We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
What is the equilibrium expression for the following reaction? PCl; (g) PC13 (9) + Cl2 (g)...
Question 9 What is the equilibrium expression for the following reaction? PCls (9) PC13 (9) + Cl2 (9) Kea [PC13)(Cl2) [PC15] Κρα [PC13]' [Cl] [PC15) Keg [PC13][14] [PC15] Кед (PCI) [PC] Keg (PC13][CL] (PCI)
Consider the reaction. PC15(g) 근 PC13(g) + Cl2(g) K = 0.042 The concentrations of the products at equilibrium are PC, l = 0 10 M and [CI| = O. 1 3 M What is the concentration of the reactant, PCl, at equilibrium? PCI,I
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)...
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] =
(References The equilibrium constant (K.) for the reaction PC13 (9) + Cl2 (9) - PCs (9) equals 49 at 230°C. If 0.372 mol each of phosphorus trichloride and chlorine are added to a 2.80 L reaction vessel, what is the equilibrium composition of the mixture at 230°C? mol PCL mol Cl2 mol PCL Submit Answer Try Another Version 5 Item attempts remaining
Consider the following reaction where Kc = 1.20x10-2 at 500 K: PCl; (g)PCl3 (g) + Cl2 (g) A reaction mixture was found to contain 0.125 moles of PCls (2),4.99-102 moles of PCl3 (g), and 4.1 Indicate True (T) or False (F) for each of the follow ー▼ I. In order to reach equilibrium PC15(g) must be produced in a ▼ 2. In order to reach equilibrium Kc must decrease ▼ 3. In order to reach equilibrium PC13 must be consumed...
Calculate AH for the reaction PC13(1) + Cl2(g) → PC15(s) given the following data: Equation ΔΗ P4(s) + 6 Cl2(g) → 4 PC13(1) -1280 kJ P4(s) + 10 C12(g) → 4 PC1:($) -1774 kJ pt pt AH= 1 pt
Choose the correct equilibrium expression for the following reaction: H_2 + Cl_2 2 HCl K_eq = [H_2]/[CHl]^2 [Cl_2] K_eq = [Cl_2]/[HCl]^2[H_2] K_eq = [HCl]^2/[HCl]^2
Consider the following reaction, Kc= 0.030 at 1000°С. Starting with the original concentration of PC15 of 0.100M, what are the concentrations of PC15, PC13 and Cl2 at equilibrium? PC15(9) --> PC13 (9) + Cl2 (9) O A [PC15)= 0.0418, [PC13)= 0.0418, C121= 0.0418 OB. [PC15)= 0.0418, [PC13]= 0.0418, (C12)= 0.0582 oc [PC15)= 0.0582, [PC13)= 0.0582. [C121= 0.0582 OD. [PC15)= 0.0582, [PC13]= 0.0418, C121= 0.0418
The equilibrium constant, K, for the following reaction is 3.52*10 at 528 K PC13(g) =PC12(g) + Cl2(g) An equilibrium mixture of the three gases in a 9.42 L container at 528 K contains 0.244 M PCIE. 9.27X10-2M PCI, and 9.27-10-2M Cly. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 17.1 L? M [PC15] [PC13] = [Cl] M M