Find the entropy change of the universe when 2 mol of water freezes at -10 degrees celsius.

To find the entropy change of the universe when 2 mol of water freezes at -10 degrees Celsius, we need to consider the changes in entropy of the water and its surroundings.

The entropy change of the system, which is the water in this case, can be calculated using the equation:

ΔS_system = n × ΔS_fusion

where ΔS_system is the change in entropy of the system, n is the number of moles of substance (in this case, 2 mol of water), and ΔS_fusion is the molar entropy of fusion.

The molar entropy of fusion for water (ΔS_fusion) can be found in a table of thermodynamic data. It is approximately 22.02 J/mol·K. However, make sure to check the specific reference or table you are using for accurate values.

Now, to calculate the entropy change of the surroundings, we can use the equation:

ΔS_surroundings = -ΔH/T

where ΔS_surroundings is the change in entropy of the surroundings, ΔH is the enthalpy change of the system (which is the heat released or absorbed during the process), and T is the temperature in Kelvin.

Since we are dealing with the freezing of water, we can assume that ΔH is the heat released during the process. The enthalpy of fusion for water (ΔH_fusion) is approximately -6.01 kJ/mol. However, when using this value, be sure to convert it into the proper unit (J/mol) and ensure the sign is correct (negative for exothermic processes).

Step 1: Convert the temperature from Celsius to Kelvin:
-10 degrees Celsius + 273.15 = 263.15 Kelvin

Step 2: Calculate the entropy change of the system:
ΔS_system = (2 mol) × (22.02 J/mol·K) = 44.04 J/K

Step 3: Calculate the entropy change of the surroundings:
ΔS_surroundings = -(-6.01 kJ/mol) / 263.15 K
ΔS_surroundings = 6.01 × 10^3 J/mol / 263.15 K ≈ 22.85 J/K

Step 4: Calculate the entropy change of the universe:
ΔS_universe = ΔS_system + ΔS_surroundings
ΔS_universe = 44.04 J/K + 22.85 J/K ≈ 66.89 J/K

Therefore, the entropy change of the universe when 2 mol of water freezes at -10 degrees Celsius is approximately 66.89 J/K.