- 1 When Acetic Acid Reacts With Water These Equilibrium Concentrations Are Found Ch Cooh 0 20mol L Ch3c00 0 00 1 (54.82 KiB) Viewed 79 times
- 1 When Acetic Acid Reacts With Water These Equilibrium Concentrations Are Found Ch Cooh 0 20mol L Ch3c00 0 00 2 (47.22 KiB) Viewed 79 times
- 1 When Acetic Acid Reacts With Water These Equilibrium Concentrations Are Found Ch Cooh 0 20mol L Ch3c00 0 00 3 (68.56 KiB) Viewed 79 times
- 1 When Acetic Acid Reacts With Water These Equilibrium Concentrations Are Found Ch Cooh 0 20mol L Ch3c00 0 00 4 (68.88 KiB) Viewed 79 times
3. Write the equilibrium expression and calculate the Ka for Ha (g) + CS: (g) CH. (8) + 2 H₂S (8) Concentrations at equilibrium: [CH]-[H₂S)-1.3 x 10¹ mol/L [H₂]=4.4 x 10¹ mol/L [CS₂] = 1.1 x 10¹ mol/L Answer (3 marks) 4. Calculate the [NO] for 2 NQ (g) + O₂(g) NO₂ (g) If [NO]=4.00 x 10¹ mol/L and [0₂] -2.00 x 10 mol/L at equilibrium, and the equilibrium constant is 2.34 x 10³ at 500°C. (3 marks) Answer ICE Table 5. 1.0 mole of hydrogen gas and 1.0 mole of fluorine gas are sealed in a 1.0 L flask and allowed to react at 450°C. At equilibrium 1.56 moles of hydrogen fluoride gas are present. Calculate the K., for the reaction. H (8) Fi (8) 2HF (8) Answer (4 marks) H₂(g)+F₂ (g)E 2HF (g) H2(g) mol/L F2(g) mol/L HF(g) mol/L 1 с E
6. At certain temperatures, a mixture of hydrogen gas and chlorine gas was prepared by placing 0.200 moles of hydrogen gas and 0.200 moles of chlorine gas into a 2.00 L flask. After a period of time the equilibrium was established. 2HCE (8) H₂(g) + Cl₂ (8) It was determined that, at equilibrium, the chlorine gas concentration had dropped to 0.020 mol/L What is Kea for this reaction? Answer (4 marks) H₂ (8)+Cl₂ (8)E 2HCl(g) H2mol/L C12(g) mol/L HCI(g) mol/L 1 7. Calculate the concentration of product gases at equilibrium when 0.240 mol/L COF₂ (g) decomposes into carbon monoxide and fluorine gas. The K, for this reaction is 1.60 X 10°. COF₂ (g) CO(g) + F₂ (g) Answer (4 marks) COF₂ (g) E CO(g) +F₂ (8) COF2(g)mol/L CO(g) mol/L F2(g) mol/L 1 с E VE с
8. Given the reaction below, what was the initial mass of HI (g) placed in a sealed 1.50 L flask which resulted in 0.025 mol of each of H₂ (g) and 2 (g) being found in the flask at equilibrium? H₂ (g) + 12 (g) 2 HI (g) K = 50.5 Answer (4 marks) H₂(8)+ 12(8) 2 HI(g) K = 50.5 HI(g)mol/L H2(g) mol/L 12(g) mol/L I C E 9. If 0.100 mol/L of CO₂ (g) and H₂(g) are injected into a sealed system, what are the concentrations of H₂ (g) and H₂O (g) at equilibrium given that Ke 0.64 at 900K? CO₂ (g)+ H₂ (8) CO (g) + H₂O (8) Answer (4 marks) CO₂(g)+ H₂(g) É CO(g)+ H₂O(g) CO2(g)mol/L H2(g)mol/L CO(g)mol/L H2O(g)mol/L I с E