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

The entropy of the reaction is found using the Gibbs free energy equation G = H - TS. By rearranging the equation to solve for S and substituting the given values, the entropy is found to be 160.3 J/(mol. K).

Step-by-step explanation:

To find the entropy (ΔS) of the reaction in which calcium carbonate decomposes into calcium oxide and carbon dioxide, we can use the Gibbs free energy equation:

G = H - TS

where G is the Gibbs free energy, H is the enthalpy, T is the temperature in Kelvin, and S is the entropy. Since we know the values of G and H, and the temperature (T) is given as 25.0°C, we can solve for S after converting the temperature to Kelvin by adding 273.15 to the Celsius temperature.

The equation rearranges to:

S = (H - G) / T

We have:

• H = 178.3 kJ/mol
• G = 130.5 kJ/mol
• T = 25.0 °C + 273.15 = 298.15 K

Now, substitute the given values into the equation:

S = (178.3 kJ/mol - 130.5 kJ/mol) / 298.15 K

S = (47.8 kJ/mol) / 298.15 K

S = 0.1603 kJ/K·mol

Converting kJ to J (1 kJ = 1000 J):

S = 160.3 J/(K·mol)

Therefore, the correct answer is:

C. 160.3 J/(mol·K)

Answer 2

C.160.3J/mol

Step-by-step 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=(47.8)/(298)=0.1603 KJ/mol`

`S=0.1603* 1000=160.3J/mol` (1KJ=1000J)

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