No idea how to do the question need help 2NO2 (g) <-> N2O4 (g) What is the value

Question

No idea how to do the question need help

2NO2 (g) <-> N2O4 (g)

What is the value of G° for the reaction at 298 K if H° = -58.03 kJ and S° = -176.6 J/K? Assuming enthalpy and entropy are temperature independent, at what temperature is G°= 0? Is G° positive or negative above this temperature?

Predict the direction in which the reaction will shift to reach equilibrium at 25 °C for the following mixture:

pNO2 = 0.21 atm, pN2O4 = 0.50 atm Calculate the equilibrium constant for this reaction at 25 °C

Explanation / Answer

Recall that

dG = dH - T*dS

so

dH = -58.03 kJ

dS = -176.6 J/K so -0.1766 kJ/K

substitute dG knowing that T = 298K

dG = (-58.03) - 298*(-0.1766 ) = -5.4032 kJ

if we assume dH and dS are always constant

then

dG = 0

so

dH - T*dS = 0

dH = T*dS

T = dH/dS < 0

T = (-58.03)/(-0.1766)

T= 328.59569 K

the value of G must be always POSITIVE above this tempearture

Q = PN2O4 / (PNO2)^2

Q = (0.5)/(0.21^2) = 11.37

clearly, Q > 1, so this favours much more N2O4

b)

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(5.4032 /(8.134*298))

K = 1.002

then, the shift goes toward more NO2 production

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