1. Phosphorous pentachloride gives an equilibrium mixture ofPCl 5 , PCl 3 and Cl

Question

1. Phosphorous pentachloride gives an equilibrium mixture ofPCl5, PCl3 and Cl2 whenheated                              PCL5(g)       ------>   PCl3(g)     +    Cl2(g)                                                    <------ A 1.00 L vessel contains and unknown amount of PCl5and 0.020 mol each of PCl3 and Cl2 atequilibrium at 250 degree celsius. How many moles ofPCl5 are in the vessel if Kc for thisreaction is 0.0415 at 250 degree celsius. 2. For the same reaction, if the initial concentration ofPCl5 is 1.00 mol/L, what is the equilibrium compositionof the gaseous mixture at 160 degree celsius? The equilibriumconstan Kc at 160 degree celsius is 0.0211. 1. Phosphorous pentachloride gives an equilibrium mixture ofPCl5, PCl3 and Cl2 whenheated                              PCL5(g)       ------>   PCl3(g)     +    Cl2(g)                                                    <------ A 1.00 L vessel contains and unknown amount of PCl5and 0.020 mol each of PCl3 and Cl2 atequilibrium at 250 degree celsius. How many moles ofPCl5 are in the vessel if Kc for thisreaction is 0.0415 at 250 degree celsius. 2. For the same reaction, if the initial concentration ofPCl5 is 1.00 mol/L, what is the equilibrium compositionof the gaseous mixture at 160 degree celsius? The equilibriumconstan Kc at 160 degree celsius is 0.0211.                                                    <------ A 1.00 L vessel contains and unknown amount of PCl5and 0.020 mol each of PCl3 and Cl2 atequilibrium at 250 degree celsius. How many moles ofPCl5 are in the vessel if Kc for thisreaction is 0.0415 at 250 degree celsius. 2. For the same reaction, if the initial concentration ofPCl5 is 1.00 mol/L, what is the equilibrium compositionof the gaseous mixture at 160 degree celsius? The equilibriumconstan Kc at 160 degree celsius is 0.0211.

Explanation / Answer

Part 1 convert the moles to molarity PCl3= 0.020 mol/1.00 l = 0.020 M Cl2 = 0.020 M Kc=[PCl3][Cl2]/[PCl5] [PCl5] =[PCl3][Cl2]/Kc [PCl5] = 0.020 * 0.020 / 0.0415 = 0.009638 M Part 2 PCl5= 1.00 mol / 1.00 L = 1.00 M             PCl5------> PCl3 + Cl2 Intial      1                  0           0 change  -x                 x          x    change should be multiplied by thecoefficients of the compounds from the equation equilb.   1-x             x         x    add intial and change Kc=[PCl3][Cl2]/[PCl5] 0.0211 = (x)(x)/(1-x) 0.0211 - 0.0211x = x2 x2 + 0.0211x - 0.0211 = 0 solve this by using x= -b ± (b2 + 4ac) where a = 1,b = 0.0211, c = -0.0211 therefore x = -0.3124 or x = 0.2702 therefore when x = -0.3124 the change for PCl5 = 1 - x = 1 + 0.3124 = 1.03124which is not possible as the compound would be used in reaction toproduce chlorine therfore the correct answer is at equlibrium PCl3= 0.7298 Cl2 = 0.7298 PCl5 = 1- 0.2702 = 0.7298 x2 + 0.0211x - 0.0211 = 0 solve this by using x= -b ± (b2 + 4ac) where a = 1,b = 0.0211, c = -0.0211 therefore x = -0.3124 or x = 0.2702 therefore when x = -0.3124 the change for PCl5 = 1 - x = 1 + 0.3124 = 1.03124which is not possible as the compound would be used in reaction toproduce chlorine therfore the correct answer is at equlibrium PCl3= 0.7298 Cl2 = 0.7298 PCl5 = 1- 0.2702 = 0.7298 PCl3= 0.7298 Cl2 = 0.7298 PCl5 = 1- 0.2702 = 0.7298

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