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Consider the reaction 2NO2(g)→N2O4(g). ΔG∘ =-5.40 kJ Calculate ΔG at 298 K if the partial pressures of NO2?
IMO this is a bit garbled (I’m not blaming you for that!).
I think is is how it’s supposed to be solved.
2NO2(g) → N2O4(g)
Kp = P(products)/P(reactants) = (P(N2O4))/((P(NO2))^2)
Kp = (1.64 atm)/((0.38 atm)^2) = 11.36 atm^-1
Next we need to convert Kp to Kc:
Kp = Kc(RT)^Δngas
Kc = Kp/((RT)^Δngas)
Δngas = Mole of gas (product) – Mole of Gas (Reactant) = 1 – 2 = -1
R = 0.08206 L atm K^-1 mol^-1
T = 298 K
Kc = Kp/((RT)^Δngas) = (11.36 atm^-1)/(((0.08206 L atm K^-1 mol^-1)*(298 K))^-1)
Kc = 277.8 L mol-1
Now we use this to calculate ∆G.
∆G = -RT(lnKeq)
R = 0.008314 kJ mol^-1 K^-1
T = 298 K
Keq = Kc = 277.8
∆G = -RT(lnKeq)
∆G = -(0.008314 kJ mol^-1 K^-1)*(298 K)*(ln(277.8))
∆G = 13.94 kJ mol^-1
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Calculate the enthalpy of the reaction: 2NO(g) + O2(g – Consider the reaction 2NO2(g)→N2O4(g). ΔG∘ =-5.40 kJ Calculate ΔG at 298 K if the partial pressures of NO2? Calculate ΔG at 298 K if the partial pressures of NO2 and N2O4 are 0.38atm and 1.64atm , respectively. I solved for the pressure as 1.64-.38^2. then R= 8.314, T is 298 .Calorimetric studies show that the reaction is exothermic. 2NO2(g) N2O4(g) + 14.1 kcal. Based on this information, which one–if any–of the following additional changes would increase the molar concentration at equilibrium of N2O4(g)? decrease the pressure increase the temperature decrease the concentration of NO2(g) stir the reaction mixture2NO2 (g) Equilibrium N2O4 (g)NO2 and N2O4 undergo the reaction shown. When a sealed container of NO2 reaches chemical equilibrium, which must be true?f The maximum number of molecules has been reached.g No N2O4 is present.h The rates of the forward and reverse reactions are equal.j No chemical
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