthalpy change that occurs when one mole of gaseous ions is dissolved to form an infinitely dilute solution of one mole of aqueous ions. There is a force of attraction between the ions and the polar water molecules so the process is exothermic and the value for ΔH hyd º is negative.

The ΔH hyd º becomes less exothermic as ionic radius increases (down a group). The force of attraction is less as the distance between the water molecule and the ions increases. The ΔH hyd º becomes more exothermic across a period because charges increase and size decreases.

APPLICATION/SKILLS Be able to construct energy cycles from hydration, lattice and solution enthalpy. For example dissolution of solid NaOH or NH 4 Cl in water.

GUIDANCE Values for lattice enthalpies (section 18), enthalpies of aqueous solutions (section 19), and enthalpies of hydration (section 20) are given in the data booklet.

EXAMPLE PROBLEM Use an energy cycle to calculate the enthalpy of solution of NaCl from data sections 18 and 20 in the IB data booklet. Compare your value with the value in section 19 of the data booklet and comment on the disagreement between the 2 values.

The enthalpy of solution is solved by adding the lattice enthalpy of the ionic solid (section 18) and the separate hydration enthalpies of the ions (found in section 20 of the IB data booklet). ΔH sol º = ΔH lat º (NaCl) + ΔH hyd º (Na + ) + ΔH hyd º (Cl - ) = +790 – 424 – 359 kJ/mol = +7 kJ/mol The value obtained in section 19 is kJ/mol. The % inaccuracy = (7-3.88)/3.88 x 100 = 80% The disagreement between the two values illustrates a general problem when a small numerical value is calculated from the difference of two large numerical values.

Citations Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.