The enthalpy of vaporization of benzene is $+35.3 \ kJ/mol$ at its boiling point,$80^{\circ} C$. The entropy change in the transition of vapour to liquid at its boiling point is

What happens to the entropy of water when it freezes to form an ice cube?

What is the change in entropy in $J \ K^{-1} \ mol^{-1}$ for the conversion of $1 \ mol$ of ice to water at $0 \, ^\circ C$? For the process $H_2O_{(s)} \rightarrow H_2O_{(l)}$ at $0 \, ^\circ C$,$\Delta H_{fus} = 6 \, kJ \ mol^{-1}$.

$18 \ g$ of ice is converted into water at $0 \ ^\circ C$ and $1 \ atm$. The entropies of $H_2O_{(s)}$ and $H_2O_{(l)}$ are $38.2$ and $60 \ J/mol \ K$ respectively. The enthalpy change for this conversion is ..... $J/mol$.

Standard entropies of $X_2, Y_2$ and $XY_5$ are $70, 50$ and $110 \ J \ K^{-1} \ mol^{-1}$ respectively. The temperature in Kelvin at which the reaction $\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5$ with $\Delta H = -35 \ kJ \ mol^{-1}$ will be at equilibrium is . . . . . . (Nearest integer).

For the vaporization of water at $1 \, \text{atm}$ pressure,the liquid water and water vapor are in equilibrium. What is the temperature in $K$?
$H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ [at $1 \, \text{atm}$ pressure] [$\Delta S = 120 \, J K^{-1}$ and $\Delta H = +45.0 \, kJ$]