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At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9)...
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) + Cl (9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.4 moles of pure NOCl in a 2.0-L flask [NOCI) - t t M t [NO] - M t [Cl] - M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI] - M [NO] - M (Cl) - M c....
63. At 35°C, K = 1.6 x 10-6 for the reaction 2NOCI(9) = 2NO(g) + Cl2 (9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.0 moles of pure NOCI in a 2.0-L flask Answer b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask Answer c. 2.0 moles of NOCI and 1.0 mole of Cl, in a 1.0-L flask
15. This question has multiple parts. Work all the parts to get the most points. At 35°C, K-1.1 x 10° for the reaction 2 NOCI(g) 근 2N0(g) + Cl2 (g) Calculate the concentrations of all species at equilibrium for each of the following original mixtures a3.8 moles of pure NOCl in a 2.0-L flask [NOCI] = NO] [C12] = b1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask NOCI NO Numeric input field 3.8 mole of...
Please explain and show steps! Thanks! At 35°C, K = 1.6×10-5 for the following reaction 2 NOCI(g)2 NO(g) + Cl2(g) Calculate the concentrations of all species at equilibrium for each of the following original mixtures (a) 2.0 mol of pure NOCl in a 1.7-L flask [NOCI] [Cl2l (b) 7.2 mol of NO and 3.6 mol of Cl2 in a 1.2-L flask [NOCI] [NO] [Cl2l (c) 1.7 mol of NOCI and 1.7 mol of NO in a 2.0-L flask [NOCI] [NO]...
At 35°C, K = 1.6 10-5 for the reaction 2 NOCI(g) 52 NO(g) + Cl2(8) If 3.6 mol NO and 1.8 mol Cl2 are placed into a 1.0-L flask, calculate the equilibrium concentrations of all species. [NOCI) - [NO] - м [C12] - Submit Answer Try Another Version 1 item attempt remaining
14. For the reaction: 2NOCI(g) 2NO(g) + Club), K = 1.6 x 10. What are the equilibrium concentrations of each species it tions of each species if 1.0 mole of NOCI is initially placed in an empty 2.0 L flask? 14. For the reaction: 2NOCI(B) = 2NO(g) + Cl2(g), K = X 1.6*10 What are the equilibrium concentrations of each species if 1.0 mole of NOCI is initially placed in an empty 2.0 L flask?
At some temperature, the Kc for the reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g) is 1.6 x 10^-5. Calculate the concentrations of all species at equilibrium if 1.0 moles of pure NOCl is initially placed in a 2.0 L flask
Consider the reaction 2 NO + Cl2 → 2 NOCI A possible mechanism is: Cl2 ⇄ 2 Cl (fast) Cl + NO → NOCl (slow) If this is correct, the rate law will be: A. rate = k [NO]0.5[Cl] B. rate = k [NO]2[Cl2] C. rate = k [NO][Cl2] D. rate = k [NO][Cl]2 E. rate = k [NO][Cl2]0.5
Practice Exercise 1 At 500 K, the reaction 2 NO(8) + Cl2(8) = 2 NOCI(8) has K, = 51. In an equilibrium mixture at 500 K, the partial pres- sure of NO is 0.125 atm and Cl, is 0.165 atm. What is the par- tial pressure of NOCI in the equilibrium mixture? (a) 0.13 atm (b) 0.36 atm (c) 1.0 atm (d) 5.1 x 10-5 atm (e) 0.125 atm Practice Exercise 2 At 500 K, the reaction PC15(8) = PC13(8) + Cl2(8) has...
At 500.0 K, gaseous NOCl decomposed to NO and Cl2 according the following reaction: 2NOCl(g) ⇌ 2NO(g) + Cl2(g) If 1.00 mole of NOCl is placed in a 1.00 liter contained, and at equilibrium, the concentration is 0.91M, what is the equilibrium constant, Kc? 4. At 500.0 K, gaseous NOCI decomposed to NO and Cl2 according the following reaction: 2NOCI(g) = 2NO(g) + Cl2(g) If 1.00 mole of NOCI is placed in a 1.00 liter contained, and at equilibrium, the...