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 $\Delta E^o$ of combustion of $2-$methylpropene is $-X \ kJ \ mol^{-1}$. The value of $\Delta H^o$ is

$A$ gas is reversibly expanded from the same initial state to the same final volume using isobaric,isothermal,and adiabatic processes. The correct order of the work done by the system on the surroundings in the three different methods is

For a pure substance,if $T_B$ is the melting point and $T_A$ is the freezing point,which graph correctly represents the relationship between the change in entropy $(\Delta S)$ and temperature $(T)$?

$2 \ mol$ of $Hg_{(g)}$ is combusted in a fixed volume bomb calorimeter with excess of $O_2$ at $298 \ K$ and $1 \ atm$ into $HgO_{(s)}$. During the reaction,temperature increases from $298.0 \ K$ to $312.8 \ K$. If heat capacity of the bomb calorimeter and enthalpy of formation of $Hg_{(g)}$ are $20.00 \ kJ \ K^{-1}$ and $61.32 \ kJ \ mol^{-1}$ at $298 \ K$,respectively,the calculated standard molar enthalpy of formation of $HgO_{(s)}$ at $298 \ K$ is $X \ kJ \ mol^{-1}$. The value of $|X|$ is. . . . . [Given : Gas constant $R = 8.3 \ J \ K^{-1} \ mol^{-1}$]

Three moles of an ideal gas are expanded isothermally from a volume of $300 \ cm^3$ to $2.5 \ L$ at $300 \ K$ against a constant external pressure of $1.9 \ atm$. The work done in joules is: