For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) <--> PCl3(g) + Cl2(g) If 4.56 g of PCl5 is placed in a 2.63 L bulb at 390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013 bar
For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) <--> PCl3(g) + Cl2(g) If...
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%
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...
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?
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.
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.
Consider the following reaction: PCl5(g) PCl3(g) + Cl2(g) If 4.91×10-3 moles of PCl5, 0.218 moles of PCl3, and 0.393 moles of Cl2 are at equilibrium in a 10.8 L container at 662 K, the value of the equilibrium constant, Kp, is _____
Consider the following reaction: PCl5(g)⇌PCl3(g)+Cl2(g) a. Initially, 0.62 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.20 mol of PCl3 in the flask. What is the equilibrium concentration of PCl5? Express your answer to two significant figures and include the appropriate units. b. What is the equilibrium concentration of Cl2? Express your answer to two significant figures and include the appropriate units. c. What is the numerical value of the equilibrium constant, Kc, for...
The equilibrium constant, K, for the following reaction is 3.40×10-2 at 527 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 10.5 L container at 527 K contains 0.271 M PCl5, 9.60×10-2 M PCl3 and 9.60×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 19.5 L? [PCl5] = M [PCl3] = M [Cl2] = M
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 __________________________________
The equilibrium constant, K, for the following reaction is 2.35×10-2 at 517 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 11.3 L container at 517 K contains 0.269 M PCl5, 7.94×10-2 M PCl3 and 7.94×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 5.27 L? [PCl5] = M [PCl3] = M [Cl2] =...