金門貸款率利計算,車貸利率試算表

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標題:

equilibrium chem

發問:

ZnO(s) + CO <--> Zn(g) + CO2has equilibrium constant 1 atm at 500K find eqm partial pressure of zinc vapour in vesselif 1 mole of each of CO + CO2 are brought into contact with ZnO(s) at 500K and at total pressure 1 atm.N2 + 3H2 <--> 2NH3have reaction at pressure 45 atm. the reaction mixture... 顯示更多 ZnO(s) + CO <--> Zn(g) + CO2 has equilibrium constant 1 atm at 500K find eqm partial pressure of zinc vapour in vessel if 1 mole of each of CO + CO2 are brought into contact with ZnO(s) at 500K and at total pressure 1 atm. N2 + 3H2 <--> 2NH3 have reaction at pressure 45 atm. the reaction mixture have 8% volume of ammonia, Find Kp for reaction. eqm. pressure in a system originally consisting NH4HS (s) is 0.77 atm find Kp for NH4HS <--> NH3 +H2S what would be the Kp of ammonia if 0.1 atm of this gas were added (No change in temp.)

最佳解答:

1. ZnO(s) + CO(g) ? Zn(g) + CO2(g) At eqm: Let no. of moles of Zn = y mol Then, no. of moles of CO = (1 - y) mol and no. of moles of (1 + y) mol Total no. of moles = y + (1 - y) + (1 + y) = (2 + y) mol PCO = 1 x [(1 - y)/(2 + y)] = (1 - y)/(2 + y) atm PZn = 1 x [y/(2 + y)] = y/(2 + y) atm PCO2 = 1 x [(1 + y)/(2 + y)] = (1 + y)/(2 + y) atm Kp = (PZn)(PCO2)/(PCO) [y/(2 + y)] x [(1 + y)/(2 + y)] / [(1 - y)/(2 + y)] = 1 y(1 + y) / (1 - y)(2 + y) = 1 y(1 + y) = (1 - y)(2 + y) y + y^2 = 2 - y - y^2 2y^2 + 2y - 2 = 0 y^2 + y - 1 = 0 y = {-1 + √[12 - 4x1x(-1)]}/2 y = 0.618 Eqm partial pressure of Zn = 0.618/(2 + 0.618) = 0.236 atm = = = = = 2. The answer depends on the initial mole ratio of N2- and H2. Let 1 mol and 3 mol be the initial number of moles of N2 and H2- respectively. N2 + 3H2 ? 2NH3 At eqm: Let no. of moles of NH3 = 2y mol Then no. of moles of N2 = (1 - y) mol and no. of moles of H2 = (3 - 3y) mol Total no. of moles = 2y + (1 - y) + (3 - 3y) = (4 - 2y) mol 2y/(4 - 2y) = 8% 2y = 0.32 - 0.16y 2.16y = 0.32 y = 0.148 PNH3 = 45 x [2(0.148)/(4 - 2x0.148)] = 3.60 atm PN2 = 45 x [(1 - 0.148)/(4 - 2x0.148)] = 10.35 atm PH2 = 45 x [(3 - 3x0.148)/(4 - 2x0.148)] = 31.05 atm Kp = (PNH3)^2/(PN2)(PH2)^3 = (3.6)^2 /(10.35)(31.05)^3 = 4.18 x 10^-5 atm^-2 = = = = = 3. NH4HS(s) ? NH3(g) + H2S(g) At eqm: PNH3 = PH2S = 0.77/2 = 0.385 atm Kp = (PNH3) x (PH2S) = (0.385)^2 = 0.148 atm^2 After 0.1 atm of NH3 was added, let y atm of partial pressure of NH3 is decreased in the new eqm. At the new eqm: PNH3 = (0.385 + 0.1 - y) atm = (0.485 - y) atm PH2S = (0.385 - y) atm Kp = (PNH3) x (PH2S) (0.485 - y)(0.385 - y) = 0.148 y^2 - 0.87y + 0.0387 = 0 y = [0.87 - √(0.87^2 - 4x0.0387x1)/]/2 y = 0.047 Partial pressure of NH3 at eqm = (0.485 - 0.047) = 0.438 atm

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