Answer:
∆G = -326,413.514 J/mol or -326.4 KJ/mol
Step-by-step explanation:
The enthalpy of formation of oxygen is zero because the enthalpy of formation of an element in its most stable form = zero.
∆Hrxn = n∆Hf(products) - m∆Hf(reactants)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation, ∆Hf = standard enthalpy of formation, ∆Hrxn= standard enthalpy of reaction.
Using the information from the data above:
∆Hrxn = [ (3(∆Hf(O2)) – [2(∆Hf(O3)]
∆Hrxn = [3(0)] – [2(+142.7)]
∆Hrxn = 0 – [+285.4]
∆Hrxn = -285.4 KJ
∆Hrxn = - 285,400 J ( 1 KJ= 1000 J )
Then,
∆Srxn = n∆S(products) - m∆S(reactants)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation, ∆S = standard entropy of product or reactant, ∆Srxn= standard entropy of reaction.
Using the values above:
∆Srxn = [ 3(∆S(O2)) – (2(∆S(O3))]
∆Srxn = [ 3(+205.14) ]- [2(+238.93)]
∆Srxn = [+615.42] – [+477.86]
∆Srxn = +615.42 – 477.86
∆Srxn = + 137.56 J/molK
Using,
∆G = ∆H - T∆S
where ∆G = Gibbs free energy change = ?
∆H = enthalpy change = -285,400 J/mol
∆S = entropy change = +137.56 J/molK
T = Temperature in Kelvin = 25 + 273.15 = 298.15 K
∆G = -285,400 – [ 298.15 ( +137.56 ) ]
∆G = -285,400 – 41,013.514
∆G = -326,413.514 J/mol
∆G = -326.4 KJ/mol
The negative value of ∆G means that the reaction is spontaneous at the given temperature.