Calculate the increase of entropy in J/K when 8.6 mole of C
2
H
5
OH melts at (-114)
o
C and 1
atm. ΔH
fus
= 5.02 kJ/mol
To calculate the increase in entropy when 8.6 moles of C2H5OH melts at -114°C and 1 atm, we need to use the equation:
ΔS = ΔH / T
where ΔS is the change in entropy, ΔH is the enthalpy change (ΔHfus), and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = -114 + 273.15
T(K) = 159.15 K
Next, we need to convert the enthalpy change from kJ/mol to J/mol:
ΔH(J/mol) = ΔH(kJ/mol) * 1000
ΔH(J/mol) = 5.02 kJ/mol * 1000
ΔH(J/mol) = 5020 J/mol
Now, we can substitute the values into the equation to calculate the change in entropy:
ΔS = ΔH / T
ΔS = 5020 J/mol / 159.15 K
Calculating this, you will find that the increase in entropy is approximately:
ΔS ≈ 31.5 J/K