Calculate the standard entropy change for the following reaction at 25 C. Mg(OH)_2 (s) +2HCl(g) –> MgCl_2 (s) + 2H_2O (g)

Question

Question:

Calculate the standard entropy change for the following reaction at 25 C. Mg(OH)_2 (s) +2HCl(g) –> MgCl_2 (s) + 2H_2O (g)

Entropy of Reaction
With the standard adjusted response under certain condition, the entropy of the response is understood utilizing
ΔS∘rxn=∑nΔS∘products−∑nΔS∘reactants, where n is the quantity of moles.
The S∘product and S∘reactants at 25∘C are as of now known qualities.

Explanation:
Consider the accompanying standard entropy for every reactant and item with the synthetic response underneath.
Mg(OH)2(s)+2HCl(g)→MgCl2(s)+2H2O(g)
where
• S∘ of Mg (OH)2(s)=63.18 J/mol⋅K
• S∘ of HCl(g)=186.77 J/mol⋅K
• S∘ of MgCl2(s)=89.62 J/mol⋅K
• S∘ of H2O(g)=188.7 J/mol⋅K
Presently consider illuminating the difference in standard entropy utilizing
ΔS∘rxn=∑nΔS∘products−∑nΔS∘reactants
Substituting every single given worth. Consequently,
ΔS∘rxn=((1 mole)(89.62 Jmol⋅K)+(2 mole)(188.7 Jmol⋅K))−((1 mole)(63.18 Jmol⋅K)+(2 mole)(186.77 Jmol⋅K))
ΔS∘rxn=30.3 J/K

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1. Entropy of Reaction
With the standard adjusted response under certain condition, the entropy of the response is understood utilizing
ΔS∘rxn=∑nΔS∘products−∑nΔS∘reactants, where n is the quantity of moles.
The S∘product and S∘reactants at 25∘C are as of now known qualities.
Consider the accompanying standard entropy for every reactant and item with the synthetic response underneath.
Mg(OH)2(s)+2HCl(g)→MgCl2(s)+2H2O(g)
where
• S∘ of Mg (OH)2(s)=63.18 J/mol⋅K
• S∘ of HCl(g)=186.77 J/mol⋅K
• S∘ of MgCl2(s)=89.62 J/mol⋅K
• S∘ of H2O(g)=188.7 J/mol⋅K
Presently consider illuminating the difference in standard entropy utilizing
ΔS∘rxn=∑nΔS∘products−∑nΔS∘reactants
Substituting every single given worth. Consequently,
ΔS∘rxn=((1 mole)(89.62 Jmol⋅K)+(2 mole)(188.7 Jmol⋅K))−((1 mole)(63.18 Jmol⋅K)+(2 mole)(186.77 Jmol⋅K))
ΔS∘rxn=30.3 J/K