What is the entropy change for the vapor-
ization of 2.2 mol H2O(ℓ) at 100◦C and 1 atm?
�H = 40700 J/mol.
dG = dH - TdS
at equilibrium dG = 0; therefore,
dH = TdS
Substitute and solve for dS.
Well, let's see if I can put a funny spin on entropy!
The entropy change for the vaporization of 2.2 mol H2O(ℓ) at 100°C and 1 atm is not just dry... it's steamy! The heat of vaporization is 40700 J/mol, which means there's a lot of energy involved in turning that liquid water into gaseous water vapor. And when water goes from liquid to gas, it's like it's doing a magic trick - it's disappearing right before our eyes! So, the entropy change here is like a disappearing act, making it quite entertaining. Just imagine all that water going poof, and you've got yourself a steamy entropy change!
To find the entropy change for the vaporization of 2.2 mol H2O(ℓ) at 100°C and 1 atm, we can use the formula:
ΔS = ΔH / T
Where:
ΔS is the entropy change
ΔH is the enthalpy change or heat of vaporization
T is the temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T = 100°C + 273.15
T = 373.15 K
Now we can calculate the entropy change:
ΔS = (40700 J/mol) / (373.15 K)
ΔS = 109.17 J/(mol·K)
Therefore, the entropy change for the vaporization of 2.2 mol H2O(ℓ) at 100°C and 1 atm is 109.17 J/(mol·K).
To calculate the entropy change for the vaporization of H2O, you can use the formula:
ΔS = ΔH / T
Where:
ΔS is the entropy change
ΔH is the enthalpy change (given as 40700 J/mol)
T is the temperature in Kelvin (given as 100°C, which needs to be converted to Kelvin)
The temperature needs to be converted to Kelvin by adding 273.15 as follows:
T(K) = T(°C) + 273.15
Let's calculate it step by step:
1. Convert the temperature from Celsius to Kelvin:
T(K) = 100°C + 273.15 = 373.15 K
2. Now, substitute the given values into the formula:
ΔS = 40700 J/mol / 373.15 K
3. Perform the division to obtain the final answer:
ΔS ≈ 109.14 J/(mol·K)
Therefore, the entropy change for the vaporization of 2.2 mol H2O(ℓ) at 100°C and 1 atm is approximately 109.14 J/(mol·K).