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.

Answer 1

	[NO2]	[NO2] [NO]	[N2O]	[O2]
I	I 0.0560	0.294	0.184	0.377
C	+x	+x	-x	-x
E	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