At elevated temperatures, BrF_5 establishes the following equilibrium. 2BrF_5(g) Br_2(g) + 5F_2(g) The equilibrium concentrations of...
Phosphorus pentachloride decomposes according to the chemical equation PCl_5(g) doubleheadarrow PCl_3(g) + Cl_2(g) K_c = 1.80 at 250 degree C 0.239 mol sample of PCl_5(g) is injected into an empty 3.05 L reaction vessel held at 250 degree C. Calculate the concentrations of PCl_5(g) and PCl_3(g) at equilibrium.
Please help Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the reaction: PCl_5(g) PCl_3(g) + Cl_2(g) At 250 degree C, 0.250 M PCl_5 is added to a flask. If K_C = 1.80, what are the equilibrium concentrations of each gas? A) [PCl_5] = 2.27 M, [PCl_3] = 2.02 M, [Cl_2] = 2.02 M B) [PCl_5] = 0.0280 M, [PCl_3] = 0.222 M, [Cl_2] = 0.222 M C) [PCl_5] = 1.25 M, [PCl_3] = 0.474 M, [Cl_2] =...
PCl_5 decomposes according to the equation PCl_5(g) doubleheadarrow PCl_3(g) + Cl_2(g) At 700 K, a 2 liter flask initially contained only 0.4600 mol PCl_3(g) and 0.6875 mol PCl_5(g). After reaching equilibrium, 0.00800 mol Cl_2(g) was measured in the flask. Calculate the equilibrium concentrations of the three gases and K_c for the reaction. For the reaction, 2 H_2S(g) 2 H_2(g) + S_2 (g), K_0 = 1.7 times 10^-7 at 800.0 degree C If the initial concentration of H_2S in a closed...
What molar ratio Write a balanced chemical equation which corresponds to the following equilibrium constant question K = [NO^-_2] [H_3O^+]/[HNO_2] HNO_2(aq) + H_2O(l) NO^-_2(aq) + H_3O^+(aq) NO^-_2(aq) + H_3O^+(aq) HNO_2(aq) + H_2O(l) NO^-_2 (aq) + H_3O^+(aq) HNO_2(aq) + H_2O(l) NO^-_2 (aq) + H_3O^+(aq) HNO_2(aq) H^+(aq) + OH^-(aq) H_2O(l) HNO_2(aq) NO^-_2 (aq) + H_3O^+(aq) For the equilibrium PCl_5(g) PCl_3(g) + Cl_2(g), K_c = 4.0 at 228 degrees C. If pure PCl_5 is placed in a 1,00-L container and allowed to come...
An decrease in temperature increases the reaction rate because temperature affects the equilibrium constant of the reaction a smaller fraction of the collisions have the correct orientation of molecules. the activation energy of the reaction will decrease. less collisions will have enough energy to exceed the activation energy. the activation energy of the reaction will increase. The equilibrium constant, K_p, for the reaction H_2(g) + I_2(g) doubleheadarrow 2HI(g) is 10.0 at 450 degree C. A rigid cylinder at that temperature...
The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 k. PCl_5 (g) irreversible PCl_3 (g) + Cl_2 (g) At equilibrium mixture of the three gases in a 1.00 L. flask at 500 K contains 0.203 M PCl_5, 4.93 times 10^-2 M PCl_5 and 4.93 times 10^-2 M Cl_2. What will be the concentration of the three gases once equilibrium has been reestablished, if 2.81 times 10^-2 mol of PCl_3 (g) is added to the flask?...
For the reaction 2 HBr(g) + Cl_2(g) rightarrow 2 HCl(g) + Br_2(g) a. Write the equilibrium constant expression for the reaction. b. Using the following G degree values, calculate Delta G degree for the reaction. HBr(g) = -53.22 kJ/mol HCl(g) = -95.27 kJ/mol Cl_2(g) = 0 kJ/mol Br_2(g) = 3.14 kJ/mol c. Calculate the equilibrium constant K_eq be at 298 K. d. Does this equilibrium lie more with reactants or products?
The following reaction was allowed to come to equilibrium at 35 degree C. The initial molar concentration for SO_3 is 0.675 M ([SO_3| = 0.675 M) and the initial molar concentration for CO_2 is 0.444 M (|CO_2] = 0.444 M). After the reaction reached equilibrium the concentration of CO_2 now equals 0.214 M ((CO_2] = 0.214 M). What is the K_c value for the reaction? 0.00987 0.0342 7.31 11.8 16.7 At 35 degree C the equilibrium constant value (K_c) for...
Write the equilibrium constant expressions for the following systems CaCO_3(s) CaO(s) + CO_2(g) 3Fe(s) + 4H_2O(g) Fe_3O4(s) +4H_2(g) 2HgO(s) 2Hg(l) + O_2(g)
The equilibrium constant K_c for the reaction H_2 (g) + Br_2(g) 2 HBr(g) is 2.180 times 10^6 at 730 degree C. Starting with 2.20 moles of HBr in a 21.6-L reaction vessel, calculate the concentrations of H_2, Br_2 and HBr at equilibrium. [H_2] = [Br_2] = [HBr] =