fus = 12.5 and vap = 93.2 both j/k x mol
Determine the entropy change when 7.30 mol of HCL(g) condenses at atmospheric pressure.
my first answer was -91.3 but it keeps telling me that is incorrect and that delta Svap if scaled at 1.00 mol but needs to be scaled to 7.30 mol
Please help
To determine the entropy change when 7.30 mol of HCL(g) condenses at atmospheric pressure, we need to calculate the change in entropy (ΔS) using the formula:
ΔS = ΔSvap × n
Where:
- ΔSvap is the molar entropy of vaporization of HCL(g), given as 93.2 J/K mol.
- n is the number of moles of HCL(g), which is 7.30 mol in this case.
Since the molar entropy of vaporization is given for 1.00 mol, we need to scale it to match the number of moles involved. To do this, we simply multiply ΔSvap by the factor of 7.30 (the number of moles):
ΔS = 93.2 J/K mol × 7.30 mol
Calculating the entropy change:
ΔS = 679.16 J/K
Therefore, the entropy change when 7.30 mol of HCL(g) condenses at atmospheric pressure is 679.16 J/K. Make sure to double-check your calculations to ensure they are correct.