Question

For the reaction: N2O4(g)⇌2NO2(g)

Kc= 0.513 at 500 K
Part A
If a reaction vessel initially contains an N2O4
concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K?
Express your answers in moles per liter to three significant figures separated by a comma.

Answer 1

Using an ICE table and solving the quadratic equation derived from the equilibrium expression for the reaction N2O4(g) ⇌ 2NO2(g), at 500 K with Kc=0.513, the equilibrium concentrations for N2O4 and NO2 are found to be approximately 0.0333 M each.

Step-by-step explanation:

For the reaction N2O4(g) ⇌ 2NO2(g) with the equilibrium constant Kc = 0.513 at 500 K, we can set up an ICE table to solve for the changes in concentration and determine the equilibrium concentrations. Initially, the concentration of N2O4 is 0.0500 M, and NO2 is not present.

• Initial: [N2O4] = 0.0500 M, [NO2] = 0 M

• Change: [N2O4] decreases by x, [NO2] increases by 2x because of the stoichiometry of the reaction.

• Equilibrium: [N2O4] = 0.0500 - x, [NO2] = 2x

The equilibrium expression is Kc = ([NO2]2)/[N2O4].

Plugging in the equilibrium concentrations gives

0.513 = (2x)2 / (0.0500 - x).

Solving this quadratic equation for x gives the change in concentration of N2O4 and allows us to calculate the equilibrium concentrations for N2O4 and NO2.

After solving, the equilibrium concentrations are approximately:

[N2O4] = 0.0333 M, [NO2] = 0.0333 M (noting that due to the 1:2 stoichiometry, the [NO2] is twice the amount of x).