For the process
[tex]\[2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g),\][/tex]
[tex]\[\Delta S = -187.9 \, \text{J/K}\][/tex]
and
[tex]\[\Delta H = -198.4 \, \text{kJ}\][/tex]
at 297.0 K are known.

What is the entropy change of this reaction?

A. [tex]\(-386.3 \, \text{J/K}\)[/tex]
B. [tex]\(-198.4 \, \text{kJ/K}\)[/tex]
C. [tex]\(-187.9 \, \text{J/K}\)[/tex]
D. [tex]\(-139.3 \, \text{kJ}\)[/tex]



Answer :

Sure! Let's solve the problem step-by-step.

We are given:
- Entropy change (ΔS) = -187.9 J/K
- Enthalpy change (ΔH) = -198.4 kJ
- Temperature (T) = 297.0 K

The question we need to answer involves determining the entropy (ΔS) of the reaction, and it mentions the Gibbs Free Energy (ΔG) equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]

The options provided may initially seem confusing, but they include different possible values for entropy and some are irrelevant to the formula we need to solve for Gibbs Free Energy (ΔG).

Since we are asked for the entropy (ΔS), we actually already have the value given directly in the problem:

[tex]\[ \Delta S = -187.9 \, \text{J/K} \][/tex]

Therefore, we do not need to perform additional calculations to find ΔS because it has been provided for us.

So, the correct answer is:

[tex]\[ \Delta S = -187.9 \, \text{J/K} \][/tex]

Therefore, the entropy change for the reaction is:

[tex]\[ -187.9 \, \text{J/K} \][/tex]