반응형 일반화학/[15장] 화학 평형194 0.500 mol PCl5 was placed into a 0.250 L 0.500 mol PCl5 was placed into a 0.250 L Consider the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) At a certain temperature, 0.500 mol of PCl5 was placed into a 0.250 L container and allowed to react. At equilibrium, the container held 0.100 mol of PCl5. What is the value of K for this reaction? --------------------------------------------------- 초기 [PCl5] = 0.500 mol / 0.250 L = 2.00 M ICE 도표를 작성하면, ..... 2022. 12. 29. 이론적 평형상수. H2(g) + I2(g) ⇌ 2HI(g) at 700 K is 49 이론적 평형상수. H2(g) + I2(g) ⇌ 2HI(g) at 700 K is 49 The reaction H2(g) + I2(g) ⇌ 2HI(g) has a K of 49.0 at a certain temperature. Determine the value of K for the reaction below at the same temperature: HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g) The equilibrium constant K for the reaction H2(g) + I2(g) ⇌ 2HI(g) at 700 K is 49. What is the equilibrium constant for the reaction HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g) at the.. 2022. 12. 29. When 4.0 mol of NH3(g) is introduced into a 2.0 L When 4.0 mol of NH3(g) is introduced into a 2.0 L When 4.0 mol of NH3(g) is introduced into a 2.0 L 2NH3(g) ⇌ N2(g) + 3H2(g) NH3(g) at equilibrium decreases to 2.0 mol 평형에서 모든 화학종의 농도? --------------------------------------------------- 초기 [NH3] = 4.0 mol / 2.0 L = 2.0 M [그림] 2NH3(g) ⇌ N2(g) + 3H2(g) 반응의 일반적인 ICE 도표와 평형 상수식. ............. 2NH3(g) ⇌ N2(g) + 3H2(g) 평형(M): 2.0–2x ....... x .......... 2022. 12. 29. 0.200 mol of H2S is placed in an empty 1.00 L container 0.200 mol of H2S is placed in an empty 1.00 L container The equilibrium constant, K, is 4.20×10^(-6) for the following reaction: 2H2S(g) ⇌ 2H2(g) + S2(g) Find the equilibrium concentrations of all entities when 0.200 mol of H2S(g) is placed in a 1.00 L container. --------------------------------------------------- [H2S] = 0.200 mol / 1.00 L = 0.200 M ICE 도표를 작성하면, ............... 2H2S(g) ... ⇌ ... 2022. 12. 29. 4.00 mol H2(g), 4.00 mol F2(g), 6.00 mol HF(g) 4.00 mol H2(g), 4.00 mol F2(g), 6.00 mol HF(g) When hydrogen reacts with fluorine, hydrogen fluoride is formed according to the following equation: H2(g) + F2(g) ⇌ 2HF(g) The equilibrium constant, K, is 1.15×10^2 at SATP. Calculate the concentrations of all entities at equilibrium if 4.00 mol H2(g), 4.00 mol F2(g), and 6.00 mol HF(g) are initially placed into a 2.00 L container. ----------------.. 2022. 12. 28. 2NO2(g) ⇌ N2O4(g) K = 1.15 NO2 0.650 mol/L 2NO2(g) ⇌ N2O4(g) K = 1.15 NO2 0.650 mol/L In a sealed container,nitrogen dioxide is in equilibrium with dinitrogen tetroxide.2NO2(g) ⇌ N2O4(g) ... K = 1.15 at 55°CFind the equilibrium concentrationsof nitrogen dioxide and dinitrogen tetroxideif the initial concentration of nitrogen dioxide is 0.650 mol/L. ICE 도표를 작성하면, ............. 2NO2(g) . ⇌ . N2O4(g)초기(M) . 0.650 ....... 2022. 12. 28. H2(g) + I2(g) ⇌ 2HI(g) Kc 100.0 H2 1.0 mol 10.0 L H2(g) + I2(g) ⇌ 2HI(g) Kc 100.0 H2 1.0 mol 10.0 L 평형상수 응용. 평형에 도달했을 때 HI의 농도 H2(g) + I2(g) ⇌ 2HI(g) ... Kc = 100.0 온도 T에서 H2 1.0 mol과 I2 1.0 mol이 10.0 L 용기에서 반응하여 평형에 도달하였을 때, HI의 농도(M)는? 단, 온도는 T로 일정. --------------------------------------------------- [H2] = [I2] = 1.0 mol / 10.0 L = 0.10 M = C(초기농도) Kc = [HI]^2 / [H2][I2] = (2x)^2 / (C–x)^2 ( 참고 https://ywpop.tistory.com/6300 ) (2x)^2 / (0.10.. 2022. 12. 15. 암모니아 르샤틀리에 N2(g) + 3H2(g) ⇌ 2NH3(g) ★ 암모니아 르샤틀리에 N2(g) + 3H2(g) ⇌ 2NH3(g) --------------------------------------------------- ▶ 참고: 르샤틀리에 원리 [ https://ywpop.tistory.com/2685 ] --------------------------------------------------- N2(g) + 3H2(g) ⇌ 2NH3(g) 평형 상태에서, ① 압력을 낮추면, 역반응 진행. > 압력이 증가하려면, 기체 몰수 증가 쪽으로. ---> Yes ( 참고 https://ywpop.tistory.com/2686 ) ② 압력을 높이면, 정반응 진행. > 압력이 감소하려면, 기체 몰수 감소 쪽으로. ---> Yes ③ 질소를 주입하면, 정반응 진행. > 반응물이.. 2022. 12. 15. CO(g) + H2O(g) ⇌ CO2(g) + H2(g) 500℃ equilibrium constant 3.9 CO(g) + H2O(g) ⇌ CO2(g) + H2(g) 500℃ equilibrium constant 3.9 An important step in the industrial production of hydrogen is the reaction of carbon monoxide with water: CO(g) + H2O(g) ⇌ CO2(g) + H2(g) a) Use the law of mass action to write the equilibrium expression for this reaction. b) At 500℃, the equilibrium constant for this reaction is 3.9. Suppose that the equilibrium partial pressures of .. 2022. 12. 12. NH4HS(s) ⇌ NH3(g) + H2S(g) Kp = 0.108 at 25℃ NH4HS(s) ⇌ NH3(g) + H2S(g) Kp = 0.108 at 25℃ Solid ammonium hydrogen sulfide, NH4HS, dissociates appreciably even at room temperature forming ammonia, NH3, and hydrogen sulfide, H2S, gases. What is the total pressure at equilibrium if solid NH4HS is placed in an evacuated container and allowed to reach equilibrium? Kp = 0.108 at 25℃. --------------------------------------------------- ICE 도표를 작성.. 2022. 12. 8. 이전 1 ··· 4 5 6 7 8 9 10 ··· 20 다음 반응형
