Answer:
-1.2 × 10² J/K
Explanation:
The molar mass of butane is 58.12 g/mol. The moles corresponding to 202 g are:
202 g × (1 mol/58.12 g) = 3.48 mol
The heat of fusion of butane is 4.7 kJ/mol. The heat (Q) associated with the freezing of 3.48 moles of butane is:
Q = -ΔHf × n
Q = -4.7 kJ/mol × 3.48 mol
Q = -16 kJ
The freezing point is -138.0°C + 273.15 = 135.2 K. The change in entropy when 3.48 moles of butane freeze at 135.2 K is:
ΔS = Q/T
ΔS = -16 × 10³ J/135.2 K
ΔS = -1.2 × 10² J/K
