For the reaction, X(s) $$\rightleftharpoons$$ Y(s) + Z(g), the plot of $$\ln {{pz} \over {{p^\theta }}}$$ versus $${{{{10}^4}} \over T}$$ is given below (in solid line), where p_z is the pressure (in bar) of the gas Z at temperature T and $${{p^\theta }}$$ = 1 bar.


(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K^$$-$$1 mol^$$-$$1)

The value of $$\Delta$$S^$$\theta$$ (in J K^$$-$$1 mol^$$-$$1) for the given reaction, at 1000 K is _________.

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.

At $$298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$$ mol^-1,

$$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$

Assume that, the enthalpies and the entropies are temperature independent.

The surface of copper gets tarnished by the formation of copper oxide. $${N_2}$$ gas was passed to prevent the oxide formation during heating of copper at $$1250$$ $$K.$$ However, the $${N_2}$$ gas contains $$1$$ mole % of water vapor as impurity. The water vapor oxidises copper as per the reaction given below : $$2Cu\left( s \right) + {H_2}O\left( g \right) \to C{u_2}O\left( s \right) + {H_2}\left( g \right)$$

$${P_{H2}}$$ is the minimum partial pressure of $${H_2}$$ (in bar) needed to prevent the oxidation at $$1250$$ $$K.$$ The value of $$\ln \left( {{P_{H2}}} \right)$$ is ________.

Given: total pressure $$=1$$ bar, $$R$$ (universal gas constant ) $$=$$ $$8J{K^{ - 1}}\,\,mo{l^{ - 1}},$$ $$\ln \left( {10} \right) = 2.3.\,$$ $$Cu(s)$$ and $$C{u_2}O\left( s \right)$$ are naturally immiscible.

At $$1250$$ $$K:2Cu(s)$$ $$ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to C{u_2}O\left( s \right);$$ $$\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}$$

$${H_2}\left( g \right) + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to {H_2}O\left( g \right);$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};$$ ($$G$$ is the Gibbs energy)