How to Calculate the Concentrations of NO2 and N2O4 at Equilibrium?

[Mathman23]

Hi all,

I have the following equilibrium reaction $\mathrm{2NO_{2} \leftrightharpoons N_{2} O_{4}}$ which has a tempeture of
$\mathrm{t = 100 \ ^\circ C.}$
The reactions takes place in a canister with a volume of 0,50 Liters, where the pressure at equilibrium is 1,6 bar.

Futhermore the equilibrium constant is found to be $\mathrm{K_{c} = 5,0 M^{-1}}$

a/ First I write the mass action expression $K_{c} = \frac{\mathrm{[N_{2}O_{4}]}}{\mathrm{[NO_{2}]^2}}$

b/ Second I calculate the amount of mol substance then the reaction is in equilibrium:

$\mathrm{n_{equ}} = \frac{\mathrm{1,6bar \cdot 0,5 L}}{\mathrm{0,0831 \frac{bar \cdot L}{mol \cdot K}\cdot \ {373,0 K}}}= 2,6 \times 10^{-2} \ \mathrm{mol}$

c/ How do I calculate $[\mathrm{NO_{2}}] \ \mathrm{and} \ [\mathrm{N_{2} O_{10}}]$ ??

Many Thanks in advance.

Sincerely
Fred
Denmark

 

[Zefram]

Hi Fred,

Great question! To calculate the concentrations of NO2 and N2O4 at equilibrium, we first need to set up an ICE table. This stands for Initial, Change, and Equilibrium.

For the initial values, we know that the initial pressure of NO2 is 1.6 bar, and the initial pressure of N2O4 is 0 bar (since it is not present initially). We also know that the volume is 0.50 L and the temperature is 100°C, which we can convert to Kelvin by adding 273.15. This gives us an initial concentration of NO2 of 0.02 M (using the ideal gas law PV = nRT).

In the change row, we can set up the change in pressure for each species. Since the reaction is 2NO2 ⇌ N2O4, the change in pressure for NO2 will be -2x, and the change in pressure for N2O4 will be +x (where x is the change in pressure at equilibrium).

In the equilibrium row, we can add the initial and change values to get the equilibrium pressures. This gives us a final pressure of NO2 of 1.6 - 2x and a final pressure of N2O4 of x.

Now, we can use the equilibrium constant expression (Kc = [N2O4]/[NO2]^2) and plug in our equilibrium pressures to solve for x. This will give us the equilibrium pressures for NO2 and N2O4, which we can then convert to concentrations using the ideal gas law.

I hope this helps! Let me know if you have any further questions.

 

[ravenprp]

Hello Fred,

To calculate the concentrations of NO2 and N2O4, we can use the ideal gas law and the given equilibrium constant. The ideal gas law states that PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Rearranging this equation, we can solve for n: n=PV/RT. Plugging in the given values, we get n= (1.6 bar)(0.5 L)/(0.0831 bar L/mol K)(373 K) = 0.0129 mol. This is the total number of moles of gas at equilibrium, which is equal to the sum of the moles of NO2 and N2O4.

Next, we can use the equilibrium constant expression that you have correctly written (Kc=[N2O4]/[NO2]^2) to solve for the concentrations of NO2 and N2O4. Plugging in the values for Kc and the total number of moles (0.0129 mol), we get 5.0 M^-1 = x/(0.0129-x)^2, where x is the concentration of NO2. Solving for x, we get x= 1.08 M for [NO2]. Then, we can use the given volume of the canister (0.50 L) to calculate the concentration of N2O4: [N2O4]= n/V = 0.0129 mol/0.50 L = 0.0258 M.

I hope this helps clarify how to calculate the concentrations of NO2 and N2O4 at equilibrium. Let me know if you have any further questions. Good luck with your studies!