Chemistry Problem: N2O4 (g) <==> 2 NO2 (g) - Calculate Keq

[afcwestwarrior]

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.


right I am doing A, so if u can help, throw a formula at me, i will be working on this prolem and i will post my work within these 20 minutes

 

[nate808]

a. To calculate Keq, we can use the equation Keq = [NO2]^2/[N2O4].
Given that at equilibrium, there is 0.1 mole of NO2 present and 1.00 mole of N2O4 was initially placed in a 5.0 liter container, we can plug in these values into the equation to get Keq = (0.1)^2/(1.00) = 0.01.

b. The dissociation of N2O4 is endothermic. This is because when the container is placed in an ice bath, the temperature decreases and the equilibrium shifts towards the reactants (N2O4) to maintain a constant temperature. This results in the colorless gas (N2O4) being favored over the reddish-brown gas (NO2).
To calculate the heat of reaction, we can use the equation ΔH = -RTln(Keq). Plugging in the values, we get ΔH = -(8.314 J/mol*K)(298 K)ln(0.01) = 11.4 kJ/mol.

c. NO2 is formed in Denver during the winter because the lower temperatures promote the dissociation of N2O4 into NO2. In the summer, the higher temperatures favor the formation of N2O4 and thus, less NO2 is formed.

d. At sea level, the temperatures are generally higher than in Denver. This means that the equilibrium will shift towards the formation of N2O4 rather than NO2. Therefore, the brown cloud is less likely to form at sea level compared to Denver.

e. I expect NO2 to be more reactive. This is because it is a single bonded molecule, making it more unstable and prone to reacting with other molecules. N2O4, on the other hand, has a double bond which is more stable and less reactive.

 

[Blueshift5]

A) To calculate Keq, we can use the formula Keq = [products]/[reactants], where the concentrations of the products and reactants are at equilibrium. In this case, the products are 2 moles of NO2 and the reactant is 1 mole of N2O4. Therefore, Keq = [0.1]^2 / [1] = 0.01.

B) When the gas is placed in an ice bath, the colorless N2O4 is favored over the reddish-brown NO2. This means that the reaction is endothermic, as energy is required to break the bonds in N2O4 and produce NO2. The heat of reaction can be calculated using the formula ΔH = -RTln(Keq), where R is the gas constant and T is the temperature in Kelvin. Plugging in the values, we get ΔH = -(8.314 J/mol*K)(298 K)ln(0.01) = 107.2 kJ/mol. This confirms that the reaction is endothermic.

C) In the winter, the temperature in Denver is cold, which means that the equilibrium constant (Keq) for the dissociation of N2O4 to NO2 is lower. This is because the reaction is endothermic, so at lower temperatures, less NO2 is formed. In the summer, the higher temperatures favor the formation of NO2, so it is not formed as much.

D) At sea level, the temperature is warmer than in Denver, so the equilibrium constant for the dissociation of N2O4 to NO2 would be higher. This means that more NO2 would be formed, so it is possible that the brown cloud could still form.

E) I would expect NO2 to be more reactive than N2O4. This is because NO2 has an unpaired electron, making it a free radical and more reactive. N2O4, on the other hand, does not have an unpaired electron and is more stable.