Question

For benzene, C₆H₆, the heat of fusion at its normal melting point of 6 °C is 10.0 kJ/mol.

The entropy change when 1.86 moles of liquid C₆H₆ freezes at 6 °C, 1 atm is

? J/K.

Answer 1

Given,

Temperature, T = 6 ^o C = 279 K

The heat of fusion = 10 KJ/mol

We know,

The change in entropy, S = Heat of fusion / Temperature

= (10 KJ/mol) / 279 K

= (10 * 10³ J/mol) / 279 K

= 35.84 J/mol K

Now, for 1.86 moles, the change in entropy, (S)_1.86 = 1.86 mol * 35.84 J/mol K

= 66.66 J/K