5 points T 14 CM 008 My The equilibrium reaction between din

-5 points T 14 CM 008 My The equilibrium reaction between dinitrogen tetraoxide, N 04(g), and nitrogen dioxide, NO2(g). can be written as the following. N204(g) 2 NO2(g) Initially. 0.32 mol N204 were introduced into a 0.77 L vessel at 21°C. At equilibrium, 0.22 mol N204 were present. what is the value of Ke for this reaction at this temperature? Supporting Materials Periodic Table l Supplemental Data constants and Factors Additional Materials

Solution

1)

Initial concentration of N2O4 = number of moles of N2O4 / volume

= 0.32 mol / 0.77 L

= 0.416 M

Equilibrium concentration of N2O4 = number of moles of N2O4 / volume

= 0.22 mol / 0.77 L

= 0.286 M

Let\'s prepare the ICE table

[N2O4] [NO2]

initial 0.416

change -1x +2x

equilibrium 0.416-1x +2x

To find x, we will use the equilibrium concentration provided in the question

[N2O4] = 0.286

0.416-1x = 0.286

x = 0.13

Equilibrium constant expression is

Kc = [NO2]^2/[N2O4]

Kc = (+2x)^2/(0.416-1x)

Kc = (+2*0.13)^2/(0.416-1*0.13)

Kc = 0.236

Answer: 0.236

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