CCl4(g) equilibrium C(s) + 2Cl(g) Kp=0.86 if 2.50 ATM of CCl4 is added to a flask and allowed to reach equilibrium. what are the final pressure of CCl4 and Cl2 at equilibrium
CCl4(g) equilibrium C(s) + 2Cl(g) Kp=0.86 if 2.50 ATM of CCl4 is added to a flask...
The following reaction has equilibrium constant of Kp = 11.5 atm at 300 oC: PCl5(g) PCl3 (g) + Cl2 (g). A flask was charged with pure PCl5 (g) and allowed to achieve the equilibrium, at which partial pressure of PCl5 (g) was 1.50 atm. Find (a) total pressure at equilibrium (b) partial pressures of PCl3 (g) and Cl2(g), (c) initial pressure of PCl5.
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...
2 NOCl(g) ⇄ 2 NO(g) + Cl2(g) Kp = 7.2× 10-6 1.50 atm of NOCl(g) is placed in a container and the system is allowed to reach equilibrium. Calculate the equilibrium pressure of Cl2(g) at equilibrium. 4.8 × 10-2 atm 8.5 × 10-3 atm 3.2 × 10-2 atm 2.4 × 10-2 atm 1.6 × 10-2 atm
The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) > CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.939 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm
Ke & Equilibrium Pressures and in • The equil. constant, kp, for the following rxn is 10.5 @ 350k 2 CH₂Cl2 cg) - Chucgs + CCl4 cgi If an equil. mixture of the 3 gases in a 13 L container @ 350 K contains CH₂Cl2 e a pressure of 0.558 atm . CHu o a pressure of 0.272 atm, the equil. PP of coly is atm. L Le Châtelier Calculations : [ ] The equil. constant, K, for the following...
1. At a particular temperature, K = 2.50 for the reaction: SO2 (g) + NO2 (g) ⇄ SO3 (g) + NO (g). If all four gases had initial concentrations of 1.00 M, calculate the equilibrium concentrations of SO2. 2. At a particular temperature, Kp = 0.25 for the reaction: N2O4 (g) ⇄ 2 NO2 (g). A flask containing only N2O4 at an initial pressure of 4.5 atm is allowed to reach equilibrium. a. Calculate the equilibrium partial pressure of N2O4....
1. The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <----> PCl3(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.00 atm at 500 K. PPCl5 = atm PPCl3 = atm PCl2 = atm 2. The equilibrium constant, Kp, for the following reaction is 0.215 at 673 K: NH4I(s) <----> NH3(g) + HI(g) Calculate the equilibrium partial pressure of HI when...
The equilibrium constant for the reaction CCl4(g) <--> C(s) + 2Cl2(g) at 700ºC is 0.85. Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.60 atm at 700ºC. Pressure = atm
The equilibrium constant in terms of pressures, Kp, for the reaction NH3(g)+ HI(g) NH4I(s) at 400 °C is 4.65. (a) If the partial pressure of ammonia is PNH, 0.881 atm and solid ammonium iodide is present, what is the equilibrium partial pressure of hydrogen iodide at 400 °C? PHI atm (b) An excess of solid NH,I is added to a container filled with NH3 at 400 °C and a pressure of 1.17 atm. Calculate the pressures of NH(g) and HI(g)...
A flask is charged with 1.800 atm of N2O4(g) and 1.00 atm of NO2(g) at 25 ∘C , and the following equilibrium is achieved: N2O4(g)⇌2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.519 atm . 1A) What is the partial pressure of N2O4 at equilibrium? 1B) Calculate the value of Kp for the reaction. 1C) Calculate the value of Kc for the reaction.