Part B please For the equilibrium reaction: PCI (g) =PC13(g) + Cl2(g); Kc = 0.00183 a. Write the concentration equilibr...
For the equilibrium reaction: PCis(g) = PC13(g) + Cl2(g); Kc = 0.00183 a. Write the concentration equilibrium equation for the reaction. [1 mark] b. If 2.0 gram of PCis(g) is introduced into a 1.5 L flask what will be the equilibrium concentrations of PCls, PC13 and Cl2? [4marks]
QUESTION 55 - WRITE YOUR ANSWER IN THE SPACE PROVIDED [5 marks] For the equilibrium reaction: PC1s(g) = PC13(g) + Cl2(g); Kc = 0.00183 a. Write the concentration equilibrium equation for the reaction. [1 mark] b. If 1.0 gram of PCls(g) is introduced into a 1.5 L flask what will be the equilibrium concentrations of PCs, PCl; and Cl2? [4marks]
Consider the reaction. PCI; (g) = PCI,(g) + Cl2(g) Kc = 0.0420 The concentrations of the products at equilibrium are [PC13] = 0.160 M and [Cl2] = 0.160 M. What is the concentration of the reactant, PCs, at equilibrium? [PCI31= C [PCIS] = M
The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g) = PC15(g) Calculate the equilibrium concentrations of reactant and products when 0.389 moles of PC13 and 0.389 moles of Cl2 are introduced into a 1.00 L vessel at 500 K [PC13] = [Cl] = [PC15] - The equilibrium constant, Kc, for the following reaction is 5.10x10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibrium concentration of HCl when 0.467 moles of NHACI(S)...
ampere The reaction, PCls (g) 5 PC13(g) + Cl2(g), has Kc = 4.20 x 102. If 0.1050 mol of PCL5, 0.0450 mol of Cl2 and 0.0450 mol of PCl3 is placed in a 0.5000-L flask at 250°C, what are the equilibrium concentrations of PCl3 and Cl, if equilibrium (PCI5] = 0.2065 M? A) [PC13] = 0.00350 M; (Cl2] = 0.00350 M B) [PC13] = 0.0900 M; [Cl2] = 0.0900 M -=51,85 (66450) (19450) C) [PC13] = 0.2100 M; [Cl2] =...
Phosphorus pentachloride decomposes according to the chemical equation PC15(g) PC13(g) + Cl2 (g) Kc = 1.80 at 250° C A 0.475 mol sample of PCI, (g) is injected into an empty 4.90 L reaction vessel held at 250° C. Calculate the concentrations of PCI, (g) and PCI,(g) at equilibrium. [PCls] [PCl3]
Consider the equilibrium between PCls, PClj and Cl2. PCIs(g)-PC13(g) + Cl2(g) K 0.251 at 571 K The reaction is allowed to reach equilibrium in a 15.2-L flask. At equilibrium, [PCIs]-9.75x102 M, [PCl,]-o.156 M and [C12] = 0.156 M. (a) The equilibrium mixture is transferred to a 7.60-L flask. In which direction will the reaction proceed to reach equilibrium? (b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 7.60-L flask. [PCI5] [C12] =
The equilibrium constant, K., for the following reaction is 1.20 10-2 at 500 K. PCI(8) =PC13(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.363 moles of PC13() are introduced into a 1.00 L vessel at 500 K M [PC131 = [PC13] - [Cl] M M
At a certain temperature, the equilibrium constant, Kc , is 0.00401 for the reaction Cl2(g)−⇀↽−2Cl(g) A. If 3.05 g Cl2Cl2 is placed in a 3.00 L flask at this temperature, what are the equilibrium concentrations of Cl2 and Cl? B. Following the establishment of equilibrium in part A, the volume of the flask is suddenly increased to 4.50 L while the temperature is held constant. What are the new equilibrium concentrations of Cl2 and Cl? C. Following the establishment of...
please write clear and complete. 7. The equilibrium constant at a particuiar temperature for the reaction: H2(g) + Br2(g) -> 2HBr(g) is 1.5 x 105 The value of Kc for HBr(g) -1 /2 H2(g) + 1 /2 Br2(g) is PCla(g) + Cl2(g) For the equilibrium: PCIs(g)-- Kc = 4.0 at 228°C. If 0.35 g of pure PC15 is added to a 2.0 L flask what will be the equilibrium concentration of Cl2(g). (you'll need to carry extra sig figs to...