Asked by Amy
For the reaction N2O4(g)=2NO2(g), the value of K at 25 degrees XCelsius is 7.19*10^-3. Calculate N2O4 at equilibrium when NO2 =2.20 *10^122 mol/L?
Answers
Answered by
DrBob222
I guess that's Kc? If NO2 at equilibrium is 2.20E122 M, then
K = (NO2)^2/(N2O4)
Substitute and solve for N2O4.
K = (NO2)^2/(N2O4)
Substitute and solve for N2O4.
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