Question

Use the thermodynamic data below to determine the standard free energy change, ∆G°, at 25° C for the reaction:2O3(g) 3O2(g)∆H°ffor O3(g) = +142.7 kJ/molS° for O3(g) = +238.93 J/mol.KS° for O2(g) = +205.14 J/mol.K g

Answer 1

∆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.