Question

Given that ΔHvap is 52.6 kJ/mol, and the boiling point is 83.4°C, one mole of this substance is vaporized at 1.00 atm.

c. determine the δgvap at 125°C. the answer should be reported in kj/mol.

Answer 1

To determine the ΔGvap at 125°C, we can use the Clausius-Clapeyron equation and the given values of ΔHvap and boiling point. Plugging in the values, we find that ΔGvap is -0.898 kJ/mol.

Step-by-step explanation:

To determine the ∆Gvap at 125°C, we need to calculate the change in Gibbs free energy for the vaporization of one mole of the substance. We can use the Clausius-Clapeyron equation:

ln (P₂/P₁) = (-∆Hvap/R)((1/T₂) - (1/T₁))

where P₁ and T₁ are the pressure and temperature at which the ∆Hvap is given, and P₂ and T₂ are the pressure and temperature at which we want to find ∆Gvap. Rearranging the equation to solve for ∆Gvap gives:

∆Gvap = -RTln(P₂/P₁)

Plugging in the given values, we have:

∆Gvap = -(8.314 J/mol·K)(398.15 K - 353.15 K)ln(1.00 atm/0.917 atm)

∆Gvap = -898 J/mol = -0.898 kJ/mol