Question

Calculate ΔS when one mole of steam at 100 ℃ is converted into ice at 0 ℃. Average specific heat of water = 4.184 x 103 JK-1 kg-1; heat of vaporization at boiling point = 2490.6 x 103 J kg-1; heat of fusion at freezing point = 333.555 x 103 J kg-1.

Answer 1

$\Delta$S = $\Delta$_condensationS + $\Delta$_coolingS + $\Delta$_freezingS = - $\Delta$_vaporizationH/T_vaporization + $\Delta$_coolingS +

- $\Delta$_meltingH/T_melting = ((18g/mol)/(1000 g*kg-1))*(-(2490.6*10³ J*kg-1)/(373.15 K) +

+ $\intop$_373.15K^273.15K[(4.184*10³ J*K-1*kg-1)/T]dT - (333.555*10³ J*kg-1)/(273.15 K)) = ((18g/mol)/(1000 g*kg-1))*(-7895.6699 + 4184*ln(273.15/373.15) J*K^-1*kg^-1) = -165.616 J*K^-1*mol^-1

We have 1 mole of water so $\Delta$S = (1 mol)*(-165.616 J*K^-1*mol^-1) = -165.616 J*K^-1