This temperature represents the freezing
point of ethanol. (Which means delta G =
0) What is the entropy change for the vaporization of 3.3 mol H2O(ℓ) at 100◦C and 1 atm?
∆H= 40700 J/mol.
Answer in units of J/K
dG = 0 = dH - TdS.
Plug in 40,700 J/mol for dH and 373 for T, solve for dS, and that's for 1 mol. Correct for 3.3 mol.
To calculate the entropy change (∆S) for the vaporization of 3.3 moles of H2O(ℓ) at 100°C and 1 atm, we can use the equation:
∆S = ∆H / T
Where:
- ∆H is the enthalpy change of the process (given as 40700 J/mol),
- T is the temperature in Kelvin.
To convert the given temperature of 100°C to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
T(K) = 100°C + 273.15 = 373.15 K
Now, we can substitute the values into the equation to calculate ∆S:
∆S = 40700 J/mol / 373.15 K
Calculating:
∆S ≈ 109.21 J/K
Therefore, the entropy change (∆S) for the vaporization of 3.3 mol H2O(ℓ) at 100°C and 1 atm is approximately 109.21 J/K.