Consider the reaction, 2 BrCl (g) = Br2(g) + Cl2(g) at t = 25°C. This reaction...
At 2935 oC the equilibrium constant for the reaction: 2 BrCl(g) Br2(g) + Cl2(g) is KP = 0.732. If the initial pressure of BrCl is 0.00845 atm, what are the equilibrium partial pressures of BrCl, Br2, and Cl2? p(BrCl) = p(Br2) = p(Cl2) =
At 400K, the equilibrium constant for the reaction Br2(g) + Cl2(g) ⇌ 2BrCl(g) is KP = 7.0. A closed vessel at 400K is charged with 1.00 atm of Br2(g), 1.00 atm of Cl2(g), and 2.00 atm of BrCl(g). Use Q to determine which of the statements below is true. A. The equilibrium partial pressure of BrCl(g) will be less than 2.00 atm B. The reaction will go to completion since there are equal amounts of Br2 and Cl2 C. At...
Consider the reaction Cl2(g) + Br2(g) <=> 2 BrCl(g), which is endothermic as written. What would be the effect on the equilibrium position of adding Cl2(g)?
Br2 (g) + Cl2 (g) ↔ 2 BrCl (g) A reaction mixture at equilibrium contains [Br2] = 0.0993 M, [Cl2] = 0.0967 M, and [BrCl] = 0.0285 M. Calculate the equilibrium constant Kc.
At 400 K, the equilibrium constant for the reaction Br2 (g) + Cl2 (g) <=> 2BrCl (g) is Kp = 7.0. A closed vessel at 400 K is charged with 1.00 atm of Br2 (g), 1.00 atm of Cl2 (g), and 2.00 atm of BrCl (g). What is the equilibrium pressure of Br2? A. 0.86 atm B. The equilibrium partial pressures of Br2 will be the same as the initial value. C. The equilibrium partial pressure of Br2 will be...
23) At 400 K, the equilibrium constant for the reaction Br2(g) + Cl2 (9) 2BrCl (9) is Kp = 70. A closed vessel at 400 K is charged with 1.00 atm of Br2 (g) 1.00 atm of Cl2 (g), and 2.00 alm of BrCl (g). Use Q to determine which of the statements below is true. A) At equilibrium, the total pressure in the vessel will be less than the initial total pressure. B) The equilibrium partial pressures of Br2....
Consider the following reaction: Br2(g)+Cl2(g)⇌2BrCl(g) Kp=1.11×10−4 at 150 K. A reaction mixture initially contains a Br2 partial pressure of 760 torr and a Cl2 partial pressure of 740 torr at 150 K. Part A Calculate the equilibrium partial pressure of BrCl.
Consider this equilibrium reaction at 400 K. Br2(g) + Cl2(g) = 2 BrCl(g) K. = 7.0 If the composition of the reaction mixture at 400 K is [BrCl] = 0.005275 M, [Br] = 0.001971 M, and [CL] = 0.0003620 M, what is the reaction quotient, Q?
Consider this equilibrium reaction at 400 K. Br2(g)+Cl2(g)↽−−⇀2BrCl(g)?c=7.0 If the composition of the reaction mixture at 400 K is [BrCl]=0.00675, [Br2]=0.00249, and [Cl2]=0.000458, what is the reaction quotient, ?Q? ?= How is the reaction quotient related to the equilibrium constant, Kc, for this reaction?
Consider the equilibrium 4. N2(g) 02(g) Br2(g) 2NOBr (g) Calculate the equilibrium constant Kp for this reaction, give the following information (298.15 K) NO (g) +1/2Br2(g) NOBr(g) Ke 4.5 2 NO (g)N2(g) 02(g) Ke 3.0 x 102 5. For the BrCl decomposition reaction 2BrCl(g) Br2(g Cl2(g) Initially, the vessel is charged at 500 K with BrCl at a partial pressure of 0.500 atm. At equilibrium, the partial pressure of BrC is 0.040 atm. Calculate Kp value at 500K
Consider the...