Question

For the reaction H₂(g) + Cl₂(g) → 2HCl(g), ΔG° = -190.2 kJ and ΔS° = 20.0 J/K at 278 K and 1 atm. The maximum amount of work that could be done by this reaction when 2.07 moles of H₂(g) react at standard conditions at this temperature is?

Answer 1

The maximum amount of work that could be done by the reaction H₂(g) + Cl₂(g) → 2HCl(g) when 2.07 moles of H₂(g) react at 278 K and 1 atm is -393.314 kJ.

Step-by-step explanation:

The question asks about the maximum work that could be done by the reaction H₂(g) + Cl₂(g) → 2HCl(g) when 2.07 moles of H₂(g) react at standard conditions, which can be calculated using the given ΔG° value. Since ΔG° represents the maximum non-expansion work that can be obtained from a reaction at standard conditions, the work done (w_max) is equal to the Gibbs free energy change (ΔG°) for the reaction. However, since the given ΔG° value is for 1 mole of H₂ reacting, we need to adjust it for 2.07 moles. Thus, w_max for 2.07 moles at 278 K would be 2.07 moles times -190.2 kJ/mole, which is equal to -393.314 kJ. This is the maximum amount of work that can be done by the reaction under the specified conditions.