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4. The equilibrium constant (KP) for the following reaction is 0.113 at 298 K, which 2 corresponds to a standard free-energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2 = 0.122 atm and PN2O4 = 0.453 atm. Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction at equilibrium, take the gas constant (R) value as 8.314 J/K.mol. N2O4(g)  2NO2(g)

Respuesta :

Answer:

[tex]\Delta G[/tex] for the reaction is -3.059 kJ/mol.

Explanation:

Reaction equilibrium: [tex]N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)[/tex]

Reaction quotient (Q) for this reaction at [tex]P_{NO_{2}}=0.122atm[/tex] and [tex]P_{N_{2}O_{4}}=0.453atm[/tex] is:          

                                [tex]Q=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}[/tex]  

                                    = [tex]\frac{(0.122)^{2}}{0.453}[/tex]

                                    = 0.0329

We know, [tex]\Delta G=\Delta G^{0}+RTlnQ[/tex]   , where [tex]\Delta G^{0}[/tex] is standard free energy change, R is gas constant and T is temperature in kelvin scale.

Here, [tex]\Delta G^{0}[/tex] = 5.40 kJ/mol

So, [tex]\Delta G=(5.40\times 10^{3}J/mol)+[8.314J.mol^{-1}.K^{-1}\times 298K\times ln(0.0329)][/tex]

             = -3059 J/mol

             = -3.059 kJ/mol