標題:
PHYSICAL CHEM QUESTION
發問:
1.when 1 mole of pure ethyl alcohol is mixed with 1 mole of acetic acid at room temperature, the equilibrium mixture contains 2/3 mole each of ester and water.a)What is the equilibrium const.? (4)b)How many moles of ester are formed at equilibrium when 3 moles of alcohol is mixed with 1mole of acid? All... 顯示更多 1.when 1 mole of pure ethyl alcohol is mixed with 1 mole of acetic acid at room temperature, the equilibrium mixture contains 2/3 mole each of ester and water. a)What is the equilibrium const.? (4) b)How many moles of ester are formed at equilibrium when 3 moles of alcohol is mixed with 1mole of acid? All substances are liquid at the reaction temperature. (0.903 mol) 2.at 27 degree celsius and 1 atmospere,N2O4 is 20% dissociated into NO2.Find a)Kp (0.167 atm) b)the per cent dissociation at 27 degree celsius and a total pressure of0.90 atmospere c)what is the excent of dissociation in a 69 gramme sample of N2O4 confined in a 20 dm^3 vessel at 27 degree celsius?(19.13%) 3.if nitrogen tetroxide is 35% dissociated into nitrogen dioxide under a pressure of 2 atm,calculatethe equilibrium const in terms of partial prossures and the percentage dissociation of the nitrogen totroxide at the same temperature but under 1 atmospheric pressure.(Kp=1.117 atm , 46.7%) 括號內為答案
最佳解答:
1a) Kc = [ester][water]/{[alcohol][acid]} = (2/3 x 2/3)/(1/3 x 1/3) = 4 b) Suppose that x moles of ester and water are formed, then: No. of moles of alcohol = 3 - x, no. of moles of acid = 1 - x Hence: x2/[(3 - x)(1 - x)] = 4 x2 = 4x2 - 16x + 12 3x2 - 16x + 12 = 0 x = 0.903 or 4.43 (rejected since x < 1) 2a) Suppose that initialy P atm of N2O4 is injected, then at eq.: Partial pressure of N2O4 = 0.8P Partial pressure of NO2 = 2 x 0.2P = 0.4P Hence 1.2P = 1 atm, i.e. P = 5/6 atm Hence Kp = (0.4P)2/(0.8P) = 0.2P = 1/6 atm = 0.167 atm b) Again, suppose initial P atm of N2O4 injected, also at eq.: Partial pressure of N2O4 = (P - x) atm Partial pressure of NO2 = 2x atm Total pressure = (P + x) = 0.9 atm With: 4x2/(P - x) = Kp = 1/6 12x2 = P - x = 0.9 - 2x 12x2 + 2x - 0.9 = 0 x = -0.370 (rej.) or 0.203 P = 0.9 - x = 0.697 So % of diss. = 0.203/0.697 x 100% = 29.1% c) 69 g of N2O4 contains 69/92 = 0.75 moles By P = nRT/V, initial pressure of N2O4 = 0.75 x 8.314 x 300/0.02 = 93533 Pa = 0.923 atm So let, at eq.: Partial pressure of N2O4 = (0.923 - x) atm Partial pressure of NO2 = 2x atm Thus, 4x2/(0.923 - x) = Kp = 1/6 24x2 = 0.923 - x 24x2 + x - 0.923 = 0 x = 0.176 So % of diss. = 0.176/0.923 x 100% = 19.13% 3) Let initial pressure of P atm of N2O4 injected, then at eq.: Partial pressure of N2O4 = 0.65P Partial pressure of NO2 = 0.7P Total = 1.35P = 2 atm, giving P = 1.481 atm So Kp = 0.49P/0.65 = 1.117 atm Now, for the second part, again, initial pressure of P atm of N2O4 injected then at eq.: Partial pressure of N2O4 = P - x Partial pressure of NO2 = 2x Total = P + x = 1 atm Also: 4x2/(P - x) = 1.117 4x2 = 1.117(1 - 2x) 4x2 + 2.234x - 1.117 = 0 x = 0.318 P = 0.682 % diss = 0.318/0.682 x 100% = 46.7%
其他解答:
PHYSICAL CHEM QUESTION
