Identify the state functions and path functions out of the following: enthalpy, entropy, heat, temperature, work, free energy.
State functions are those values which depend only on the state of the system and not on how it is reached e.g., enthalpy, entropy, temperature and free energy. Path functions are those values which depend on the path of the system. e. $g$, heat and work.
The molar enthalpy of vaporisation of acetone is less than that of water. Why?
Amount of heat required to vaporise one mole of a liquid at constant temperature and under standard pressure ( 1 bar ) is called its molar enthalpy of vaporisation $\Delta_{\text {vap }} H^{\mathrm{s}}$. Molar enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in $\mathrm{H}_2 \mathrm{O}$ molecule.
Which quantity out of $\Delta_r G$ and $\Delta_r G^{\text {S }}$ will be zero at equilibrium?
Gibbs energy for a reaction in which all reactants and products are in standard state. $\Delta_r G^{\mathrm{s}}$ is related to the equilibrium constant of the reaction as follows
$$\begin{aligned} \Delta_r G & =\Delta_r G^{\mathrm{s}}+R T \ln K \\ \text{At equilibrium,}\quad 0 & =\Delta_r G^{\circ}+R T \ln K \quad \left(\because \Delta_r G=0\right)\\ \text{or}\quad \Delta_r G^{\mathrm{s}} & =-R T \ln K \\ \Delta_r G^{\mathrm{s}} & =0 \text { when } K=1 \end{aligned}$$
For all other values of $K, \Delta_r G^{\mathrm{s}}$ will be non-zero.
Predict the change in internal energy for an isolated system at constant volume.
For isolated system there is no transfer of energy as heat, i.e., $q=0$ and there is no transfer of energy as work. i.e., $\mathrm{W}=0$. According to the first law of thermodynamics
$$\begin{aligned} \Delta U & =q+W \\ \Delta U & =0+0=0 \end{aligned}$$
Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
The two conditions under which heat becomes independent of path are
(i) when volume remains constant
(ii) when pressure remains constant
Explanation
(i) At constant volume By first law of thermodynamics, $\Delta U=q+W$ or $q=\Delta U-W$. But $W=-p \Delta V$ Hence, $q=\Delta U+p \Delta V$. But as volume remains constant $\Delta V=0$
$\therefore q_v=\Delta V$ but $\Delta U$ is a state function.
Hence, $q_v$ is a state function.
(ii) At constant pressure As we know, $q_p=\Delta U+p \Delta V$. But $\Delta U+p \Delta V=\Delta H$.
$\therefore q_p=\Delta H$. As $\Delta H$ is a state function therefore, $q_p$ is a state function.