At 500 K the reaction PCl5(g) <----> PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?
At 500 K the reaction PCl5(g) <----> PCl3(g) + Cl2(g) has Kp = 0.497. In an...
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 Kp for the reaction PCl5(g) <---> PCl3(g) + Cl2(g) is 1.15 at 25 degrees Celsius. The reaction starts with a mixture of 0.177 atm PCl5, 0.223 atm PCl3, and 0.111 atm Cl2. When this mixture comes to equilibrium at 25 degrees Celsius, what are the equilibrium pressures of each component?
Practice Exercise 1 At 500 K, the reaction 2 NO(8) + Cl2(8) = 2 NOCI(8) has K, = 51. In an equilibrium mixture at 500 K, the partial pres- sure of NO is 0.125 atm and Cl, is 0.165 atm. What is the par- tial pressure of NOCI in the equilibrium mixture? (a) 0.13 atm (b) 0.36 atm (c) 1.0 atm (d) 5.1 x 10-5 atm (e) 0.125 atm Practice Exercise 2 At 500 K, the reaction PC15(8) = PC13(8) + Cl2(8) has...
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.
For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5. PCl5(g) PCl3(g) + Cl2(g) Suppose that 2.210 g of PCl5 is placed in an evacuated 535 mL bulb, which is then heated to 600. K. (a) What would be the pressure of PCl5 if it did not dissociate? (b) What is the partial pressure of PCl5 at equilibrium? (c) What is the total pressure in the bulb at equilibrium? (d) What is the degree of dissociation of...
The gas-phase reaction for the production of PCl5 from PCl3 and Cl2 at 250°C has an equilibrium constant Kp of 24.2. 1st attempt Part 1 (1 point)See Periodic Table If the initial pressure of PCl5 in a reaction vessel is 1.41 atm, what is the partial pressure of PCl5 when equilibrium is achieved? atm
The equilibrium partial pressures for the reaction Cl2 (g) + PCl3 (g) ↔ PCl5 (g) at 300 K are PCl2 = 0.75 atm, PPCl3 = 0.45 atm, and PPCl5 = 0.73 atm. The value of Kp is __________. A. 0.15 B. 0.048 C. 4.7 D. 2.16
The equilibrium PCl5(g) ⇄ PCl3(g) + Cl2(g) is established at 250oC. At equilibrium the partial pressures of the components are 0.020 Atm (PCl5), 1.28 Atm (PCl3), and 1.28 Atm (Cl2). If the partial pressure of Cl2 is suddenly increased to 2.15 Atm, what is the partial pressure of PCl5 after equilibrium has been reestablished?
The equilibrium constant (Kp) for the interconversion of PCl5 and PCl3 is 0.0121 at certain temperature. A vessel is charged with PCl5 giving an initial pressure of 0.123 atm. PCl5 (g) ⥦ PCl3 (g) + Cl2 (g) a) Calculate equilibrium partial pressure of PCl3 b) Calculate total pressure at equilibrium
The dissociation of PCl5(g) to PCl3(g) and Cl2(g) is an endothermic reaction. PCl5(g) PCl3(g) + Cl2(g) What is the effect on the position of equilibrium of each of the following changes? a) Compressing the gaseous mixture _________________________________ b) Decreasing the temperature _________________________________ c) Adding Cl2(g) to the equilibrium mixture __________________________________ d) Removing PCl5 (g) from the equilibrium mixture __________________________________