To calculate the entropy change, we need to consider three steps for this process:
1. Melting the ice at 273 K
2. Heating the water from 273 K to 373 K
3. Vaporizing water at 373 K
For each step, we'll calculate the entropy change individually and then add them up to find the total entropy change.
1. Melting the ice at 273 K:
The entropy change for melting ice can be calculated using the following formula:
ΔS_melting = ΔH_melting / T_melting
Where ΔH_melting is the molar heat of fusion at 273 K, which is given as 6 kJ/mol, and T_melting is the temperature, 273 K. Plugging in the values, we get:
ΔS_melting = (6 kJ/mol) / (273 K) = 0.02196 kJ/K per mol
2. Heating the water from 273 K to 373 K:
For this step, we can use the formula for the entropy change of a substance being heated at constant pressure:
ΔS_heating = C_p * ln(T2/T1)
Where C_p is the molar heat capacity, which is given as 75.2 J/K per mol, T1 is the initial temperature (273 K), and T2 is the final temperature (373 K).
First, calculate the natural logarithm of the ratio of the final and initial temperatures:
ln(T2/T1) = ln(373 K / 273 K) = 0.3127
Now, plug this value and the molar heat capacity into the formula for the entropy change during heating:
ΔS_heating = (75.2 J/K per mol)(0.3127) = 23.504 J/K per mol
However, since we need the answer in kJ/K, we will convert the units as follows:
ΔS_heating = 23.504 J/K per mol * (1 kJ / 1000 J) = 0.023504 kJ/K per mol
3. Vaporizing water at 373 K:
The entropy change for vaporizing water can be calculated using the following formula:
ΔS_vaporizing = ΔH_vaporizing / T_vaporizing
Where ΔH_vaporizing is the molar heat of vaporization at 373 K, which is given as 40.60 kJ/mol, and T_vaporizing is the temperature, 373 K. Plugging in the values, we get:
ΔS_vaporizing = (40.60 kJ/mol) / (373 K) = 0.10882 kJ/K per mol
Now, we can add up all the entropy changes for the three steps:
Total entropy change = ΔS_melting + ΔS_heating + ΔS_vaporizing
= 0.02196 kJ/K per mol + 0.023504 kJ/K per mol + 0.10882 kJ/K per mol
= 0.154284 kJ/K per mol
The total entropy change accompanying conversion of 1 mole of ice at 273 K and 1 atm pressure into steam at 373 K and 1 atm pressure is 0.154284 kJ/K per mol.
Calculate the entropy change accompanying conversion of 1 mole of ice at 273 K and 1 atm pressure into steam at 373 K and 1 atm pressure, given that at 273 K, the molar heat of fusion of ice is 6 kJ/mol and at 373 K, the molar heat of vapourization of water is 40.60 kJ/mol. Also assume that the molar heat capacity in the temperature range 373-273 K remains constant as 75.2 J/K per mol.
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