Question

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 298 K, \[{\Delta _f}{H^0}(Sn{O_2}(s))\] =-581.0 kJ \[{\rm{mo}}{{\rm{l}}^{ - 1}}\] , \[{\Delta _f}{H^0}(C{O_2}(g))\]=-394 kJ \[{\rm{mo}}{{\rm{l}}^{ - 1}}\]
\[{S^0}({\rm{Sn}}{{\rm{O}}_{\rm{2}}}(s)) = 56.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}},{S^0}({\rm{Sn}}(s)) = 52.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\]
\[{S^0}({\rm{C}}(s)) = 6.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}},{S^0}({\rm{C}}{{\rm{O}}_2}(g)) = 210.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\]
Assume that the enthalpies and the entropies are temperature independent.

Answer 1

Hint: Here, we have to use the Gibbs free energy equation to calculate the temperature at which the reduction takes place. For that, we have to calculate the change of enthalpy and entropy first. Then, using the equation of free energy we can calculate the temperature.

Complete Step by Step Solution:
The reduction reaction of cassiterite is,
\[{\rm{Sn}}{{\rm{O}}_{\rm{2}}}({\rm{s}}) + {\rm{C(s)}} \to {\rm{Sn(s)}} + {\rm{C}}{{\rm{O}}_{\rm{2}}}({\rm{g}})\]

Let's first calculate the change in enthalpy (\[\Delta H\] ). We know that, \[\Delta H\]is the difference between the enthalpy of formation of reactants and products.
So, \[\Delta H = \Delta {H_f}{\rm{(Product}}) - \Delta {H_f}{\rm{(Reactant}})\]

Given, \[{\Delta _f}{H^0}(Sn{O_2}(s))\] =-581.0 kJ \[{\rm{mo}}{{\rm{l}}^{ - 1}}\] , \[{\Delta _f}{H^0}(C{O_2}(g))\]=-394 kJ \[{\rm{mo}}{{\rm{l}}^{ - 1}}\]
So, \[\Delta H = - 394\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} - ( - 581){\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} = 187\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} = 187000\,{\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]

Now, we have to find out the entropy change. The entropy change is also the difference between the entropy of product and reactant.
So, \[\Delta S = {S_p} - {S_r}\].

Given values are,
\[{S^0}({\rm{Sn}}{{\rm{O}}_{\rm{2}}}(s)) = 56.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}},{S^0}({\rm{Sn}}(s)) = 52.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\]
\[{S^0}({\rm{C}}(s)) = 6.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}},{S^0}({\rm{C}}{{\rm{O}}_2}(g)) = 210.0\,{\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\]

The above values are to be used to calculate entropy change.

\[\Delta S = \left( {52 + 210} \right) - \left( {56 + 6} \right)\]
\[\Delta S = 262 - 62\]
\[\Delta S = 200\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]

Now, we have to use the free energy formula , \[\Delta G = \Delta H - T\Delta S\] ,Here, \[\Delta G\] stands for free energy change, \[\Delta H\] is for change in enthalpy, T is for temperature and \[\Delta S\] stands for entropy change.
So, \[\Delta G = \Delta H - T\Delta S\]
\[0 = 187000 - T \times 200\]
\[T = \dfrac{{187000}}{{200}}\]
\[T = 935\,{\rm{K}}\]
Therefore, at 935 K, the reduction takes place.

Note: It is to be noted that Gibbs free energy tells us the maximum work done in a system keeping the pressure and temperature constant. It is symbolically represented by the letter G. At equilibrium, the \[\Delta G\] value is zero.