Question

Question #7782e

Answer 1

The molar enthalpy of solidification of pure sulfur is -4.123 kJ/mol.

Explanation:

If you want to get the correct sign for the molar entropy of solidification, you must use the formulas

heat lost by sulfur = -heat gained by the calorimeter

and

moles × molar heat = -mass × specific heat capacity × change in temperature

In symbols, these become

`nΔ_sH = -mcΔT`

Here are the numbers we have to insert into the formulas.

`n ="moles of S" = 89.77 color(red)(cancel(color(black)("g S"))) × "1 mol"/(32.06 color(red)(cancel(color(black)("g S")))) = "2.800 mol"`

Since your numbers are all given to 4 significant figures, we must use the mass of water to 4 significant figures.

The density of water at 32.40 °C is 0.9943 g/mL.

Hence,

`m = "mass of water" = 250.0 color(red)(cancel(color(black)("mL water"))) × "0.9943 g water"/(1 color(red)(cancel(color(black)("mL water")))) = "248.6 g water"`

The specific heat capacity of water is

`c = "4.184 J·°C"^"-1""g"^"-1"`

`ΔT = T_"final" – T_"initial" =" 43.50 °C – 32.40 °C = 11.10 °C"`

This gives us

`"2.800 mol" × Δ_sH = "-248.6" color(red)(cancel(color(black)("g"))) × "4.184 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 11.10 color(red)(cancel(color(black)("°C"))) = "-11 546 J"`

`Δ_sH = "-11 546 J"/"2.800 mol" = "-4123 J/mol" = "-4.123 kJ/mol"`

∴ The molar enthalpy of solidification of sulfur is -4.123 kJ/mol.

The negative sign tells us that the sulfur loses heat energy when it solidifies.

NOTE: There is either an error in your data or in my calculations, because the molar entropy of solidification of sulfur is -1.727 kJ/mol.