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number 8 please QUESTIONS For the equilibrium N204() 2 NO2(8), at 298 K, Kp = 0.15....
12 please 91TU 45.80 10-2 d. 172 e 50.8 QUESTION 12 Nitrogen trifluoride decomposes to form nitrogen and fluorine gases according to the following equation: 2 NF3(g) N2(g) + 3 F2(g) When 2.2400000000000002 mol of NF3 is placed in a 4.5-L container and allowed to come to equilibrium at 800 K, the mixture is found to contain 0.0354 mol of N. What is the value of Ko at this temperature? (R=0.082057 L .atm/mol - K) a. 1.9099999999999977 x 10-3 b....
Nitrogen trifluoride decomposes to form nitrogen and fluorine gases according to the following equation: 2 NF3(g) N2(g) + 3 F2(g) When 2.82 mol of NF3 is placed in a 2.50-L container and allowed to come to equilibrium at 800 K, the mixture is found to contain 0.0297 mol of N2. What is the value of Kp at this temperature? (R = 0.082057 L ⋅ atm/mol ⋅ K) a. 1.83 × 10–3 b. 4.43 × 10–7 c. 1.91 × 10–3 d....
Consider the decomposition of N204(g) into NO2(g). N204(g) = 2 NO2(g) kp = 47.9 at 400 K Suppose that 1.00 atm of N204 decomposes and reaches equilibrium at 400 K. Determine the partial pressure of NO2 at equilibrium. 1.96 atm 0.922 atm 1.92 atm 0.960 atm 0.979 atm
At a particular temperature, Kp-0.39 for the reaction below. N204(g) 2 NO2(g) (a) A flask containing only N204 at an initial pressure of 4.8 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. N2O4 NO2 atm (b) A flask containing only NO2 at an initial pressure of 9.6 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. No2 (e) From your answers to parts (e) and (b), does it matter...
N2(g) + 3 H2(g) ⇌ 2 NH3(g) KP = 6.78 x 105 at 298 K (determined using atm) A 7.5 x 101 L container being held at 298 K is charged with the three gases present in the above equation. Once finished, the initial partial pressure of N2 was 0.59 atm, the initial partial pressure of H2 was 0.45 atm, and the initial partial pressure of NH3 was 0.11 atm. The gas mixture was then allowed to reach equilibrium. Use...
I need help with all of them. Please and thank you! 8. A What is the pH of a solution prepared by dissolving 2.25 mol of chloroacetic acid, (HC2H2O2Cl) in 1.50 L of water? For chloroacetic acid, K. = 1.4 x 10. A) 1.69 B) 2.27 C) 11.7 D) 1.34 E) 2.55 10. Hydroxylamine, HONH, readily forms salts such as hydroxylamine hydrochloride which are used as antioxidants in soaps. Hydroxylamine has Kb of 9.1 x 107. What is the pH...
1 2 Part A Calculate Kp at 298 K for the reaction NO(g) + + O2(g) → NO2 (g) assuming that AH is constant over the interval 298-600 K. Kp = 2.31x106 Submit Previous Answers ✓ Correct Part B Calculate Kp at 477 K for this reaction assuming that AHR is constant over the interval 298-600 K. PO AQ * o o ? Kp = | Submit Previous Answers Request Answer Revie For the reaction 2CH4 (g) = C2H2(g) +...
QUESTION 10 [CLO-6] At 900 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2502 + O2(g) ==== 2503 (g) At equilibrium, the partial pressure of SO2 is 35.0 atm and that of O2 is 15.9 atm. The partial pressure of SO3 is atm 4.21 x 10-3 6.20 x 10-4 82.0 40.2 192
4. An equilibrium reaction, 2 NO.(g) NO.(g), has a K, = 2.50 and a total pressure at equilibrium of 2.50 atm. 2 N0219) = N204(9) PT = 2.50 am a. Calculate the equilibrium partial pressures for each gas. KD: 2.50 Ke N204 = 2.50 M le gut) PNO 2 + PN2O4 = 2.51 PNO 2 NO2(g) = N20419) 7x22 X-2.50 Pavé = 62(0.1))2:0.04 atm W IN x = 10.0x2 PNA = (0.1) = 0.1 atmi ē -2% tx Ex: 2.50...
(1). The equilibrium constant, Kp, for the following reaction is 1.80×10-2 at 698K. 2HI(g) =H2(g) + I2(g) If an equilibrium mixture of the three gases in a 15.5 L container at 698K contains HI at a pressure of 0.399 atm and H2 at a pressure of 0.562 atm, the equilibrium partial pressure of I2 is atm. (2). Consider the following reaction: PCl5(g) =PCl3(g) + Cl2(g) If 1.17×10-3 moles of PCl5, 0.217 moles of PCl3, and 0.351 moles of Cl2 are at...