Calculate the entropy change of the universe (in J/K) when 12.2 g of chlorine are melted in a laboratory at 24.8 oC. Report your answer in scientific notation to three significant figures.
Melting Point (°C) -101.0
Boiling Point (°C) -34.1
ÄHo(Fusion) (kJ/mol) 6.40
ÄHo(Vaporization) (kJ/mol) 20.41
which ones am i supposed to use...?
To calculate the entropy change of the universe during the melting of chlorine, you need to consider the following:
1. The change in entropy of the system (ΔS_sys) during the melting process.
2. The change in entropy of the surroundings (ΔS_surr) due to the heat transfer between the system and the surroundings.
In this case, the chlorine is the system, and the laboratory is the surroundings. To calculate these entropy changes, you can use the equation:
ΔS_univ = ΔS_sys + ΔS_surr
Now let's break down the calculations:
1. ΔS_sys - The change in entropy of the system during the melting process can be calculated using the equation:
ΔS_sys = ΔH_fusion / T
where ΔH_fusion is the enthalpy of fusion, and T is the temperature in Kelvin.
Given: ΔH_fusion = 6.40 kJ/mol
First, we need to convert grams of chlorine to moles. Chlorine (Cl2) has a molar mass of approximately 35.45 g/mol:
Moles of Cl2 = 12.2 g / 35.45 g/mol
Next, we need to convert from moles to kJ:
ΔH_fusion = 6.40 kJ/mol * (moles of Cl2)
Finally, we convert to J/K by dividing by the temperature:
ΔS_sys = ΔH_fusion / T (in Kelvin)
2. ΔS_surr - The change in entropy of the surroundings can be calculated using the equation:
ΔS_surr = -(ΔH_fusion / T_surr)
where T_surr is the temperature of the surroundings in Kelvin.
For our calculation, T_surr is given: 24.8 °C = 24.8 + 273.15 K
Now that you have the equations and the necessary values, you can plug in the numbers and calculate the entropy change of the universe.