Question

The ionization energy for potassium is 419 kj mol-1. The electron affinity for bromine is 325 kj mol-1. Use these values and Hess's law to calculate the enthalpy for the following reaction: k(g) + br(g) → k(g) + br-(g), ΔH = ? Express your answer with the appropriate units.

Answer 1

The enthalpy change for the reaction K(g) + Br(g) → K+(g) + Br-(g) is calculated by adding the ionization energy of potassium (absorbed energy) to the negative electron affinity of bromine (released energy), resulting in an enthalpy change of 94 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy change ΔH for the reaction K(g) + Br(g) → K+(g) + Br-(g), we will consider the processes involved using Hess's Law, which states that the total enthalpy change during the complete course of a chemical reaction is the same, whether the reaction is made in one step or several steps.

The ionization energy for potassium is the energy required to remove one electron from a neutral potassium atom to form a K+ ion:
K(g) → K+(g) + e- ΔH = 419 kJ mol-1
The electron affinity for bromine is the energy released when an electron is added to a neutral bromine atom to form a Br- ion:
Br(g) + e- → Br-(g) ΔH = -325 kJ mol-1
For the overall reaction, the ΔH will be the sum of the absorption of energy for ionization and the release of energy for electron affinity:
ΔH = Ionization Energy (K) - Electron Affinity (Br)
ΔH = 419 kJ mol-1 - 325 kJ/mol-1
ΔH = 94 kJ mol-1
Therefore, the enthalpy change for the reaction K(g) + Br(g) → K+(g) + Br-(g) is 94 kJ mol-1.