Question

Calculate the standard entropy change for the reaction: 2 hydrogen sulfide(g) plus sulfur dioxide (g) goes to form 3 sulfur(s) plus 2 water(g). Given S(s) = 31.88 J/mol. Look up the other values in the text appendix. No units required. Just the numerical answer

Answer 1

Answer: The standard entropy change of the given reaction is -186.51

Step-by-step explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

`\Delta S^o_(rxn)=\sum [n* \Delta S^o_((product))]-\sum [n* \Delta S^o_((reactant))]`

For the given chemical reaction:

`2H_2S(g)+SO_2(g)\rightarrow 3S(s)+2H_2O(g)`

The equation for the entropy change of the above reaction is:

`\Delta S^o_(rxn)=[(3* \Delta S^o_((S(s))))+(2* \Delta S^o_((H_2O(g))))]-[(2* \Delta S^o_((H_2S(g))))+(1* \Delta S^o_((SO_2(g))))]`

We are given:

`\Delta S^o_((S(s)))=31.88J/K.mol\\\Delta S^o_((H_2O(g)))=188.825J/K.mol\\\Delta S^o_((H_2S))=205.79J/K.mol\\\Delta S^o_((SO_2))=248.22J/K.mol`

Putting values in above equation, we get:

`\Delta S^o_(rxn)=[(3* (31.88))+(2* (188.825))]-[(2* (205.79))+(1* (248.22))]\\\\\Delta S^o_(rxn)=-186.51J/K`

Hence, the standard entropy change of the given reaction is -186.51