At a given temperature, 0.0524 mol NO_2 (g) is placed in a 1.00 L flask. After r

Question

At a given temperature, 0.0524 mol NO_2 (g) is placed in a 1.00 L flask. After reaching equilibrium, the concentration of NO_2 (g) is 3.9 times 10^-3 M. What is K_c for the reaction below? 2 NO_2 (g) = N_2O_4 (g) A gaseous mixture of NO_2O_4 is in equilibrium. If the concentration of N_2O_4 is 5.3 times 10^5 M, what is the concentration of NO_2? 2 NO_2 (g) = N_2O_4 (g) K_c = 170 The equilibrium constant, K_c, for the following reaction is 1.0 Times 10^-3 at 1500 K. If 0.570 M N_2 and 0.570 M O_2 are allowed to equilibrate at 1500 K, what is the concentration of NO? N_2 (g) + O_2 (g) = 2 NO (g)

Explanation / Answer

0.0524 mol of NO2 in 1 L so

[NO2] = 0.0524

in equilbirium

[NO2] = 3.9*10^-3

find KC

Kc = [N2O4]/[NO2]^2

in equilibrium

[NO2] = 0.0524 - 2x

[N2O4] = 0+x

then

[NO2] = 0.0524 - 2x = 3.9*10^-3

x = (3.9*10^-3-0.0524)/-2 = 0.02425

[N2O4] =0.02425

then

Kc = (N2O4)/(NO2^2) = (0.02425)/( 3.9*10^-3)^2

Kc = 1594.345

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