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In: Chemistry

A steel reaction vessel at 25 oC initially contained pure NO2 at a concentration of 2.0...

A steel reaction vessel at 25 oC initially contained pure NO2 at a concentration of 2.0 x 10-4 mol/L. Some of the NO2 reacted to form N2O4 as described by the following equilibrum equation:

2NO2(g) <----------> N2O4(g)

At equilibrium, the concentration of N2O4 was found to be 6.01 x 10-6 mol/L. What is the concentration-based equilibrium constant (Kc) for this reaction at 25 oC?

I know the correct answer is 170, but I'd like to know how to get there. Thanks!

Expert Solution

2NO₂(g) N₂O₄(g)

initial conc(mol/L) 2.0x10^-4 0

change -2a +a

Equb conc 2.0x10^-4 - 2a a

Given Equilibrium concentration of N₂O₄ = a = 6.01 x 10^-6 mol/L

So Equilibrium concentration of NO₂ is = 2.0x10^-4 - 2a

= 2.0x10^-4 - (2x6.01 x 10^-6)

= 1.88x10 ^-4 mol/L

Equilibrium constant ,

queen_honey_blossom answered 1 day ago

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