The reaction N2O4 - 2 NO2 in equilibrium has the following equilibrium constants

Question

The reaction N2O4 - 2 NO2 in equilibrium has the following equilibrium constants at the given temperatures: T= 350 K Keq = 4.07. T= 500 K Keq = 2.04. Assuming delta H at standard state is constant over this temperature range, calculate the change in reaction entropy, delta S at standard state, at 400 K.

2. (20 points) The reaction N2O4 2 NO2 in equilibrium has the following equilibrium constants at the given temperatures: Kc,-4.07 Keq = 2.04 T = 350 K T= 500 K Assuming Horn is constant over this temperature range, calculate the change in reaction entropy, AS n, at 400 K.

Explanation / Answer

According to van't Hoff's equation,

lnK2/K1 = (deltaH0r / R) [ 1/T1) - (1/T2)]

ln(4.07/2.04) = (deltaH0rxn / 0.008314)*[(1/350) - (1/500)]

deltaH0rxn = 6.70 kJ

deltaS0rxn = deltaH0rxn / T = 6.70 / 400 = 0.0167 kJ/K = 16.7 J/K

NOTE: I think the reaction is endothermic, and hence K value should be high at high temperature.

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