A student ran the following reaction in the laboratory at 280 K: 2CH2Cl2(g) CH(g) + CC14(8)...
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 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 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 =
A student ran the following reaction in the laboratory at 380 K: CH4(g) + CCl4(g) 2CH2Cl2(g) When she introduced 4.26×10-2 moles of CH4(g) and 5.94×10-2 moles of CCl4(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 5.11×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
1. A student ran the following reaction in the laboratory at 632 K: 2HI(g) ->H2(g) + I2(g) When she introduced 0.362 moles of HI(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 3.55×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc = 2. A student ran the following reaction in the laboratory at 616 K: CO(g) + Cl2(g) -> COCl2(g) When she introduced 0.131 moles of CO(g) and 0.161 moles...
A student ran the following reaction in the laboratory at 425 K: PCl5(g) --> PCl3(g) + Cl2(g) When she introduced 4.59 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.94×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
A student ran the following reaction in the laboratory at 557 K: CO(g) + Cl2(g) = COCl2(g) When she introduced 1.06 moles of CO(g) and 1.09 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of COCl2(g) to be 1.02 M. Calculate the equilibrium constant. Ko she obtained for this reaction. Ke=
A student ran the following reaction in the laboratory at 541 K: COC12(E) CO(g) + Cl2(g) When she introduced 1.13 moles of COC12(e) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.83x10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction.
A student ran the following reaction in the laboratory at 529 K: CoCl2(g) P CO(g) + Cl2(g) When she introduced 1.68 moles of CoCl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.61*10-2 M. Calculate the equilibrium constant, Ke she obtained for this reaction. Kc =