For the process 2SO2(g) O2(g) Right arrow. 2SO3(g), Delta. S = –187. 9 J/K and Delta. H = –198. 4 kJ at 297. 0 K are known. What is the entropy of this reaction? Use Delta. G = Delta. H – TDelta. S. –386. 3 J/K –198. 4 kJ/K –187. 9 J/K –139. 3 kJ.

Entropy is the amount of heat or temperature inaccessible for the system to work and convert the chemical or thermal energy into mechanical energy.

How to calculate the entropy?

Given,

[tex]\rm \Delta S[/tex] = [tex]\rm -187.9 \;J/K[/tex]
[tex]\rm \Delta H[/tex] = -198.4 kJ
Temperature (T) = 297.0 K

The balanced reaction can be shown as,

[tex]\rm 2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex]

The Gibbs free energy can be calculated as:

[tex]\rm \Delta G = \Delta H - \rm T\Delta S[/tex]

Here,[tex]\rm \Delta G[/tex] = gibbs free energy, [tex]\rm \Delta H[/tex] = change in enthalpy, T = temperature and [tex]\rm \Delta S[/tex] = change in entropy.

Substituting values in the equation:

[tex]\begin{aligned}\rm \Delta G &= -198400 \;\rm kJ -297 \times (-187.9\;\rm kJ/mol)\\\\\rm \Delta G &= -198400 + 55806.3\\\\\rm \Delta G &= - 142593.7 \;\rm kJ\end{aligned}[/tex]

Therefore, [tex]\rm -187.9 \;J/K[/tex] is the entropy [tex](\rm \Delta S)[/tex] of the reaction.

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