The reaction NO_2 (g) + NO (g) doublesidearrow N_2O (g) + O_2 (g) reached equili

ID: 483779 • Letter: T

Question

The reaction NO_2 (g) + NO (g) doublesidearrow 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 NO_2 concentration had become 0.118 M. What is the value of Kc for the reaction at this temperature? Also determine the equilibrium concentrations of N_2O, O_2, and NO.

Explanation / Answer

[NO2]

[NO2] [NO]

[N2O]

[O2]

I 0.0560

0.294

0.184

0.377

-x

0.0560 + x

0.294 + x

0.184 – x

0.377 – x

[NO2] = 0.118 M = 0.0560 + x at equilibrium.

Solving

x = 0.062 M.

Using the ICE table ------------

that the equilibrium concentrations are;

[NO] = 0.356 M,

[N2O] = 0.122 M

And

[O2] = 0.315 M.

Kc = [N2O] [O2] /[N2O] [NO] = (0.122)(0.315) /(0.118)(0.356) =0.915-----ANSWER