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Lilit [14]
3 years ago
8

What is the value for ΔSºreaction for the following reaction, given the standard entropy values?

Chemistry
2 answers:
Alex3 years ago
8 0

<u>Answer:</u> The \Delta S^o of the reaction is 179Jmol^{-1}K^{-1}

<u>Explanation:</u>

Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.

Mathematically,

\Delta S_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}]

For the given chemical equation:

Al_2O_3(s)+3H_2(g)\rightarrow 2Al(s)+3H_2O(g)

We are given:

\Delta S^o_{Al_2O_3}=51Jmol^{-1}K^{-1}\\\Delta S^o_{H_2}=131Jmol^{-1}K^{-1}\\\Delta S^o_{Al}=28Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O}=189Jmol^{-1}K^{-1}

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{H_2O})+(3\times \Delta S^o_{H_2O})]-[(1\times \Delta S^o_{Al_2O_3})+(3\times \Delta S^o_{H_2})]

\Delta S^o=[(2\times 28)+(3\times 189)]-[(1\times 51)+(3\times 131)]=179Jmol^{-1}K^{-1}

Hence, the \Delta S^o of the reaction is 179Jmol^{-1}K^{-1}

Diano4ka-milaya [45]3 years ago
6 0
Substituting the values:

51 + 3(131) = ΔH + 2(28) + 3(189)
ΔH = -225 J/mol

When written outside of the equation, this becomes 225 J/mol
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