At 2010K, the equilibrium constant K_c for the following reaction is 4.0 times 1

Question

At 2010K, the equilibrium constant K_c for the following reaction is 4.0 times 10^-4 N_2(g) + O_2(g) e 2 NO(g) If the equilibrium concentrations opf N_2 and O_2 are 0.05 mol/L and 0.6 mpl/L, respectively, at 2010k, what is the equilibrium concentration of NO? At 2010K, the equilibrium constant K_c for the following reaction is 4.0 times 10^-4 N_2(g) + O_2(g) e 2 NO(g) If initial concentrations of N_2 and O_2 are 0.3 mpl/L and o.4 mol/L, respectively, at 2010K, what is the equilibrium concentration of NO? One tenth (0.1) mol N_2O_4(g) is placed in a 1,00 L flask at a given temperature. After reaching equilibrium, the concentration of NO_2(g) is 0.04M. What is K_c for the reaction below? N_2O_4(g) e 2NO_2(g)

Explanation / Answer

3) k=[NO]^2 / [N2][O2]

4*10^-4 *0.05*0.6 = [NO]^2

[NO]=0.00346

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4) N2    +   O2 ->   2NO
0.3-x 0.4-x 2x

k=(2x)^2 / (0.3-x)(0.4-x) =4*10^-4

10000x^2 = 0.12-0.3x-0.4x + x^2

9999x^2 +0.7x-0.12=0

-> x =0.00343

2x = 0.00686 mol/L

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5) N2O4 -> 2NO2

0.1-x 2x

2x = 0.04
x =0.02

k=[NO2]^2/[N2O4] = (0.04)^2 / 0.08
=0.02

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