發問:
1.when 1 mole of pure ethyl alcohol is mixed with 1 mole of acetic acid at room temperature, the equilibrium mixture contains 2/3 mole each of ester and water.a)What is the equilibrium const.? (4)b)How many moles of ester are formed at equilibrium when 3 moles of alcohol is mixed with 1mole of acid? All... 顯示更多 1.when 1 mole of pure ethyl alcohol is mixed with 1 mole of acetic acid at room temperature, the equilibrium mixture contains 2/3 mole each of ester and water. a)What is the equilibrium const.? (4) b)How many moles of ester are formed at equilibrium when 3 moles of alcohol is mixed with 1mole of acid? All substances are liquid at the reaction temperature. (0.903 mol) 2.at 27 degree celsius and 1 atmospere,N2O4 is 20% dissociated into NO2.Find a)Kp (0.167 atm) b)the per cent dissociation at 27 degree celsius and a total pressure of0.90 atmospere c)what is the excent of dissociation in a 69 gramme sample of N2O4 confined in a 20 dm^3 vessel at 27 degree celsius?(19.13%) 3.if nitrogen tetroxide is 35% dissociated into nitrogen dioxide under a pressure of 2 atm,calculatethe equilibrium const in terms of partial prossures and the percentage dissociation of the nitrogen totroxide at the same temperature but under 1 atmospheric pressure.(Kp=1.117 atm , 46.7%) 括號內為答案
最佳解答:
1a) Kc = [ester][water]/{[alcohol][acid]} = (2/3 x 2/3)/(1/3 x 1/3) = 4 b) Suppose that x moles of ester and water are formed, then: No. of moles of alcohol = 3 - x, no. of moles of acid = 1 - x Hence: x2/[(3 - x)(1 - x)] = 4 x2 = 4x2 - 16x + 12 3x2 - 16x + 12 = 0 x = 0.903 or 4.43 (rejected since x < 1) 2a) Suppose that initialy P atm of N2O4 is injected, then at eq.: Partial pressure of N2O4 = 0.8P Partial pressure of NO2 = 2 x 0.2P = 0.4P Hence 1.2P = 1 atm, i.e. P = 5/6 atm Hence Kp = (0.4P)2/(0.8P) = 0.2P = 1/6 atm = 0.167 atm b) Again, suppose initial P atm of N2O4 injected, also at eq.: Partial pressure of N2O4 = (P - x) atm Partial pressure of NO2 = 2x atm Total pressure = (P + x) = 0.9 atm With: 4x2/(P - x) = Kp = 1/6 12x2 = P - x = 0.9 - 2x 12x2 + 2x - 0.9 = 0 x = -0.370 (rej.) or 0.203 P = 0.9 - x = 0.697 So % of diss. = 0.203/0.697 x 100% = 29.1% c) 69 g of N2O4 contains 69/92 = 0.75 moles By P = nRT/V, initial pressure of N2O4 = 0.75 x 8.314 x 300/0.02 = 93533 Pa = 0.923 atm So let, at eq.: Partial pressure of N2O4 = (0.923 - x) atm Partial pressure of NO2 = 2x atm Thus, 4x2/(0.923 - x) = Kp = 1/6 24x2 = 0.923 - x 24x2 + x - 0.923 = 0 x = 0.176 So % of diss. = 0.176/0.923 x 100% = 19.13% 3) Let initial pressure of P atm of N2O4 injected, then at eq.: Partial pressure of N2O4 = 0.65P Partial pressure of NO2 = 0.7P Total = 1.35P = 2 atm, giving P = 1.481 atm So Kp = 0.49P/0.65 = 1.117 atm Now, for the second part, again, initial pressure of P atm of N2O4 injected then at eq.: Partial pressure of N2O4 = P - x Partial pressure of NO2 = 2x Total = P + x = 1 atm Also: 4x2/(P - x) = 1.117 4x2 = 1.117(1 - 2x) 4x2 + 2.234x - 1.117 = 0 x = 0.318 P = 0.682 % diss = 0.318/0.682 x 100% = 46.7%
其他解答:
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