Using the Born Haber Cycle

Summary

In this informative video, Chris Harris from Allert Tutors guides viewers through the process of using Born-Haber cycles to calculate the second ionization energy of calcium. He explains the significance of each component of the cycle, discusses potential pitfalls, and emphasizes the importance of understanding the enthalpy of formation versus lattice enthalpy. By working through a detailed example with relevant data, Chris highlights how to manipulate values and maintain energy conservation. He concludes with insights on how ion size and charge impact the exothermic nature of lattice formation, making the video a valuable resource for understanding ionic compounds.

Takeaways

• 😀 The video introduces Born-Haber cycles, explaining how to calculate various aspects of the cycle.
• 🔍 The speaker highlights potential pitfalls in calculations and uses an example that includes several challenges.
• 📊 The video focuses on determining the second ionization energy of calcium as a key aspect of the Born-Haber cycle.
• 🔄 The Born-Haber cycle allows for alternative pathways to calculate enthalpy changes, especially when direct routes are obstructed.
• 📉 The enthalpy of formation for calcium dichloride is -185 kJ/mol, indicating that it is an exothermic process.
• ⚛️ The enthalpy of atomization for fluorine must be doubled (from +79 to +158 kJ/mol) since the reaction involves two fluorine atoms.
• ⚡ The first ionization energy of calcium is +590 kJ/mol, which is crucial for the calculations within the cycle.
• 💡 Electron affinity for fluorine is -670 kJ/mol, reflecting the energy released when fluorine gains electrons.
• 🧮 The final calculations show that the second ionization energy of calcium is +140 kJ/mol, confirming that it is endothermic.
• 🔋 The video emphasizes that the charge and size of ions significantly affect the lattice enthalpy of formation, with higher charges resulting in more exothermic reactions.

Q & A

What is the primary focus of the video?

The video focuses on using Born-Haber cycles to calculate various aspects, particularly the second ionization energy of calcium.

What type of data does the speaker use in the Born-Haber cycle example?

The speaker uses enthalpy values, including the enthalpy of formation, atomization, ionization energy, and electron affinity for various substances.

How does the speaker suggest identifying the second ionization energy of calcium in the Born-Haber cycle?

The second ionization energy of calcium is identified as a gap in the cycle, which the speaker marks to show what needs to be calculated.

What is the enthalpy of formation for calcium dichloride mentioned in the video?

The enthalpy of formation for calcium dichloride is given as -185 kJ/mol.

Why does the speaker multiply the enthalpy of atomization of fluorine by two?

The speaker multiplies the enthalpy of atomization of fluorine by two because two fluorine atoms (F2) are needed to form two fluoride ions (2F-).

What is the significance of changing the signs of enthalpy values when using the Born-Haber cycle?

Changing the signs of enthalpy values indicates whether the process is endothermic or exothermic, which is essential for accurately applying the cycle.

What is the final calculated value for the second ionization energy of calcium?

The final calculated value for the second ionization energy of calcium is +140 kJ/mol.

What does the speaker emphasize about energy conservation in the Born-Haber cycle?

The speaker emphasizes that the total energy change in the cycle should equal zero, as energy cannot be created or destroyed.

How do ion size and charge affect lattice enthalpy of formation?

The larger the charge on the ions and the smaller the ionic radius, the more exothermic the lattice formation will be due to stronger attractions between the ions.

What comparison does the speaker make regarding the exothermic reactions of different ions?

The speaker compares lithium and fluorine, which are smaller ions with higher charge density, to potassium and bromide, indicating that the smaller ions will result in a more exothermic reaction.