Solid CO2 (dry ice) sublimes at 195 K and its heat of sublimation ΔHsubl = 25.2 kJ/mol.

Question

A block of dry ice with a mass of 23.6 g sublimes. What is the entropy change, ΔS, of the CO₂? (Enter your answer in J/K).

**Heat is equal to T * &DeltaS. Be sure the units are correct.

Answer 1

Molar mass CO2 = 44 g / mol

23.6 g CO2 x 1 mol CO2 / 44 g = 0.5363 moles CO2

0.05363 moles CO2 x 25.2 kJ / mol = 13.52 kJ of heat

Heat = T∆S

Heat = 13.52 kJ

T = 195K

∆S = ?

Solving for ∆S..

∆S = heat / T = 13.52 kJ / 195K

∆S = 0.06933 kJ / K

To convert this to J / K we have..

0.06933 kJ/K x 1000 J / kJ = 69.3 J / K