Q5.3-63E - Chemistry (OpenStax)

Question

Calculate \({\bf{\Delta {\rm H}}}_{{\bf{298K}}}^{\bf{^\circ }}\) for the process

\({\bf{Sb(s)  + }}\frac{{\bf{5}}}{{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{SbC}}{{\bf{l}}_{\bf{5}}}\left( {\bf{g}} \right)\)

from the following information:

\({\bf{Sb(s)  + }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{SbC}}{{\bf{l}}_{\bf{3}}}\left( {\bf{g}} \right){\bf{                 \Delta {\rm H}}}_{{\bf{298K}}}^{\bf{^\circ }}{\bf{ =   - 314kJ}}\)

\({\bf{SbC}}{{\bf{l}}_{\bf{3}}}\left( {\bf{g}} \right){\bf{ + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g) }} \to {\bf{ SbC}}{{\bf{l}}_{\bf{5}}}\left( {\bf{g}} \right){\bf{               \Delta {\rm H}}}_{{\bf{298K}}}^{\bf{^\circ }}{\bf{ =   - 80kJ}}\)

Step-by-Step Solution

Answer

For the reaction \({\rm{Sb(s)  + }}\frac{5}{2}C{l_2}(g) \to {\rm{SbC}}{{\rm{l}}_{\rm{5}}}\left( {\rm{g}} \right)\), the value of change in enthalpy is equal to -394 kJ. 

Step 1: Hess’s law

To solve the problem, we are going to use Hess’s law. To do so, we will simply perform mathematical operations on both the given equations to simplify them in the form of the required one. 

Hess’s law: The total enthalpy changes tends to become equal when a reaction occurs by multiple paths and the first and last conditions are equal.

Step 2: Enthalpy change

For solving the question using Hess’s law, follow the below-mentioned steps. 

\({\bf{Sb(s)  + }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{SbC}}{{\bf{l}}_{\bf{3}}}\left( {\bf{g}} \right){\bf{                 \Delta {\rm H}}}_{{\bf{298K}}}^{\bf{^\circ }}{\bf{ =   - 314kJ}}\)

\({\bf{SbC}}{{\bf{l}}_{\bf{3}}}\left( {\bf{g}} \right){\bf{ + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g) }} \to {\bf{ SbC}}{{\bf{l}}_{\bf{5}}}\left( {\bf{g}} \right){\bf{               \Delta {\rm H}}}_{{\bf{298K}}}^{\bf{^\circ }}{\bf{ =   - 80kJ}}\)

multiplying equation (1) by 1 and equation (2) by 1 and then adding them,

\({\bf{Sb(s)  + }}\frac{{\bf{5}}}{{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{SbC}}{{\bf{l}}_{\bf{5}}}\left( {\bf{g}} \right)\)                        = -394kJ 

Hence, for the reaction \({\rm{Sb(s)   + }}\frac{5}{2}C{l_2}(g) \to {\rm{SbC}}{{\rm{l}}_{\rm{5}}}\left( {\rm{g}} \right)\), the value of change in enthalpy equal to -394 kJ.