Cesium chloride is formed according to the following equation :

$\mathrm{Cs}\left(\mathrm{s}\right)+0.5{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to \mathrm{CsCl}\left(\mathrm{s}\right)$.

The enthalpy of sublimation of $\mathrm{Cs}$, enthalpy of dissociation of chloride, ionization energy of $\mathrm{Cs}$ & electron affinity of chlorine are $81.2, 243.0, 375,7$ and $-348.3 \mathrm{kJ} {\mathrm{mol}}^{-1}$. The energy change involved in the formation of $\mathrm{CsCl}$ is $-388.6 \mathrm{kJ} {\mathrm{mol}}^{-1}$. Calculate the lattice energy of $\mathrm{CsCl}$. (in $\mathrm{kJ} {\mathrm{mol}}^{-1}$) enolved.

$∆{\mathrm{H}}_{\mathrm{fussion}}\left(\mathrm{ice}\right)=1.436 \mathrm{kcal}/\mathrm{mole}$

A sample of the sugar D-ribose$\left({\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5\right)$ of mass $0.727 \mathrm{g}$ was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature was raised by $0.910 \mathrm{K}$. In a separate experiment in the same calorimeter, the combustion of $0.825 \mathrm{g}$ of benzoic acid, for which the internal energy of combustion is $-3251 \mathrm{kJ} {\mathrm{mol}}^{-1}$, gave a temperature rise of $1.940 \mathrm{K}$. Calculate the internal energy of combustion of D-ribose in $\mathrm{kJ} {\mathrm{mol}}^{-1}$. Give answer in 2 decimal places.

Using the following values, calculate the standard enthalpy change of the above reaction in $\mathrm{kJ} {\mathrm{mol}}^{-1}$.

${∆}_{\mathrm{f}}{\mathrm{H}}^{°}\left({\mathrm{H}}_2\mathrm{O}, \mathcal{l}\right)=-285.85 \mathrm{kJ}·{\mathrm{mol}}^{-1}$

${∆}_{\mathrm{f}}{\mathrm{H}}^{°}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6,\mathrm{aq}\right)=-1263.1 \mathrm{kJ}·{\mathrm{mol}}^{-1}$

${∆}_{\mathrm{f}}{\mathrm{H}}^{°}\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11},\mathrm{aq}\right)=-2238.3 \mathrm{kJ}·{\mathrm{mol}}^{-1}$

Calculate work done in adiabatic compression of one mole of an ideal gas (monoatomic) from an initial pressure of $1\mathrm{atm}$ to final pressure of $2\mathrm{atm}$. Initial temperature $=300\mathrm{K}$

(a) If process is carried out reversibly

(b) If process is carried out irreversible against $2\mathrm{atm}$ external pressure.

Compute the final volume reached by gas in two cases.