For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3. Calculate [NO2] at equilibrium...

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For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3. Calculate [NO2] at equilibrium...

For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3.

Calculate [NO2] at equilibrium when [N2O4]=6.90×10−2mol/L.

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Answer 1

Answer #1

N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3.

[N2O4]=6.90×10−2mol/L.

K = [NO2]^2/[N2O4]

7.19*10^-3 = [NO2]^2/6.9*10^-2

[NO2]^2 = 7.19*10^-3 *6.9*10^-2

[NO2]^2 = 0.00049611

[NO2] = 0.0223M >>>>>answer

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Answer 2

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For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3. Calculate [NO2] at equilibrium...

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Add Answer to: For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3. Calculate [NO2] at equilibrium...

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For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3. Calculate [NO2] at equilibrium...