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Eduardwww [97]
3 years ago
9

Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical rea

ction:
P4O10 + 6H2O → 4H3PO4
Round your answer to zero decimal places.
Chemistry
2 answers:
Lapatulllka [165]3 years ago
8 0

<u>Answer:</u> The value of \Delta S^o for the reaction is -206 J/K

<u>Explanation:</u>

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:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

P_4H_{10}(s)+6H_2O(l)\rightarrow 4H_3PO_4(s)

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

\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)})]

We are given:

\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

Putting values in above equation, we get:

\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

Hence, the value of \Delta S^o for the reaction is -206 J/K

Aleksandr [31]3 years ago
5 0

Answer:

ΔS° = - 47.2 J/mol.K

Explanation:

  • P4O10 + 6H2O → 4H3PO4

ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10

∴ S°mH2O(l) = 69.9 J/mol.K

∴ S°mP4O10 = 231 J/mol.K

∴ S°mH3PO4 = 150.8 J/mol.K

⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231

⇒ ΔS° = - 47.2 J/mol.K

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<u>Explanation</u>:

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