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Para calcular a constante de equilíbrio, Kp, da reação N2O4 (g) → 2 NO2 (g) a 800 K, podemos usar a fórmula: \[ Kp = \frac{{[NO2]^2}}{{[N2O4]}} \] Primeiro, vamos calcular as concentrações no equilíbrio. Sabemos que a pressão de NO2 é 0,41 atm. Como a reação é N2O4 (g) → 2 NO2 (g), a pressão de N2O4 será igual a \( 7,275 \times 10^{-3} \) mol/L - 2 * 0,41 atm = \( 7,275 \times 10^{-3} \) mol/L - 0,82 atm. Agora, podemos substituir esses valores na fórmula de Kp: \[ Kp = \frac{{(0,41)^2}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] Calculando o valor de Kp, obtemos: \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 0,82)^2}} \] \[ Kp = \frac{{0,1681}}{{(7,275 \times 10^{-3} - 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