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.

The entropy of this reaction (ΔS) = - 187.9 J/K

Explanation:

. Step : Data given

ΔS = - 187.9 J/K

ΔH = - 198.4 kJ

T = 297.0 K

Step 2: The balanced equation

2SO2 (g) + O2 (g) → 2SO3 (g)

Step 3: Calculate the entropy

ΔG = ΔH - TΔS

⇒ΔG = the gibbs free energy

⇒ΔH = the change in enthalpy of the reaction = - 198.4 kJ

⇒with T = the temperature = 297.0 K

⇒with ΔS = the change of entropy = - 187.9 J/K

ΔG = - 198400 kJ - 297 * (-187.9kJ/mol)

ΔG = - 198400+55806.3

ΔG = - 142593.7 kJ = the gibbs free energy

The entropy of this reaction (ΔS) = - 187.9 J/K