Chloromethane forms by the reaction CH4(g) + Cl2(g) ⇌ CH3Cl(g) + HCl(g) at 1500K, Kp = 1.6 x 104 . In the reaction mixture, PCH4 = 0.13 atm, PCl2 = 0.035 atm, PCH3Cl = 0.24 atm, and PHCl = 0.47 atm. Is CH3Cl or CH4 forming?
Chloromethane forms by the reaction CH4(g) + Cl2(g) ⇌ CH3Cl(g) + HCl(g) at 1500K, Kp =...
A-The following reaction has a Kp=1.6 x 104 at 1500 K. CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) a. In a reaction mixture, PCH4 = 0.13atm , PCl2 = 0.035atm , PCH3Cl = 0.24atm , and PHCl = 0.47atm . Is the system at equilibrium? If not how does the reaction proceed to reach equilibrium? At equilibrium, the following stresses are applied. How is the equilibrium affected? Explain. i. HCl is removed. ii. CH3Cl is added iii. The volume is reduced...
When 20.5 g CH4 (MM = 16.04 g/mol) and excess Cl2 undergo a reaction that has a 90.5% yield, what mass of chloromethane (CH3Cl, MM = 50.48 g/mol) forms? CH4(g) + Cl2(g) --> CH3Cl(g) + HCl(g)
Write the equilibrium constant expression : CH4 (g) + Cl2 (g) = CH3Cl (g) + HCl (g)
Consider the following reaction: I2(g)+Cl2(g)⇌2ICl(g), Kp=81.9 (at 25∘C) A reaction mixture at 25∘C initially contains PI2 = 0.110 atm , PCl2 = 0.110 atm , and PICl = 0.00 atm. Find the equilibrium partial pressure of I2 at this temperature. Find the equilibrium partial pressure of Cl2 at this temperature. Find the equilibrium partial pressure of ICl at this temperature.
Consider the following reaction at 248 C and 1.00 atm: CH3Cl (g) + H2(g) --> CH4(g) + HCl (g)For this reaction, the enthalpy change at 248 C is -83.3 kJ/mol. At constant pressure the molar heat capacities (Cp) for the compounds are as follows: CH3Cl (48.5 J/mol K) ; H2 (28.9 J/mol K) ; CH4 (41.3 J/mol K) ; HCl (29.1 J/mol K) Assuming that the Cp values are independent of temperature, calculate ∆H for this reaction at 25 C.
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 equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g) --> 2CH2Cl2(g) Calculate the equilibrium partial pressures of all species when CH4 and CCl4, each at an intitial partial pressure of 0.959atm, are introduced into an evacuated vessel at 350 K. PCH4 = atm PCCl4 = atm PCH2Cl2 = atm
Consider the following reaction: I2(g)+Cl2(g) ⇌ 2ICl(g) Kp= 81.9 at 25 ∘ C. Calculate ΔGrxn for the reaction at 25 ∘ Cunder each of the following conditions. PICl= 2.65 atm ; PI2= 0.320 atm ; PCl2= 0.221 atm .
I2(g)+Cl2(g)⇌2ICl(g),Kp=81.9 (at 25∘C) The reaction between I2 and Cl2 is carried out at the same temperature, but with the following initial partial pressures: PI2=0.170atm, PCl2=0.170atm, and PICl=0.00 atm. find equibilibrum partial pressure of ICl.
Consider the following reaction: I2(g)+Cl2(g) ⇌ 2ICl(g) Kp= 81.9 at 25 ∘ C. Calculate ΔGrxn for the reaction at 25 ∘ C under each of the following conditions. PICl= 2.58 atm PI2= 0.322 atm PCl2= 0.214 atm Express your answer using one significant figure.