Question

Calculate ΔG_rxn at 298 K under the conditions shown below for the following reaction.

3 O₂(g) → 2 O₃(g) ΔG° = +326 kJ

P(O₂) = 0.41 atm, P(O₃) = 5.2 atm

Answer 1

sGoxo_at 298 Kunder the condition ssbowa below for the following reaction 30259)+2 03(5)__ AGC = +326 KJ_ Plaz): 0.41 atm P(O₃) = 5.2 atm (Gibbs free → for the calculation of A6rxn energy) for the given reaction the formula reuired is - AG - AGE+ RT in[@] AG= standared free energy (KJ /mol) = 326 kJ R E gas constant (8314_I/K-mol) I: Temperature (Kelvin) = 298 K a = Reaction auotient a Products r Reactant? _given w sGº: +326 KT convert Kilo I to Joule +326 KJ = +326 000 I 2) R = 8.314 3) I = 298 K ist we have to at different Quotient calculate pressure reaction (atm) L P= Presoure a= [Product ] [Pproduct = [03] [reactant ] . [Reactant] [02]} = [5.2]2 Scanned with [0.413 -CamScanner Tasli
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= 27.04 0:0689 [Q] =_392-333 Putting _ all this values in AG xxn formula, we get - - Agroxn = 46° + $I_dn [] = 3,26,000+(8:314 x 298 x_ln [39 2:333]) 326000 +(8-314 X 298 x5:93) = 326000 t14788 Agron = 340,788 J u m ...(convert J to KJ) A Gaxa um Gore = 340 KJ This is the DGxxn = 340.788 I or 340K I for the given reaction. 3 Scared with CamScanner

∆G = ∆G° + RT ln [Q] this is the formula for the calculation of ∆G

∆G° = 326 kJ , R = 8.314 , T = 298 k

[Q] = 392.33 ( reaction quotient )

P(O3)2 = 5.2 atm , P(O2)3 = 0.41 atm

By this we calculated ,∆G = 3,40,788 kJ / 340kJ for the following reaction