Chemistry Problem: Calculating Keq for N2O4 Gas Dissociation at 25oC

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Dinitrogen tetroxide (N2O4) dissociates into nitrogen dioxide (NO2), with the equilibrium expression Keq = [NO2]^2/[N2O4]. At 25°C, placing 1.00 mole of N2O4 in a 5.0-liter container results in an initial concentration of 0.2 M for N2O4 and 0.02 M for NO2 at equilibrium. The dissociation is confirmed to be endothermic, as cooling the gas causes it to become colorless, indicating a shift in equilibrium. In cold conditions like Denver's winter, more NO2 forms due to the temperature effect on equilibrium. At sea level, the brown cloud is less likely to form due to different atmospheric pressures affecting gas behavior.
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Dinitrogen tetroxide, N2O4, is a colorless gas that boils at 21oC. As a gas, it is extensively dissociated to NO2. As a liquid, it is partly dissociated to NO2. NO2 is a reddish-brown toxic gas that makes up part of the brown cloud in Denver during the winter months.
N2O4 (g) <==> 2 NO2 (g)
a. At 25oC, 1.00 mole N2O4 is placed in a 5.0 liter container. At equilibrium, the container has 0.1 mole NO2 present. Calculate a numerical value for Keq.
b. At 25oC, the gas inside the container is reddish-brown. When this container is placed in an ice bath, the gas is colorless. Is the dissociation of N2O4 exothermic or endothermic? Explain. Calculate the heat of reaction to confirm your answer.
c. In the high altitude of Denver, the winter is cold. Explain why NO2 is formed in Denver in the winter and not in the summer.
d. Assuming the same conditions as Denver except at sea level, would you expect the brown cloud to form? Give reasons.
e. (Extra credit: 1 point) Which structure do you expect to more reactive, N2O4 or NO2? Give reasons.


i need help on a, so if u can help me help me please, i will show you my work.
 
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so here's what i did Keq= [NO2]2/[N2O4]
 
so what do i do, do i figure out the pressure of those chemicals or do i figure out there concentrations
 
here's what i did, 1mol/ 5 liters = .2 M and .1 mole/ 5 liters = .02

then i took .02^2 NO2/ .2 M N2O4 =.002 is this right, or do have to too do something else
 
this is right but I am sure if I am suppost to use those numbers
 
The original concentratio of N2O4 is certainly 0.2M but some of it has dissociated to form the NO2. Let x=[NO2], the Keq becomes:

Keq = [x}^2/[[N2O4]-x]

substitute and determine the Keq.
 
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