Question

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

Answer 1

Answer: The $\Delta S^o$ of the reaction is $179Jmol^{-1}K^{-1}$

Explanation:

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 $179Jmol^{-1}K^{-1}$

Answer 2

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