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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)

C160.3 J/(mol.K)

D1,912 J/(mol.K)

Verified Answer :

lublana

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

dsdrajlin

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).

Learn more about entropy here: https://brainly.com/question/6364271

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