Find the entropy change of the universe when 2 moles of water freezes at -10 degrees celsius. The delta G for the freezing of water at -10 degrees celsius is -210 J/mol and the heat of fusion of ice at this temperature is 5610 J/mol.

To find the entropy change of the universe when 2 moles of water freeze at -10 degrees Celsius, we can use the equation:

ΔSuniv = ΔSsys + ΔSsurr

where ΔSuniv is the entropy change of the universe, ΔSsys is the entropy change of the system (water), and ΔSsurr is the entropy change of the surroundings.

The entropy change of the system (ΔSsys) can be calculated using the equation:

ΔSsys = ΔHsys / T

where ΔHsys is the enthalpy change of the system and T is the temperature in Kelvin.

Given that the temperature is -10 degrees Celsius, we need to convert it to Kelvin:

T = -10 + 273.15 = 263.15 K

Now we can calculate the entropy change of the system (water):

ΔSsys = ΔHsys / T
= (-210 J/mol) / (263.15 K)
= -0.798 J/(mol·K)

The entropy change of the surroundings (ΔSsurr) is equal to the heat transferred to the surroundings divided by the temperature:

ΔSsurr = -q / T

Given that the heat of fusion of ice is 5610 J/mol, the amount of heat transferred to the surroundings is:

q = ΔHfusion * moles
= (5610 J/mol) * (2 mol)
= 11220 J

Now we can calculate the entropy change of the surroundings:

ΔSsurr = -q / T
= (-11220 J) / (263.15 K)
= -42.65 J/(mol·K)

Finally, we can calculate the entropy change of the universe:

ΔSuniv = ΔSsys + ΔSsurr
= (-0.798 J/(mol·K)) + (-42.65 J/(mol·K))
= -43.448 J/(mol·K)

Therefore, the entropy change of the universe when 2 moles of water freeze at -10 degrees Celsius is approximately -43.448 J/(mol·K).