For the reaction PCl5(g) ↔ PCl3(g) + Cl2(g) the value of K = 15.4 at 497.0...
For the reaction PCl5(g) ↔ PCl3(g) + Cl2(g) the value of K = 15.4 at 497.0 K. Calculate the equilibrium partial pressures (in bar) of all species at 497.0 K if the initial pressures were P(PCl5(g)) = 0.700 bar, P(PCl3(g)) = 0.320 bar, and P(Cl2(g)) = 0.000 bar.
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For the reaction PCl5(g) ↔ PCl3(g) +
Cl2(g) the value of K = 15.4 at 497.0...
For the reaction PCl5(g) <-->PCl3(g) + Cl2(g) the value of K = 28.3 at 532.0 K....
For the reaction PCl5(g) <-->PCl3(g) + Cl2(g) the value of K = 28.3 at 532.0 K. Calculate the equilibrium partial pressures (in bar) of all species at 532.0 K if the initial pressures were P(PCl5(g)) = 0.664 bar, P(PCl3(g)) = 0.326 bar, and P(Cl2(g)) = 0.000 bar.
The equilibrium partial pressures for the reaction Cl2 (g) + PCl3 (g) ↔ PCl5 (g) at...
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
At 500 K the reaction PCl5(g) <----> PCl3(g) + Cl2(g) has Kp = 0.497. In an...
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?
Consider the reaction: PCl5 (g) ⇔ PCl3 (g) + Cl2 (g) The initial concentration of PCl5...
Consider the reaction: PCl5 (g) ⇔ PCl3 (g) + Cl2 (g) The initial concentration of PCl5 is 0.40 M and the equilibrium concentration of PCl3 is 0.33 M. Calculate the equilibrium constant Kc.
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr,...
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g)is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is PCl3(g)+Cl2(g)↽−−⇀PCl5(g Calculate the new partial pressures,pP, after equilibrium is reestablished.p pcl5 , pcl2 , ppcl3.
For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) PCl3(g) + Cl2(g) If 3.72 g...
For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) PCl3(g) + Cl2(g) If 3.72 g of PCl5 is placed in a 3.68 L bulb at 390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013 bar bar The number of significant digits is set to 3; the tolerance is +/-4%
The equilibrium PCl5(g) ⇄ PCl3(g) + Cl2(g) is established at 250oC. At equilibrium the partial pressures...
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 following reaction has equilibrium constant of Kp = 11.5 atm at 300 oC: PCl5(g) PCl3...
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.
The equilibrium constant for the following reaction is 1.42×10-3 at 179 °C. PCl5(g)<-->PCl3(g) + Cl2(g) K...
The equilibrium constant for the following reaction is 1.42×10-3 at 179 °C. PCl5(g)<-->PCl3(g) + Cl2(g) K = 1.42×10-3 at 179 °C Calculate the equilibrium constant for the following reactions at 179 °C. (a) PCl3(g) + Cl2(g)<--> PCl5(g) K = (b) 2 PCl5(g)<-->2 PCl3(g) + 2 Cl2(g) K =
1. The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <---->...
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...
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