Please show steps for ice table. 1. The following reaction is carried out with i
ID: 304117 • Letter: P
Question
Please show steps for ice table. 1. The following reaction is carried out with initial concentrations of [PC15] 4.580 M, [Cb] = 5.870 M, and [PC13] = 1.280 M. PC, (g) + Cl2 (g)?PCIs(g) Given that the equilibrium concentration of PCls is 5.847 M, calculate Kc. (10 points) 2. Consider the reaction and its equilibrium constant: 12 (9) +Cl2 (g) 2 ICI (g) Kp 81.9 A reaction mixture contains P12 = 0.114 atm, and Pici = 0.355 atm. Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed? [Hint: Calculate Q first, then compare Q to K.) (10 points) 3. Consider the following reaction at chemical equilibrium: 2 KCIO3 (s)2 KCI (s) +3 02 (9) What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture? Adding an inert gas at constant volume? (10 points)Explanation / Answer
The reaction is PCl3(g)+Cl2(g <---> PCl5(g)
KC = [PCl5]/{ [PCl3]*[Cl2]},
Preparing the ICE table for concentrations
Component initial concentration Change Eq. concentration
PCl3 1.280 -x 1.280-x
Cl2 5.870 -x 5.870-x
PCl5 4.580 x 4.580+x
Given at equilibrium, [PCl5] = 5.847= 4.580+x
Or x= 1.267M
So at equilibrium, [PCl3]= 1.280-1.267= 0.013M and [Cl2] = 5.870-1.267= 4.603M and [PCl5]= 5.847M
KC= 5.847/(0.013*4.603)= 97.71
2.
For the reaction, I2(g)+Cl2(g) <-->2ICl, the direction in which reaction proceeds is governed by
Q= reaction coefficient = (PICl)2/ {PI2*PCl2}, where P denotes the partial pressure of the component.
Q= 0.355*0.355/(0*0.114)= infinity.
If Q<K, the reaction proceeds in the forward direction towards products.
If Q>K, the reaction proceeds backwards towards reactants.
If Q= K, the reaction is at equilibrium
Hence Q>K, the reaction proceeds back wards. The reaction mixture is not at equilibrium.
3.
For the reaction 2KClO3(s)<-->2KCl(s)+3O2(g)
As per Lechatlier Principle, as the volume of reaction mixture is increased, the reaction proceeds so as to compensate the effect of increased volume. So when the volume of reaction mixture is increased, the reaction proceeds in a direction where there is a decrease in volume. So backward reaction proceeds. When the volume of reaction mixture is decreased, the opposite reaction takes place. Since Kp = (PO2)3, P denotes partial pressure, addiing an inerts gas does not effect the volume.