How much energy will it take to vaporize 45 g of liquid bromine (Br 2) if the ΔH of formation of Br 2 (g) is 31 kJ/mole?

To calculate the amount of energy required to vaporize a given amount of liquid bromine, we need to use the concept of the enthalpy of vaporization. The enthalpy of vaporization is the amount of energy required to convert one mole of a substance from its liquid state to its gaseous state.

First, we need to convert the mass of bromine from grams to moles. To do this, we divide the given mass (45 g) by the molar mass of bromine (Br₂), which is approximately 159.8 g/mol. Therefore, the number of moles is:

Number of moles = 45 g / 159.8 g/mol = 0.2811 mol (rounded to four decimal places)

Next, using the given enthalpy of formation (ΔH) of Br₂(g) as 31 kJ/mol, we can calculate the energy required to vaporize 0.2811 moles of bromine using the following equation:

Energy = ΔH × Number of moles

Energy = 31 kJ/mol × 0.2811 mol

Energy ≈ 8.7251 kJ (rounded to four decimal places)

Therefore, it would take approximately 8.7251 kJ of energy to vaporize 45 g of liquid bromine.