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

14 days is the half-life of phosphorus-32