The equilibrium constant in terms of pressures for the reaction C6H5CH2OH(g) <--> C6H5CHO(g) + H2(g) is Kp = 0.558 at 523 K.
(a) A pure sample of gaseous benzyl alcohol, C6H5CH2OH(g), is introduced into a rigid flask at a temperature of 523 K so that its original pressure is 0.126 atm. Calculate the fraction of this starting material that is converted to products at equilibrium.
(b) A second sample of benzyl alcohol is introduced into a rigid flask at a temperature of 523 K, this time at an original pressure of 4.96 atm. Again, calculate the fraction of the starting material that is converted to products at equilibrium.
While solving the problem I have expressed partial pressure of a particular compound with "p" with the structure of the compound in the subscript.
The equilibrium constant in terms of pressures for the reaction C6H5CH2OH(g) <--> C6H5CHO(g) + H2(g) is...
The equilibrium constant in terms of pressures for the reaction COCl2(g) <--> CO(g) + Cl2(g) is Kp = 5.00 at 873 K. (a) A pure sample of gaseous phosgene, COCl2(g), is introduced into a rigid flask at a temperature of 873 K so that its original pressure is 0.121 atm. Calculate the fraction of this starting material that is converted to products at equilibrium. (b) A second sample of phosgene is introduced into a rigid flask at a temperature of...
a search this course Use the References to access important valees if ded fer this qoesres The equilibrium constant in terms of pressures for the reaction is Kp-2.59-10 at 1.11 10K (a) A pure sample of gaseous hydrogen lodide, Hl(g), is introduced indo a rigid flask at a semperature of 1.1110 K so that its original converted to products at equilibrium pressure is 0.142 atm. Caliculate the fraction of this starting material Cat i (b) A second sample of hydrogen...
Some C6H5CH2OH is allowed to dissociate into C6H5CHO and H2 at 523 K. At equilibrium, [C6H5CH2OH] = 0.212 M, and [C6H5CHO] = [H2] = 5.24×10-2 M. Additional C6H5CHO is added so that [C6H5CHO]new = 9.85×10-2 M and the system is allowed to once again reach equilibrium. C6H5CH2OH(g) C6H5CHO(g) + H2(g) K = 1.30×10-2 at 523 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system...
Some C6H5CH2OH is allowed to dissociate into C6H5CHO and H2 at 523 K. At equilibrium, [C6H5CH2OH] = 0.325 M, and [C6H5CHO] = [H2] = 6.50×10-2 M. Additional C6H5CHO is added so that [C6H5CHO]new = 0.119 M and the system is allowed to once again reach equilibrium. C6H5CH2OH(g) <-------> C6H5CHO(g) + H2(g) K = 1.30×10-2 at 523 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations (M) of reactants and products after...
The equilibrium constant in terms of pressures, Kp, for the reaction of SO2 and O2 to form SO3 is 0.365 at 1.15×103 K: SO2(g) + O2(g) = 2SO3(g) A sample of SO3 is introduced into an evacuated container at 298 K and allowed to dissociate until its partial pressure reaches an equilibrium value of 0.867 atm. Calculate the equilibrium partial pressures of SO2 and O2 in the container. PSO2 = PO2 =
A sample of gaseous PCs was introduced into an evacuated flask so that the pressure of pure PCI would be 0.48 atm at 433 K. However, PCIE decomposes to gaseous PCI; and Cly, and the actual pressure in the flask was found to be 0.80 atm. Calculate Ko for the decomposition reaction below at 433 K. PCI() = PCI;(9) + Cl2(0) Also calculate K at this temperature. At a particular temperature, 11.1 mol of SO3 is placed into a 3.8-L...
The equilibrium constant, K. for the following reaction is 1.80X10-2 at 698 K: 2HI(g) P H2(g) +1,2) Calculate the equilibrium partial pressures of all species when HI(g) is introduced into an evacuated flask at a pressure of 1.83 atm at 698 K PH The equilibrium constant, K, for the following reaction is 1.04x10-2 at 548 K: NHCI() NH3(g) + HCl(g) Calculate the equilibrium partial pressure of HCl when 0.579 moles of NH CI(s) is introduced into a 1.00 L vessel...
1. The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <----> PCl3(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.00 atm at 500 K. PPCl5 = atm PPCl3 = atm PCl2 = atm 2. The equilibrium constant, Kp, for the following reaction is 0.215 at 673 K: NH4I(s) <----> NH3(g) + HI(g) Calculate the equilibrium partial pressure of HI when...
The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) > CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.939 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm
The squares in the equation are equilibrium signs. The equilibrium constant, Ky, for the following reaction is 0.110 at 298 K: NH_HS(s) NH3(g) + H2S(g) Calculate the equilibrium partial pressure of H2S when 0.416 moles of NH_HS(s) is introduced into a 1.00 L vessel at 298 K. Phys= c atm The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PC15(g) PC13(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PC15(g) is introduced into...