Question

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.

Answer 1

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

Step-by-step 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°
`_{system` = ∑ΔS°( product ) - ∑ΔS°( reactant )

ΔS°
`_{system` = [ ΔS°(CaO) + ΔS°( CO₂ ) ] - ΔS°( CaCO₃ )

we substitute

ΔS°
`_{system` = [ 39.8 J/mol.K + 213.7 J/mol.K ] - 92.9 J/mol.K

ΔS°
`_{system` = 160.6 J/mol.K

i.e, for 1 mol CaCO₃, ΔS°
`_{system` = 160.6 J/mol.K

Now, for 1.58 mol CaCO₃,

ΔS°
`_{system` = 1.58 mol × 160.6 J/mol.K

ΔS°
`_{system` = 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