Question

Calculate the change in energy (in kJ) for the reaction shown below, if you begin with 52 grams of fluorine (F2), and the ΔHrxn = −542 kJ. H₂(g) + F₂(g) → 2 HF(g)

Answer 1

The change in energy for the reaction H₂(g) + F₂(g) → 2 HF(g) when starting with 52 grams of fluorine is approximately -741.88 kJ, given that ΔHrxn is -542 kJ/mol for fluorine.

Step-by-step explanation:

To calculate the change in energy for the reaction provided, we need to know the molar mass of fluorine (F₂), which is 38 g/mol, and how many moles of fluorine 52 grams corresponds to. Using the stoichiometry of the reaction, we can then determine how much energy is released when those moles of fluorine react based on the given ΔHrxn.

First, we find the number of moles of fluorine:\[ \text{moles F}_2 = \frac{52\,\text{g}}{38\,\text{g/mol}} = 1.36842\,\text{moles} \]

Since the stoichiometry of the reaction is 1 mole of H₂ reacting with 1 mole of F₂ to give 2 moles of HF, the moles of fluorine directly determine the change in energy. Given the reaction enthalpy is ΔHrxn = -542 kJ for the reaction of 1 mole of F₂, we can now calculate the energy change for 1.36842 moles of fluorine:

\[ \text{Energy change} = 1.36842\,\text{moles} \times (-542\,\text{kJ/mol}) = -741.88\,\text{kJ} \]

Therefore, the change in energy for the reaction when beginning with 52 grams of fluorine is approximately -741.88 kJ.