At $298 \mathrm{~K}, K_p$ for reaction $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$ is 0.98 . Predict whether the reaction is spontaneous or not.
For the reaction, $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g), K_p=0.98$
As we know that $\quad \Delta_r G^{\mathrm{s}}=-2.303 R \mathrm{log} K_p$
Here, $K_p=0.98$ i.e., $K_p<1$ therefore, $\Delta_r G^{\circ}$ is positive, hence the reaction is non-spontaneous.
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in figure. What will be the value of $\Delta H$ for the cycle as a whole?
The net enthalpy change, $\Delta H$ for a cyclic process is zero as enthalpy change is a state function, i.e., $\Delta H$ (cycle) $=0$
The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$ is $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Will the standard molar entropy of $\mathrm{H}_2 \mathrm{O}(s)$ be more, or less than $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ?
The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$ is $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The solid form of $\mathrm{H}_2 \mathrm{O}$ is ice. In ice, molecules of $\mathrm{H}_2 \mathrm{O}$ are less random than in liquid water.
Thus, molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})<$ molar entropy of $\mathrm{H}_2 \mathrm{O}(l)$. The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})$ is less than $70 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
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.