Consider the following reaction 2CH,C12(g) =-CH4(g) + CC14(g) If 0.254 moles of CH2Cl2(g) 0.584 moles of...
The equilibrium constant, Kc, for the following reaction is 9.52x10-2 at 350 K: CH4(8) + CC14(8) — 2CH2Cl2(g) Calculate the equilibrium concentrations of reactants and product when 0.305 moles of CH4 and 0.305 moles of CCl4 are introduced into a 1.00 L vessel at 350 K. [CH] = [CC14] = [CH2Cl2] = A student ran the following reaction in the laboratory at 531 K: COC12(8) CO(g) + Cl2(8) When she introduced 1.87 moles of COCl2(g) into a 1.00 liter container,...
A) The equilibrium constant, Kc, for the following reaction is 10.5 at 350K. 2CH2Cl2(g) <->CH4(g) + CCl4(g) If an equilibrium mixture of the three gases in a 19.9 L container at 350K contains 0.416 mol of CH2Cl2(g) and 0.202 mol of CH4, the equilibrium concentration of CCl4 is B) 2SO3(g) <->2SO2(g) + O2(g) If 0.184 moles of SO3(g), 0.477 moles of SO2, and 0.234 moles of O2 are at equilibrium in a 15.0 L container at 1.23×103 K, the value...
a) The equilibrium constant, Kc, for the following reaction is 9.52×10-2at 350 K. CH4(g) + CCl4(g) -> 2 CH2Cl2(g) Calculate the equilibrium concentrations of reactants and product when 0.251 moles ofCH4and 0.251 moles of CCl4are introduced into a 1.00 L vessel at 350 K. [ CH4] = M [ CCl4] = M [ CH2Cl2] = M b) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) ->PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of reactant...
A student ran the following reaction in the laboratory at 293 K: 2CH2Cl2(g) CH4(g) + CCl4(g) When she introduced 6.91×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.19×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc=?
A student ran the following reaction in the laboratory at 283 K: 2CH2Cl2(g) ->CH4(g) + CCl4(g) When she introduced 7.70×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.59×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) = CH4(g) + CC14(9) Calculate the equilibrium concentrations of reactant and products when 0.377 moles of CH,Cl2 are introduced into a 1.00 L vessel at 350 K (CH2Cl2] = [CH4] = [CCl4] =
A student ran the following reaction in the laboratory at 294 K: 2CH2Cl2(g) ____> CH4(g) + CCl4(g) When she introduced 7.04×10-2 moles of CH2Cl2(g) into a 1.00 Liter container, she found the equilibrium concentration of CCl4(g) to be 3.25×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) >>CH4(g) + CCl4(g) Calculate the equilibrium concentrations of reactant and products when 0.327 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K. [CH2Cl2] = M [CH4] = M [CCl4] = M The equilibrium constant, Kc, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) >>CO(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.275 moles of COCl2(g) are...
The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) = CH4(g) + CCl4(g) Calculate the equilibrium concentrations of reactant and products when 0.294 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K. [CH2Cl2] = ? M [CH4] = ? M [CCl4] = ? M
Consider the following reaction: 2CH() CH2(g) + 3H2(g) The reaction of CH4 is carried out at some temperature with an initial concentration of (CH) = 0.088M. At equilibrium, the concentration of Hy is 0.020 M. Find the equilibrium constant at this temperature. Express your answer using two significant figures. TO AED K- Submit Request Answer