Calculate the equilibrium constant ($\mathrm{K}$) for the following reaction at $400 \mathrm{K}$.

$2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

Given that ${\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{H}}^{\mathrm{o}}=80.0 \mathrm{kJ} {\mathrm{mol}}^{-1}$and ${\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{S}}^{\mathrm{o}}=120 \mathrm{J} {\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$ at $400 \mathrm{K}$.

Use ${10}^{-4.178}=6.634\times {10}^{-5}$ 

Sodium carbonate $\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)$ can be obtained by heating sodium hydrogen carbonate, ${\mathrm{NaHCO}}_3$.

$2{\mathrm{NaHCO}}_3\left(\mathrm{s}\right)\to {\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Calculate the temperature above which ${\mathrm{NaHCO}}_3$ decomposes to form products at $1\mathrm{bar}.$

Given, enthalpy of formation values, $∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{kJmol}}^{-1}\right)$ :

${\mathrm{NaHCO}}_3\left(\mathrm{s}\right)=-947.7, {\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)=-1130.9, {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=-241.8, {\mathrm{CO}}_2\left(\mathrm{g}\right)=-393.5$

Entropy values, ${\mathrm{S}}^{\mathrm{o}}\left({\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right) : {\mathrm{NaHCO}}_3\left(\mathrm{s}\right)=102.1, {\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)=136.0, {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=188.8, {\mathrm{CO}}_2\left(\mathrm{g}\right)=213.7$.

$\left(\mathrm{i}\right) 2\mathrm{Zn}+{\mathrm{O}}_2\to 2\mathrm{ZnO}; ∆{\mathrm{G}}^{\circ }=-161 \mathrm{J}$

$\left(\mathrm{ii}\right) 2\mathrm{Zn}+2\mathrm{S}\to 2\mathrm{ZnS}; ∆{\mathrm{G}}^{\circ }=-293 \mathrm{J}$

$\left(\mathrm{iii}\right) 2\mathrm{S}+2{\mathrm{O}}_2\to 2{\mathrm{SO}}_2; ∆{\mathrm{G}}^{\circ }=-408 \mathrm{J}$

What would be $∆{\mathrm{G}}^{\circ }$ for the following reaction?

$2\mathrm{ZnS}+3{\mathrm{O}}_2\to 2\mathrm{ZnO}+2{\mathrm{SO}}_2$

The equilibrium constant for the reaction given below is $2.0\times {10}^{-7}$ at $300 \mathrm{K}$. Calculate the standard free energy change for the reaction:

${\mathrm{PCl}}_5 \left(\mathrm{g}\right)\rightleftharpoons {\mathrm{PCl}}_3 \left(\mathrm{g}\right)+{\mathrm{Cl}}_{2 }\left(\mathrm{g}\right)$

An ideal mono-atomic gas was at $1$ $\mathrm{atm}$ and $300 \mathrm{K}$. If one mole of such ideal gas is allowed to expand adiabatically against a constant pressure of $0.3$ $\mathrm{atm}$ till double its volume, determine $∆\mathrm{S}$ in ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$ for the process.

Given: $\log 2=0.3; \log 3=0.48; \log 5=0.7$

The value of ${\log }_{10} \mathrm{K}$ for a reaction $\mathrm{A}\rightleftharpoons \mathrm{B}$ is

(Given ${\mathrm{\Delta }}_{\mathrm{r}}{{\mathrm{H}}^{\mathrm{o}}}_{298 \mathrm{K}}=-54.07 \mathrm{kJ} {\mathrm{mol}}^{-1}, {\mathrm{\Delta }}_{\mathrm{r}}{{\mathrm{S}}^{\mathrm{o}}}_{298 \mathrm{K}}=10 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}$ and $\mathrm{R}=8.314 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}; 2.303\times 8.314\times 298=5705$)