The reaction NO2(g)+ NO(g) <----> N2O(g)+O2(g) reached equilibrium at a certain

Question

The reaction

NO2(g)+ NO(g) <----> N2O(g)+O2(g)

reached equilibrium at a certain high temperature. Originally,the reaction vessel contained the following initial concentrations: 0.184M N2O, 0.377M O2, 0.0560M NO2 and 0.294M NO. The concentration of NO2, the only colored gas in the mixture, was monitored by following the intensity of the color. At equilibrium,the NO2 concentration had become 0.118M. What is the value of Kc for the reaction at this temperature? Also determine the equilibrium concentrations of N2O, O2, AND NO.

Explanation / Answer

By watching the concentration of NO2 we can see that

The concentration of NO2 is increased ---->It means Reverse reaction is favoured

NO2 + NO -----> N2O + O2

0.056 0.294 0.184 0.377

0.056+x 0.294+x 0.184-x 0.377-x

At equilibrium

[NO2] = 0.118M = 0.056+x

x = 0.062 M

[NO] = 0.294 + x = 0.356 M

[N2O] = 0.184-x = 0.122 M

[O2] = 0.377-x = 0.315 M

Kc = [N2O] [O2] / [NO] [NO2]

Kc = 0.9148

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