Question & Answer: For a gaseous reaction, standard conditions are 298 K and a partial pressure…..

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction the standard change in Gibbs free energy is Δ3。--69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are PNO-0.450 atm, Po,ー0.500 atm, and PNo, 0.850 atm Number k.J/ mol

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2NO(g) + O_2(g) rlhar 2NO_2(g) the standard change in Gibbs free energy is delta G degree = -69.0 kJ/mol. What is delta G for this reaction at 298 K when the partial pressures are P_NO = 0.450 atm, P_O_2 = 0.500 atm, and P_NO_2, = 0.850 atm delta G =

Expert Answer

Apply full dG equation

dG = dG º + RT*ln(Q)

Q = P-NO2^2 /(P-NO)^2 * (p-O2)

Q = (0.85^2)/((0.45^2)(0.5)) = 7.1358

now, dGº = -69 kJ = -69000

substitute all data

dG = dG º + RT*ln(Q)

dG = -69000 +(8.314*298)*ln(7.1358)

dG = -64,131.26 J/mol

dG = -64.231 kJ/mol

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