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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.

Sagot :

[tex]\rm -187.9 \;J/K[/tex]entropy of the reaction.

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.

Learn more about entropy here:

https://brainly.com/question/4617981

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