Enthalpy of sublimation of iodine is $24 \ cal \ g^{-1}$ at $200 \ ^oC$. If specific heat of $I_{2(s)}$ and $I_{2(vap)}$ are $0.055$ and $0.031 \ cal \ g^{-1} K^{-1}$ respectively,then enthalpy of sublimation of iodine at $250 \ ^oC$ in $cal \ g^{-1}$ is

Consider the following statements:
$(A)$ Entropy of a perfect crystalline solid at absolute zero approaches zero.
$(B)$ For spontaneity of a reaction at constant temperature and pressure, $T\Delta S > \Delta H$ (where $\Delta H$ is positive).
Identify the correct answer from the options given below.

The molar heat of formation of $NH_4NO_{3(s)}$ is $-367.54 \ kJ \ mol^{-1}$ and those of $N_2O_{(g)}$ and $H_2O_{(l)}$ are $+81.46 \ kJ \ mol^{-1}$ and $-285.78 \ kJ \ mol^{-1}$ respectively at $25 \ ^oC$ and $1.0 \ atm$ pressure. Calculate $\Delta U$ at $25 \ ^oC$ for the reaction:
$NH_4NO_{3(s)} \rightarrow N_2O_{(g)} + 2H_2O_{(l)}$ (in $kJ$)

Calculate $\Delta H^{\circ}$ for the reaction, $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$, given the following reactions:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2}H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$

The standard enthalpies of formation of $1,3-butadiene(g)$,$CO_{2(g)}$,and $H_2O_{(l)}$ at $298 \ K$ are $-30$,$-94$,and $-68 \ kcal/mol$ respectively. If the magnitude of resonance enthalpies of $1,3-butadiene$ and $CO_2$ are $10$ and $20 \ kcal/mol$ respectively,the enthalpy of combustion of $1,3-butadiene(g)$ at $298 \ K$ is $........ \ kcal/mol$. (Enthalpy of vaporization of $H_2O_{(l)}$ at $298 \ K = 10 \ kcal/mol$)

Expansion of $1$ $mol$ of an ideal gas takes place from $2$ $L$ to $8$ $L$ at $300$ $K$ against a constant external pressure of $1$ $atm$. Calculate $\Delta S_{total}$ in $J$ $K^{-1}$ $mol^{-1}$.
(Given: $R = 8.3$ $J$ $K^{-1}$ $mol^{-1}$,$1$ $L$ $atm = 100$ $J$,$\ln 2 = 0.693$) (in $.5$)