The molar heat of fusion and vaporisation of benzene are 10.9 and $31.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The changes in entropy for the solid $\rightarrow$ liquid and liquid $\rightarrow$ vapour trasitions for benzene are $x$ and $y, \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. The value of $(y-x)$ (in $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ) is (At 1 atm benzene melts at $5.5^{\circ} \mathrm{C}$ and boils at $80^{\circ} \mathrm{C}$ )

Observe the following reaction,

$$ 2 A_2(g)+B_2(g) \xrightarrow{T(\mathrm{~K})} 2 A_2 B(g)+600 \mathrm{~kJ} $$

The standard enthalpy of formation $\left(\Delta_f H^{\ominus}\right)$ of $A_2 B(g)$ is

Identify the molecule for which the enthalpy of atomisation ( $\Delta_a H^{\ominus}$ ) and bond dissociation enthalpy $\left(\Delta_{\text {bond }} H^{\ominus}\right)$ are not equal

Given below are two statements :

Statement I For isothermal irreversible change of an ideal gas, $q=-w=p_{\text {ext }}\left(V_{\text {final }}-V_{\text {initial }}\right)$

Statement II For adiabatic change, $\Delta U=w_{\text {adiabatic }}$

The correct answer is :