For the reaction N2O4(g)=2NO2(g), the value of K at 25 degrees XCelsius is 7.19*10^-3. Calculate N2O4 at equilibrium when NO2 =2.20 *10^122 mol/L?
I guess that's Kc? If NO2 at equilibrium is 2.20E122 M, then
K = (NO2)^2/(N2O4)
Substitute and solve for N2O4.
To solve this problem, we will use the equilibrium constant expression and set up an ICE (Initial, Change, and Equilibrium) table.
The balanced equation for the reaction is: N2O4(g) ⇌ 2NO2(g)
The equilibrium constant expression for this reaction is given by:
Kc = [NO2]^2 / [N2O4]
We are given the equilibrium constant value (Kc = 7.19 x 10^-3) and the concentration of NO2 (2.20 x 10^122 mol/L). We need to calculate the concentration of N2O4 at equilibrium.
Let's set up the ICE table:
N2O4(g) ⇌ 2NO2(g)
Initially: x 2.20 x 10^122
Change: -x +2x
Equilibrium: x 2.20 x 10^122 + 2x
Now, we substitute the equilibrium concentrations into the equilibrium constant expression:
Kc = [NO2]^2 / [N2O4]
7.19 x 10^-3 = (2.20 x 10^122 + 2x)^2 / x
Next, we solve for x algebraically. Simplifying the equation gives:
7.19 x 10^-3 = (4.84 x 10^244 + 8.8 x 10^122 x + 4x^2) / x
Multiply both sides by x to eliminate the denominator:
7.19 x 10^-3 x = 4.84 x 10^244 + 8.8 x 10^122 x + 4x^2
Rearrange the equation to obtain a quadratic equation:
4x^2 + (8.8 x 10^122 - 7.19 x 10^-3) x + 4.84 x 10^244 = 0
Finally, solve the quadratic equation using the quadratic formula or any other appropriate method to find the value(s) of x, which represents the concentration of N2O4 at equilibrium.