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標題:

F6 Chem Eqm. constant Kp

發問:

At 50℃, the degree of dissociation of N2O4 is 0.5, when the total pressure of the equilibrium mixture is 1 atm. (a) Calculate the value of Kp. (b) If the total pressure of the system is now increased to 2 atm, what is the degree of dissociation of N2O4? Ans: (a) 1.33 atm (b) 0.378

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• 數學方程(步驟)
• 功能團體選舉有哪28個職業-

 

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最佳解答:

At 50℃, the degree of dissociation of N2O4 is 0.5, when the total pressure of the equilibrium mixture is 1 atm. (a) Calculate the value of Kp. Let n mol be the initial number of moles of N2O4. N2O4(g) ? 2NO2(g) At equilibrium: Degree of dissociation of N2O4 = 0.5 No. of moles of N2O4 = (n - 0.5n) = 0.5n mol No. of moles of NO2 = 2 x 0.5n = n mol Partial pressure of N2O4, PN2O4 = 1 x [0.5n/(0.5n + n)] = 1/3 atm Partial pressure of NO2, PNO2 = 1 x [n/(0.5n + n)] = 2/3 atm Kp = PNO2/PN2O4 = (2/3)^2/(1/3) = 1.33 atm (b) If the total pressure of the system is now increased to 2 atm, what is the degree of dissociation of N2O4? Let α be the degree of dissociation of N2O4 when total pressure is 2 atm. Let n mol be the initial number of moles of N2O4. At equilibrium: Degree of dissociation of N2O4 = α No. of moles of N2O4 = n(1 - α) mol = (n - αn) mol No. of moles of NO2 =2αn mol Total no. of moles = (n - αn) + 2αn = (n + αn) mol Partial pressure of N2O4, PN2O4 = 2 x [(n - αn)/(n + αn )] = (2 - 2α)/(1 + α) atm Partial pressure of NO2, PNO2 = 2 x [2αn/(n + αn)] = 4α/(1 + α) atm Kp = [4α/(1 + α)]^2 / [(2 - 2α)/(1 + α)] = 1.33 8α2 / [(1 + α)(1 - α)] = 1.33 (1 + α)(1 - α) = 8α2/1.33 1 - α^2 = 6α 7α^2 = 1 α = √(1/7) α = 0.378 Hence, the degree of dissociation = 0.378 2011-02-03 02:01:52 補充： The last fifth line should be: 1 - α^2 = 6α^2

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