Calculate $\Delta S_{total}$ for the following reaction at $300 \ K$.
$NH_4NO_{3(s)} \longrightarrow NH_4^+{(aq)} + NO_3^-{(aq)}$ $(\Delta H = 28.1 \ kJ \ mol^{-1}, \Delta S_{sys} = 108.7 \ J \ K^{-1} \ mol^{-1})$

An ideal gas undergoes a reversible isothermal expansion from state $I$ to state $II$ followed by a reversible adiabatic expansion from state $II$ to state $III$. The correct plot$(s)$ representing the changes from state $I$ to state $III$ is(are)
($p$ : pressure,$V$ : volume,$T$ : temperature,$H$ : enthalpy,$S$ : entropy)

For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to the final state $Z$. The final state $Z$ can be reached by either of the two paths shown in the figure. Which of the following choice$(s)$ is (are) correct? [take $\Delta S$ as change in entropy and $w$ as work done].
$(A)$ $\Delta S_{X \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z}$
$(B)$ $w_{X \to Z} = w_{X \to Y} + w_{Y \to Z}$
$(C)$ $w_{X \to Y \to Z} = w_{X \to Y} + w_{Y \to Z}$
$(D)$ $\Delta S_{X \to Y \to Z} = \Delta S_{X \to Y}$

For the reaction $Br_{2(l)} + Cl_{2(g)} \rightarrow 2BrCl_{(g)}$,the enthalpy change is $\Delta H = 30 \ kJ/mol$ and the entropy change is $\Delta S = 105 \ J/mol \cdot K$. The temperature at equilibrium is $...... \ K$.

$2.4 \ g$ coal is burnt in a bomb calorimeter in excess of oxygen at $298 \ K$ and $1 \ atm$ pressure. The temperature of the calorimeter rises from $298 \ K$ to $300 \ K$. The enthalpy change during the combustion of coal is $-x \ kJ \ mol^{-1}$. The value of $x$ is. (Nearest Integer) (Given: Heat capacity of bomb calorimeter $20.0 \ kJ \ K^{-1}$. Assume coal to be pure carbon)

For the reaction $X_2O_{4(g)} \to 2XO_{2(g)}$,given $\Delta U = 2.1 \, kcal$ and $\Delta S = 20 \, cal \, K^{-1}$ at $300 \, K$. Calculate $\Delta G$ in $kcal$.