Respuesta :
Answer: The balanced chemical equation is written below and [tex]\Delta S^o[/tex] for the reaction is -160.6 J/K
Explanation:
When calcium hydroxide reacts with sulfur dioxide, it leads to the formation of calcium sulfate and water molecule.
The chemical equation for the reaction of calcium hydroxide and sulfur dioxide follows:
[tex]Ca(OH)_2(s)+SO_2(g)\rightarrow CaSO_3(s)+H_2O(l)[/tex]
To calculate the entropy change of the reaction, we use the equation:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}][/tex]
For the given reaction:
[tex]\Delta S^o_{rxn}=[(1\times \Delta S^o_{CaSO_3(s)})+(1\times \Delta S^o_{H_2O(l)})]-[(1\times \Delta S^o_{Ca(OH)_2(s)})+(1\times \Delta S^o_{SO_2(g)})][/tex]
Taking the standard entropy change values:
[tex]\Delta S^o_{CaSO_3(s)}=101.4Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O(l)}=69.9Jmol^{-1}K^{-1}\\\Delta S^o_{Ca(OH)_2(s)}=83.4Jmol^{-1}K^{-1}\\\Delta S^o_{SO_2(g)}=248.5Jmol^{-1}K^{-1}[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(1\times (101.4))+(1\times (69.9))]-[(1\times (83.4))+(1\times (248.5))]\\\\\Delta S^o_{rxn}=-160.6J/K[/tex]
Hence, the balanced chemical equation is written above and [tex]\Delta S^o[/tex] for the reaction is -160.6 J/K
