Calculate each of these equilibrium concentrations based on the reaction below.
2 NO(g) + O2(g) 2 NO2(g) K = 1.71 1012
(a) [NO] = 0.0048 M; [O2] = 0.000057 M; [NO2] = ?
(b) [NO] = 0.0026 M; [O2] = 0.000023 M; [NO2] = ?
K= [NO2]^2/([NO2]^2 * [O2])
conc NO2= sqrt(K*NO^2 * O2)
= sqrt(1.71E12*.0048^2*.000057)
= sqrt( 2250) moles/liter
check my work, it is easy to make a math error typo on a keyboard.
This is based on the very high forward reaction K you gave. I don't remember this reaction being anywhere close to this K value, if I guessed, I would guess 1E5 for K or thereabouts at room temp.
I just checked it, so based on that formula, the other solution should come out to be something like....
sqrt(1.71E12*0.0026^2*0.000023)
=265 moles?
To calculate the equilibrium concentration of [NO2] in both cases (a) and (b), we'll use the equation for the equilibrium constant (Kc) and the given initial concentrations. Here are the steps to solve each case:
(a) [NO] = 0.0048 M; [O2] = 0.000057 M; [NO2] = ?
1. Write down the balanced equation:
2 NO(g) + O2(g) -> 2 NO2(g)
2. Identify the equilibrium constant (Kc) value:
Kc = 1.71 x 10^12
3. Set up the expression for Kc using the concentrations:
Kc = [NO2]^2 / ([NO]^2 * [O2])
4. Substitute the given concentrations into the equation:
1.71 x 10^12 = [NO2]^2 / ([NO]^2 * [O2])
Note: [NO2] is unknown, so let's define it as "x" and rewrite the equation:
1.71 x 10^12 = x^2 / (0.0048^2 * 0.000057)
5. Solve for x by rearranging the equation and applying appropriate algebraic steps:
x^2 = (1.71 x 10^12) * (0.0048^2 * 0.000057)
x^2 = 2.40 x 10^5
x ≈ 490.0 (rounded to three significant figures)
Therefore, the equilibrium concentration of [NO2] in case (a) is approximately 490.0 M.
(b) [NO] = 0.0026 M; [O2] = 0.000023 M; [NO2] = ?
Follow the same steps as above, but use the given initial concentrations for [NO] and [O2]:
1. Write down the balanced equation:
2 NO(g) + O2(g) -> 2 NO2(g)
2. Identify the equilibrium constant (Kc) value:
Kc = 1.71 x 10^12
3. Set up the expression for Kc using the concentrations:
Kc = [NO2]^2 / ([NO]^2 * [O2])
4. Substitute the given concentrations into the equation:
1.71 x 10^12 = [NO2]^2 / ([NO]^2 * [O2])
Note: [NO2] is unknown, so let's define it as "x" and rewrite the equation:
1.71 x 10^12 = x^2 / (0.0026^2 * 0.000023)
5. Solve for x by rearranging the equation and applying appropriate algebraic steps:
x^2 = (1.71 x 10^12) * (0.0026^2 * 0.000023)
x^2 = 1.40 x 10^5
x ≈ 374.2 (rounded to three significant figures)
Therefore, the equilibrium concentration of [NO2] in case (b) is approximately 374.2 M.