Dinitrogen tetraoxide decomposes to give nitrogen dioxide. Calculate the equilibrium concentration of NO2 at 100oC if the equilibrium constant is 0.200 and [N2O4] = 0.800M.
N2O4(g) ¡ê 2 NO2(g)
You have some strange symbols which I assume was supposed to be an arrow. I've corrected it. Do you mean [N2O4] = 0.800M at equilibrium? or initially? I'll assume at equilibrium.
...........N2O4(g) ==> 2NO2(g)
Kc = (NO2)^2/(N2O4
0.200 = (NO2)^2/(0.800.
Solve for (NO2).
To calculate the equilibrium concentration of NO2 at 100oC given the equilibrium constant (K) and the initial concentration of N2O4, we'll use the expression for the equilibrium constant and set up an ICE table.
The equilibrium constant expression is given by: K = [NO2]^2 / [N2O4]
Let's define:
[N2O4] = initial concentration of N2O4 = 0.800 M
[NO2] = equilibrium concentration of NO2 (to be determined)
Setting up an ICE table, we have:
N2O4(g) ¡ê 2 NO2(g)
____________________________________
Initial: 0.800 M 0 M
Change: -x +2x
Equilibrium: 0.800 - x 2x
Using the equilibrium concentrations in the expression for K, we have:
K = [NO2]^2 / [N2O4]
0.200 = (2x)^2 / (0.800 - x)
Now we can solve for x:
0.200 = 4x^2 / (0.800 - x)
Rearranging the equation:
0.200 * (0.800 - x) = 4x^2
0.16 - 0.2x = 4x^2
Rewriting the equation:
4x^2 + 0.2x - 0.16 = 0
Now we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = 4, b = 0.2, and c = -0.16. Plugging these values into the quadratic formula:
x = (-0.2 ± √(0.2^2 - 4(4)(-0.16))) / (2(4))
Simplifying:
x = (-0.2 ± √(0.04 + 2.56)) / 8
x = (-0.2 ± √2.60) / 8
x ≈ (-0.2 ± 1.6125) / 8
We're only interested in the positive root, so:
x ≈ (1.4125) / 8
x ≈ 0.1766
Therefore, the equilibrium concentration of NO2 at 100oC is approximately 0.1766 M.
To calculate the equilibrium concentration of NO2 at 100oC, we can use the equilibrium constant expression:
Kc = [NO2]^2 / [N2O4]
Given that the equilibrium constant (Kc) is 0.200 and the concentration of N2O4 ([N2O4]) is 0.800M, we can plug these values into the equation and solve for [NO2]:
0.200 = [NO2]^2 / 0.800
Rearranging the equation to solve for [NO2]:
[NO2]^2 = 0.200 * 0.800
[NO2]^2 = 0.160
[NO2] = √0.160
Taking the square root of 0.160:
[NO2] ≈ 0.400M
Therefore, the equilibrium concentration of NO2 at 100oC would be approximately 0.400M.