The reaction NO_2 (g) + NO(g) N_2O (g) + O_2 (g) reached equilibrium at a certai

Question

The reaction NO_2 (g) + NO(g) N_2O (g) + O_2 (g) reached equilibrium at a certain high temperature. Originally, the reaction vessel contained the following initial concentrations: 0.184 M N_2O, 0.377 M O_2, 0.0560 M NO_2 and 0.294 M NO. The concentration of NO_2, the only colored gas in the mixture, was monitored by following the intensity of the color. At equilibrium, the NO2 concentration had become 0.118 M. What is the value of Kc for the reaction at tin's temperature? Also determine the equilibrium concentrations of N_2O, O_2, and NO.

Explanation / Answer

initially

[N2O] = 0.184

[O2] = 0.377

[NO2] = 0.056

[NO] = 0.294

K = [N2O][O2]/([NO2][NO])

in equilibrium:

[N2O] = 0.184+x

[O2] = 0.377+x

[NO2] = 0.056 -x

[NO] = 0.294-x

and we know that

[NO2] = 0.056 -x = 0.118

-0.118+0.056 = x

x = -0.062

so:

[N2O] = 0.184+-0.062 = 0.122

[O2] = 0.377+-0.062 = 0.315

[NO2] = 0.056 +0.062 = 0.118

[NO] = 0.294+0.062= 0.356

K = [N2O][O2]/([NO2][NO]) = 0.122* 0.315/ (0.118*0.356) = 0.91482

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