Calculate ∆S universe for the process H2O (s) → H2O (l) given that
∆Hfusion = 6.02 kJ/mol and ∆S system = 22.1 J/K mol . Remember that
∆S surrounding = -∆Hsystem / T.
a) -20oC
b) 0oC
c) 20oC
To calculate ∆S universe for the given process, we need to consider the ∆S system and ∆S surrounding.
First, let's calculate ∆S surrounding using the equation ∆S surrounding = -∆H system / T.
Given ∆H fusion = 6.02 kJ/mol, we need to convert it into joules:
∆H fusion = 6.02 kJ/mol × 1000 J/kJ = 6020 J/mol.
Now, we can calculate ∆S surrounding for each temperature:
a) -20°C:
Convert -20°C to Kelvin: T = (-20 + 273) K = 253 K
∆S surrounding = -∆H system / T = -6020 J/mol / 253 K = -23.80 J/K mol
b) 0°C:
Convert 0°C to Kelvin: T = (0 + 273) K = 273 K
∆S surrounding = -∆H system / T = -6020 J/mol / 273 K = -22.07 J/K mol
c) 20°C:
Convert 20°C to Kelvin: T = (20 + 273) K = 293 K
∆S surrounding = -∆H system / T = -6020 J/mol / 293 K = -20.53 J/K mol
Therefore, the values of ∆S surrounding for each temperature are:
a) -23.80 J/K mol
b) -22.07 J/K mol
c) -20.53 J/K mol
∆S universe is calculated by adding ∆S system and ∆S surrounding:
∆S universe = ∆S system + ∆S surrounding
Given ∆S system = 22.1 J/K mol, we can calculate ∆S universe for each temperature:
a) -20°C:
∆S universe = 22.1 J/K mol + (-23.80 J/K mol) = -1.70 J/K mol
b) 0°C:
∆S universe = 22.1 J/K mol + (-22.07 J/K mol) = 0.03 J/K mol
c) 20°C:
∆S universe = 22.1 J/K mol + (-20.53 J/K mol) = 1.57 J/K mol
Therefore, the values of ∆S universe for each temperature are:
a) -1.70 J/K mol
b) 0.03 J/K mol
c) 1.57 J/K mol