Consider the reaction: CaCO3(s)CaO(s) CO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions.

the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K

Explanation:

Given the data in the question;

CaCO₃(s)       →     CaO(s)     +     CO₂(g)  

1.58 moles        1.58 moles     1.58 moles

Since 1 mole of CaCO₃ gives 1 mole of CaO and 1 mole of CO₂

Thus, 1.58 mole of CaCO₃ gives 1.58 moles of CaO and 1.58 moles of CO₂.

Now,

At 298 K, standard entropy values are;

ΔS° ( CaCO₃ ) = 92.9 J/mol.K

ΔS° ( CaO )     = 39.8 J/mol.K

ΔS° ( CO₂ )      = 213.7 J/mol.K

So,

ΔS° = ∑ΔS°( product ) - ∑ΔS°( reactant )

ΔS° = [ ΔS°(CaO) + ΔS°( CO₂ ) ] - ΔS°( CaCO₃ )

we substitute

ΔS° = [ 39.8 J/mol.K + 213.7 J/mol.K ] - 92.9 J/mol.K

ΔS° = 160.6 J/mol.K

i.e, for 1 mol CaCO₃, ΔS° = 160.6 J/mol.K

Now, for 1.58 mol CaCO₃,

ΔS° = 1.58 mol × 160.6 J/mol.K

ΔS° = 253.748 J/K

Therefore, the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K