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.
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
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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