For chemical reactions,the calculation of change in entropy is normally done:

The enthalpy change for the transition of liquid water to steam at $100\,^oC$ is $40.8\, kJ\, mol^{-1}$. The entropy change for this process is .... $J\, K^{-1}\, mol^{-1}$.

Enthalpy of vaporization of benzene is $+35.3 \ kJ \ mol^{-1}$ at its boiling point,$80^{\circ} C$. The entropy change in the transition of the vapour to liquid at its boiling point in $[J \ K^{-1} \ mol^{-1}]$ is

One mole of an ideal diatomic gas $(C_V = 5 \ cal)$ was transformed from initial $25 \ ^{\circ}C$ and $1 \ L$ to the state when temperature is $100 \ ^{\circ}C$ and volume $10 \ L$. The entropy change of the process can be expressed as $(R = 2 \ cal / mol \cdot K)$ :-

The entropy values (in $J \, K^{-1} \, mol^{-1}$) of $H_{2(g)} = 130.6$,$Cl_{2(g)} = 223.0$,and $HCl_{(g)} = 186.7$ at $298 \, K$ and $1 \, atm$ pressure. Then entropy change for the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$ is

Given the standard entropy values at $298 \ K$ and $1 \ atm$ as $H_2(g) : 130.6 \ J \ K^{-1} \ mol^{-1}$,$Cl_2(g) : 223.0 \ J \ K^{-1} \ mol^{-1}$,and $HCl(g) : 186.7 \ J \ K^{-1} \ mol^{-1}$,calculate the entropy change $(\Delta S^{\circ})$ in $J \ K^{-1} \ mol^{-1}$ for the reaction: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$