For the following reaction,
2NOBr(g) <-----> 2NO(g) + Br2(g)
the equilibrium constant Kp = 2.54×10-2 at 357 K and 3.57×10-2 at 381 K. Calculate ΔH° for the reaction, assuming no change in ΔH° between 357 K and 381 K.
ΔH° = ________ kJ mol-1
For the following reaction, 2NOBr(g) <-----> 2NO(g) + Br2(g) the equilibrium constant Kp = 2.54×10-2 at...
The equilibrium constant, Kp, for the following reaction is 6.25 at 298 K. 2NO(g) + Br2(g) 2NOBr(g) If ΔH° for this reaction is -16.1 kJ, what is the value of Kp at 187 K? Kp =
The equilibrium constant for the reaction: 2NO(g) + Br2(g) <----> 2NOBr(g) is Kc = 1.3x10^-2 at 1,000 Ka.) At this temperature, does the equilibrium favor the product or reactants?b.) Calculate Kc for 2NOBr <----> 2NO + Br2c.) Calculate Kc for NOBr <----> NO + 1/2Br2
The equilibrium constant, Kp, for the following reaction is 0.160 at 298K. 2NOBr(g) 2NO(g) + Br2(g) If an equilibrium mixture of the three gases in a 19.9 L container at 298K contains NOBr at a pressure of 0.297 atm and NO at a pressure of 0.251 atm, the equilibrium partial pressure of Br2 is ? atm.
QUESTION 13 [CLO-5] At 1000 K, the equilibrium constant for the reaction 2NO (g) + Br2 (g) ==== 2NOBr (g) is Kp = 0.013. Calculate Kp for the reverse reaction, 2NOBr (g) 2NO (g) + Br2 (9) ==== -0.013 1.1 77 0.013 0 1.6x 10-4
The equilibrium constant, Kc, for the following reaction is 6.50×10-3 at 298K. 2NOBr(g) 2NO(g) + Br2(g) If an equilibrium mixture of the three gases in a 11.1 L container at 298K contains 0.376 mol of NOBr(g) and 0.396 mol of NO, the equilibrium concentration of Br2
Consider the following reaction: 2NO(g)+Br2(g)⇌2NOBr(g) ,Kp=28.4, at 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 102 torr and that of Br2 is 134 torr . Part A What is the partial pressure of NOBr in this mixture? Express your answer in torrs to three significant figures.
Consider the reaction: 2NO(g)+Br2(g)⇌2NOBr(g) Kp=28.4 at 298 K. In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br2 is 158 torr . What is the partial pressure of NOBr in this mixture?
The equilibrium constant, Kc, for the following reaction is 7.68×10-3 at 307 K. 2NOBr(g) 2NO(g) + Br2(g) Calculate Kc at this temperature for the following reaction: NO(g) + 1/2Br2(g) NOBr(g)
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...
The equilibrium constant, Kc, for the following reaction is 5.19×10-3 at 286 K. 2NOBr(g) <-->2NO(g) + Br2(g) Calculate Kc at this temperature for the following reaction: NO(g) + 1/2Br2(g) <-->NOBr(g) Kc=?