Question

The decomposition of calcium carbonate, CaCO3(s) --> CaO(s) + CO2(g), has the following values for free energy and enthalpy at 25.0°C.

G = 130.5 kJ/mol
H = 178.3 kJ/mol

What is the entropy of the reaction? Use G = H – TS.

A. -160.3 J/(mol.K)

B. -47.8 J/(mol.K)

C. 160.3 J/(mol.K)

D. 1,912 J/(mol.K)

Answer 1

Answer

C.160.3J/mol

Explanation:

We are given that

G=$130.5KJ/mol$

$H=178.3KJ/mol$

T=$25^{\circ}=25+273=298k$

We have to find the entropy of the reaction.

$G=H-TS$

Substitute the values in the formula

$130.5=178.3-298S$

$298S=178.3-130.5=47.8$

$S=\frac{47.8}{298}=0.1603 KJ/mol$

$S=0.1603\times 1000=160.3J/mol$  (1KJ=1000J)

Hence, the entropy of the reaction=S=160.3J/mol

Answer 2

The entropy for the decomposition of calcium carbonate is -160.3 J/(mol.K) (Option A).

What is entropy?

Entropy is a thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Let's consider the decomposition of calcium carbonate.

CaCO₃(s) ⇒ CaO(s) + CO₂(g)

Then, we will convert 25.0 °C to Kelvin using the following expression.

K = °C + 273.15 = 25.0 + 273.15 = 298.2 K

Finally, we will calculate the entropy of the reaction using the following expression.

G = H - T × S

S = G - H / T

S = 130.5 kJ/mol - 178.3 kJ/mol / 298.2 K = -160.3 J/(mol.K)

where,

• G is the free energy.
• H is the enthalpy.
• T is the absolute temperature.
• S is the entropy.

The entropy for the decomposition of calcium carbonate is -160.3 J/(mol.K) (Option A).

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