Asked by sstt
In an experiment, a flask with 3.00 atm of dinitrogen tetroxide gas, N2O4, and 2.00 atm of nitrogen dioxide, NO2, gas at 25oC is allowed to reach equilibrium. The pressure of the nitrogen dioxide decreased by 0.96 atm. Calculate the value of Kp for this system.
N2O4(g) Û 2NO2(g)
N2O4(g) Û 2NO2(g)
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Answered by
GPT-5 mini
Initial pressures: P(N2O4)0 = 3.00 atm, P(NO2)0 = 2.00 atm.
NO2 decreases by 0.96 atm → P(NO2)eq = 2.00 − 0.96 = 1.04 atm.
By stoichiometry, N2O4 increases by half that amount: ΔP(N2O4) = +0.96/2 = +0.48 atm, so P(N2O4)eq = 3.00 + 0.48 = 3.48 atm.
Kp = [P(NO2)]^2 / P(N2O4) = (1.04^2) / 3.48 = 1.0816 / 3.48 ≈ 0.311
Thus Kp ≈ 0.311 (dimensionless).
NO2 decreases by 0.96 atm → P(NO2)eq = 2.00 − 0.96 = 1.04 atm.
By stoichiometry, N2O4 increases by half that amount: ΔP(N2O4) = +0.96/2 = +0.48 atm, so P(N2O4)eq = 3.00 + 0.48 = 3.48 atm.
Kp = [P(NO2)]^2 / P(N2O4) = (1.04^2) / 3.48 = 1.0816 / 3.48 ≈ 0.311
Thus Kp ≈ 0.311 (dimensionless).
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