A chemical engineer is studying the following reaction:4 HCl(g)+O2(g) 2H,0(g)+2Cl,(g)At the temperature the engineer picks, the equilibrium constant K, for this reaction is 0.095.The engineer charges ("fills") four reaction vessels with hydrogen chloride and oxygen, and lets the reaction begin. He then measures the composition of themixture inside each vessel from time to time. His first set of measurements are shown in the table below.Predict the changes in the compositions the engineer should expect next time he measures the compositions.reactionvesselcompoundexpected change in pressurepressureHC18.85 atmT increaseOI decreaseO (no change)2.6.22 atm↑ increaseOI decrease(no change)AH,03.58 atm↑ increaseOl decreaseO (no change)Cl,9 02 atmO T increaseOl decreaseO (no change)HC18.38 atm1 increaseO decreaseO (no change)6.60 atm1 increase0 decreaseO (no change)B.H,O4.95 atmTincrease0I decreaseO (no change)Cl,11.25 atmT increaseOI decrease(no change)HCI6.05 atm40 1increaseI decreaseO (no change)5.52 atmO1 increase0I decreaseO (no change)H,O4.98 atm1 increaseOI decreaseO (no change)Cl,10.42 atmO1 increaseO Į decreaseO (no change)

When the reaction quotient is less than the equilibrium constant in terms of partial pressure, the system will shift to right. When reaction quotient is more than Kp, the system will shift to left. When Qp and Kp are equal, the system is at equilibrium. Hence, the shift does not occur.The given reaction is as follows, 4HCl (g) + O2(g) → 2H2O(g) + 2Cl2(g) The expression of Qp is as follows, Qp = PH2O2 PCl22PHCl4 PO2 Given, Kp = 0.095 When Kp > Qp, products are favored. When Kp < Qp, reactants are favored. When Kp = Qp, no change occurs.For reaction vessel A, To find the value of Qp as follows, PHCl = 8.85 atmPO2 = 6.22 atmPH2O = 3.58 atmPCl2 = 9.02 atm Substituting the values in Qp expression, Qp = PH2O2 PCl22PHCl4 PO2= 3.582 9.0228.854 6.22Qp = 0.02733 Therefore, Kp > Qp. Therefore, the expected change in pressure is as follows, HCl = decreaseO2 = decreaseH2O = increaseCl2 = increase For reaction vessel B, To find the value of Qp as follows, PHCl = 8.38 atmPO2 = 6.60 atmPH2O = 4.95 atmPCl2 = 11.25 atm Substituting the values in Qp expression, Qp = PH2O2 PCl22PHCl4 PO2= 4.952 11.2528.384 6.60Qp = 0.09527 ≈ 0.095 Therefore, Kp = Qp. Therefore, the expected change in pressure is as follows, HCl = no changeO2 = no changeH2O = no changeCl2 = no change For reaction vessel C, To find the value of Qp as follows, PHCl = 6.05 atmPO2 = 5.52 atmPH2O = 4.98 atmPCl2 = 10.42 atm Substituting the values in Qp expression, Qp = PH2O2 PCl22PHCl4 PO2= 4.982 10.4226.054 5.52Qp = 0.36 Therefore, Kp < Qp. Therefore, the expected change in pressure is as follows, HCl = increaseO2 = increaseH2O = decreaseCl2 = decrease