Q5.3-64E - Chemistry (OpenStax)

Question

Calculate\({\bf{\Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}\) for the process Zn(s) + S(s) + 2O₂(g)⟶ZnSO₄(s)

from the following information:

\({\bf{Zn}}\left( {\bf{s}} \right){\bf{  +   S}}\left( {\bf{s}} \right) \to {\bf{ZnS}}\left( {\bf{s}} \right){\bf{  \Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ =   - 206}}{\bf{.0 kJ}}\)

\({\bf{ZnS}}\left( {\bf{s}} \right){\bf{  +   2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{ZnS}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right){\bf{      \Delta H}}_{{\bf{298}}}^{\bf{^\circ }}{\bf{  =   - 776}}{\bf{.8 kJ}}\)

Step-by-Step Solution

Answer

The change in enthalpy for the following reaction,\({\rm{Zn(s)  + }}{\rm{ S(s)  +   2}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{ZnS}}{{\rm{O}}_{\rm{4}}}\left( {\rm{s}} \right)\) is -982.8 kJ. 

Step 1: Hess’s law

To solve the problem, we have to follow Hess’s law and go step-by-step to find out the change in enthalpy for the given reaction. It can be illustrated as:

Step 2: Calculation of enthalpy

The reactions are:

\({\bf{Zn}}\left( {\bf{s}} \right){\bf{  +   S}}\left( {\bf{s}} \right) \to {\bf{ZnS}}\left( {\bf{s}} \right){\bf{  \Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ =   - 206}}{\bf{.0 kJ}}\) 

\({\rm{ZnS}}\left( {\rm{s}} \right){\rm{  +   2}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{ZnS}}{{\rm{O}}_{\rm{4}}}\left( {\rm{s}} \right){\rm{      \Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{  =  }} - {\rm{776}}{\rm{.8 kJ}}\)

 

From the above two reactions, if we want to find the change in enthalpy of the given reaction, we will simply perform the mathematical operation like,

\({\bf{Zn}}\left( {\bf{s}} \right){\bf{  +   S}}\left( {\bf{s}} \right) \to {\bf{ZnS}}\left( {\bf{s}} \right){\bf{  \Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ =   - 206}}{\bf{.0 kJ}}\) 

\({\rm{ZnS}}\left( {\rm{s}} \right){\rm{  +   2}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{ZnS}}{{\rm{O}}_{\rm{4}}}\left( {\rm{s}} \right){\rm{      \Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{  =  }} - {\rm{776}}{\rm{.8 kJ}}\)

multiplying equation (1) by 1 and equation (2) by 1 and adding them we will get,

 Zn(s) + S(s) + 2O₂(g)⟶ZnSO₄(s)                =-982.8kJ

Thus, the change in enthalpy for the following reaction\({\bf{Zn(s)  + S(s)   +  2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{ZnS}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right)\) is -982.8 kJ.