Chemistry - Thermodynamics - II Question with Solution | TestHub

$\Delta {\mathrm{H}}_{300}-\Delta {\mathrm{H}}_{200}=(-10)(100)$ Joule

$$\begin{array}{rl}\Rightarrow & \Delta {\mathrm{H}}_{300}=2200-1000=+21000Joule\\ \Rightarrow & \Delta {\mathrm{H}}_{300}=+21\mathrm{kJ}/\mathrm{mol}\\ & \Delta {\mathrm{S}}_{300}=\frac{21000}{300}=70\mathrm{J}/\mathrm{K}-\mathrm{mol}\\ & \Delta {\mathrm{S}}_{300}-\Delta {\mathrm{S}}_{200}=(-10)\ln (3/2)\\ \Rightarrow & \Delta {\mathrm{S}}_{200}=70+10\ln (3/2)=74.05\mathrm{J}/\mathrm{K}\end{array}$$

Steps for Marking:

$$\begin{array}{rl}& \Delta {\mathrm{H}}_{vap,300\mathrm{K}}=21,000Joule\\ & \Delta {\mathrm{S}}_{vap,300\mathrm{K}}=\frac{21,000}{300}=+70\mathrm{J}/\mathrm{K}mole\\ & \Delta {\mathrm{S}}_{200}=\Delta {\mathrm{S}}_{300}+10\ln \frac{300}{200}\\ & \Delta {\mathrm{S}}_{vapourisation,200\mathrm{K},1atm}=74.05\mathrm{J}/\mathrm{K}mole\end{array}$$