Answer: The value of [tex]\Delta S^o[/tex] for the reaction is -206 J/K
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 of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]P_4H_{10}(s)+6H_2O(l)\rightarrow 4H_3PO_4(s)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(4\times \Delta S^o_{(H_3PO_4)})]-[(1\times \Delta S^o_{(P_4O_{10})})+(6\times \Delta S^o_{(H_2O)})][/tex]
We are given:
[tex]\Delta S^o_{(H_3PO_4)}=110.5J/K.mol\\\Delta S^o_{(H_2O(l))}=69.91J/K.mol\\\Delta S^o_{(P_4H_{10})}=228.86J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(4\times (110.5))]-[(1\times (228.86))+(6\times (69.91))]\\\\\Delta S^o_{rxn}=-206.32J/K=-206J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the reaction is -206 J/K
