Question

Using the table provided below, what is the entropy change if 4.5 g of CaCO3(s) is placed in a container and allowed to decompose to CaO(s) and CO2(g) according to the following reaction?

CaCO3(s) ⇔ CaO(s) + CO2(g)

Substance	S° (J/mol × K)
CaCO3(s)	92.88
CaO(s)	39.75
CO2(g)	213.6

in J/K

Note: Report your final answer to the correct number of significant figures and include the sign (+/-).

Answer 1

Given:

Sof(CaCO3(s)) = 92.88 J/mol.K

Sof(CaO(s)) = 39.75 J/mol.K

Sof(CO2(g)) = 213.6 J/mol.K

Balanced chemical equation is:

CaCO3(s) ---> CaO(s) + CO2(g)

ΔSo rxn = 1*Sof(CaO(s)) + 1*Sof(CO2(g)) - 1*Sof( CaCO3(s))

ΔSo rxn = 1*(39.75) + 1*(213.6) - 1*(92.88)

ΔSo rxn = 160.47 J/mol.K

This is when 1 mol of CaCO3 reacts

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass(CaCO3)= 4.5 g

use:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(4.5 g)/(1.001*10^2 g/mol)

= 4.496*10^-2 mol

Now use:

delta So = ΔSo rxn * number of mol

= (160.47 J/mol.K)*(4.496*10^-2 mol)

= 7.21 J/K

Answer: 7.21 J/K