Chemistry, 30.03.2020 21:06, annagwhiteou0hrh
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
Answers: 1
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]](/tpl/images/0571/2033/52737.png)
![\Delta S^o_{rxn}=[(3\times \Delta S^o_{(S(s))})+(2\times \Delta S^o_{(H_2O(g))})]-[(2\times \Delta S^o_{(H_2S(g))})+(1\times \Delta S^o_{(SO_2(g))})]](/tpl/images/0571/2033/492b6.png)
![\Delta S^o_{rxn}=[(3\times (31.88))+(2\times (188.825))]-[(2\times (205.79))+(1\times (248.22))]\\\\\Delta S^o_{rxn}=-186.51J/K](/tpl/images/0571/2033/7cc5d.png)
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Calculate the standard entropy change for the reaction: 2 hydrogen sulfide(g) plus sulfur dioxide (g...
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