14.39. The K value for the reaction 2 NOBr(g) – 2 NO(g) + Brz(8) is 3.0...
Consider the following reaction where Kc = 6.50×10-3at 298 K: 2 NOBr (g) 2 NO (g) + Br2(g) A reaction mixture was found to contain 8.20×10-2moles of NOBr (g), 4.08×10-2moles of NO (g), and 4.11×10-2moles of Br2(g), in a 1.00 liter container. Indicate True (T) or False (F) for each of the following: ___TF 1. In order to reach equilibrium NOBr(g) must be consumed . ___TF 2. In order to reach equilibrium Kc must increase . ___TF 3. In order...
Consider the following reaction where Kc = 6.50×10-3 at 298 K: 2 NOBr (g) goes to 2 NO (g) + Br2 (g) A reaction mixture was found to contain 8.54×10-2 moles of NOBr (g), 2.30×10-2 moles of NO (g), and 4.35×10-2 moles of Br2 (g), in a 1.00 liter container. Indicate True (T) or False (F) for each of the following: 1. In order to reach equilibrium NOBr(g) must be consumed . 2. In order to reach equilibrium Kc must...
2. Determine Ke for the reaction (Answer: 9.7x10-16): 2 N2(g) + 1/2O2(g) + Brz(g) $ NOBr (g) From the following information (Hint: Manipulate the equations first, then manipulate K values): 2 NO(g) 5 N2(g) + O2(g) NO (g) + Br2 (g) $ NOBr (g) Ke=2.1x1030 Ko=1.4
Consider the reaction: 2 NO(g) + Br2(g) ↔ 2 NOBr(g) Kp = 28.4 atm-1 at 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 173 Torr and that of Br2 is 142 Torr. What is the partial pressure (in Torr) of NOBr in this mixture? PNOBr = ___ Torr
Consider the following reaction where Kc = 154 at 298 K: 2NO(g) + Br2(g) 2NOBr(g) A reaction mixture was found to contain 2.42×10-2 moles of NO(g), 4.38×10-2 moles of Br2(g) and 9.68×10-2 moles of NOBr(g), in a 1.00 Liter container. Indicate True (T) or False (F) for each of the following: 1. In order to reach equilibrium NOBr(g) must be produced. 2. In order to reach equilibrium Kc must decrease. 3. In order to reach equilibrium NO must be produced....
A student ran the following reaction in the laboratory at 225 K: 2NOBr(g) 2 2NO(g) + Brz(g) When she introduced 0.198 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 1.89x10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Br2(g) = 2NOBr(g) A reaction mixture was found to contain 2.86x10-2 moles of NO(g), 3.93x10-2 moles of Br2(g) and 8.83x10-2 moles of NOBr(g), in a 1.00 Liter container. Indicate True (T) or False (F) for each of the following: 1. In order to reach equilibrium NOBr(g) must be consumed. 2. In order to reach equilibrium K, must increase. 3. In order to reach equilibrium NO must be...
The equilibrium constant for the reaction 2 NO(g) + Br2 = 2 NOBr(9) is Ke = 2.4 x 10-2 at certain temperature. Calculate K for 2 NOBr(9) = 2 NO(g) + Brz (9) Express your answer using two significant figures. V AEC O ? K- Submit Request Answer Part C Complete previous part(s)
Consider the following reaction where Kc = 154 at 298 K. 2NO(g) + Br2(g) 2NOBr(g) A reaction mixture was found to contain 4.44×10-2 moles of NO(g), 3.68×10-2 moles of Br2(g) and 7.92×10-2 moles of NOBr(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals . The reaction A. must run in the forward direction to reach equilibrium. B. must run in the...
At 141 oC, Keq = 0.00830 for the reaction: NO(g) + 1/2 Br2(g) NOBr(g) (a) What is the value of Keq for the reaction NOBr(g) NO(g) + 1/2 Br2(g)? Keq = . (b) What is the value of Keq for the reaction 2 NO(g) + Br2(g) 2 NOBr(g)? Keq = . (c) What is the value of Keq for the reaction 2 NOBr(g) 2 NO(g) + Br2(g)? Keq = .