When heated, N2O4 dissociates as follows: N2O4 =2NO2
A flask initially contians .560 M N2O4. at this temperature, Kc is 4.00. what is the equilibrium concentration of NO2?
Set up an ICE chart, substitute into Kc, and solve.
initially:
N2O4 = .560 M
NO2 = 0
change:
NO2 = +2x
N2O4 = -x
equilibrium:
N2O4 = 0.560-x
NO2 = 2x
Solve for x, then 2x will be (NO2). Post your work if you get stuck.
i ended up with x=.28 and i multiplied it by 2 and it was not the right answer
To find the equilibrium concentration of NO2, we will use the given initial concentration of N2O4 and the value of Kc.
Let's assume that at equilibrium, the concentration of N2O4 is x M. Since the reaction N2O4 = 2NO2, the concentration of NO2 at equilibrium will be 2x M.
The equilibrium constant Kc is defined as the ratio of the product concentrations to the reactant concentrations, each raised to their respective stoichiometric coefficients. In this case, Kc = [NO2]^2 / [N2O4].
Given that Kc = 4.00, we can set up the equation:
4.00 = (2x)^2 / x
Now, let's solve for x.
First, let's square both sides of the equation:
16 = 4x^2 / x
Next, let's multiply both sides by x to eliminate the fraction:
16x = 4x^2
Now, divide both sides by 4x:
4 = x
Therefore, the equilibrium concentration of N2O4 is 4 M, and the equilibrium concentration of NO2 is 2 * 4 M = 8 M.