If $92 \ g$ of $Na$ reacts with water in an open vessel at $300 \ K$, what is the value of work done (in $J$)? $[$Assume ideal nature of the gaseous product$]$

The heat of combustion of sucrose $(C_{12}H_{22}O_{11})$ is $1350 \ kcal \ mol^{-1}$. How much heat (in $kcal$) will be released by the combustion of $17.1 \ g$ of sucrose (in $.5$)?

Match the thermodynamic processes given under Column $I$ with the expression given under Column $II$:

Column $I$	Column $II$
$A$. Freezing of water at $273 \ K$ and $1 \ atm$	$P$. $q=0$
$B$. Expansion of $1 \ mol$ of an ideal gas into a vacuum under isolated conditions	$Q$. $w=0$
$C$. Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container	$R$. $\Delta S_{sys} < 0$
$D$. Reversible heating of $H_{2(g)}$ at $1 \ atm$ from $300 \ K$ to $600 \ K$,followed by reversible cooling to $300 \ K$ at $1 \ atm$	$S$. $\Delta U=0$
	$T$. $\Delta G=0$

Enthalpy of formation of $CO_{2(g)}$,$H_2O_{(l)}$ and $C_6H_{12}O_{6(s)}$ are $-393$,$-286$ and $-1170 \ kJ \ mol^{-1}$ respectively. The quantity of heat liberated when $18 \ g$ of $C_6H_{12}O_{6(s)}$ is burnt completely in oxygen is (in $kJ$)

Products are favoured in a chemical reaction taking place at a constant temperature and pressure. Consider the following statements: $(i)$ The change in Gibbs energy for the reaction is negative. $(ii)$ The total change in Gibbs energy for the reaction and the surroundings is negative. $(iii)$ The change in entropy for the reaction is positive. $(iv)$ The total change in entropy for the reaction and the surroundings is positive. The statements which are always true are:

$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore: