QUESTION 5In the gas-phase reaction 2A+B 3C + 2D, it was found that when 1.50 mole A, 2.00mole B and 1.00 mole D were mixed and allowed to come to equilibrium at 25°C, theresulting mixture contained 1.40 mol C at a total pressure of 2.00 bar. Please draw atable showing the inital amount of each substance, the reacted amount of eachsubstance, and the final amount of each substance, and calcualte the mole fraction ofeach substance. Please pay attention to the Stoichiometry: the reacted C (which is thefinal moles of C given) divided by 3 is the amount of reacted B, which then mulitply by 2is the amount of reacted A, etc.(a) the mole fractions of each species at equalibrium (please enter your answer with 3decimals. for example, 0.4467 is written as 0.447)XA=, XB=, Xc=, XD=For the following answers, please enter numbers with 2 decimals. for example, -25.445is written as -25.4.(b) Kx =(c)Kp=(d) ArGo=(unit kJ/mol)

Write the reaction as follows: 2A+B↔ 3C+2D The temperature of the reaction, T = 25°C = 25+273 K = 298 K The initial moles of A is 1.50 mol The initial moles of B is 2.0 mol The initial moles of D is 1.0 molDraw the ICE table as follows: 2A + B ↔ 3C + 2DImol: 1.50 2.0 0 1.0Cmol: -2x -x +3x +2xEmol:1.50-2x 2.0-x +3x 1.0+2x Calculate the value of x as follows: The final moles of C is 1.40 mol 3x = 1.40 molx = 0.467 mol(a) Calculate the moles of the components at equilibrium as follows: nA = 1.50-2x= 1.50-20.467= 1.033 mol nB = 2.0-x= 2.0-0.467= 1.533 mol nC = +3x= 1.40 mol nD = 1.0+2x= 1.0+20.467 mol= 1.934 mol The total moles of the mixture, n = nA+ nB+nC + nD= 1.033 mol +1.533 mol +1.40 mol + 1.934 mol= 5.9 molNow calculate the mole fraction of each component at equilibrium as follows: xA = nAn= 1.033 mol5.9 mol= 0.175 xB = nBn= 1.533 mol5.9 mol= 0.260 xC = nCn= 1.40 mol5.9 mol= 0.237 xD = nDn= 1.934 mol5.9 mol= 0.328 Therefore, the required solutions are xA = 0.175, xB = 0.260, xC = 0.237, and xD = 0.328.(b) Calculate the equilibrium constant as follows: Kx = xD2xC3xA2xBKx = 0.32820.23730.17520.260Kx = 0.180 Therefore, the required soltuion is Kx = 0.180. (c) Write the relation between Kp and Kx as follows: Kp = KxRT∆n .......1 Here, The gas constant, R = 0.08206 Latm/molK ∆n = 2+3-2-1= 2 Substitute the corresponding values in equation(1) and calculate the value of Kp. Kp = KxRT∆nKp = 0.180×0.08206×298 K2Kp= 107.640 Therefore, the required soltuion is Kp= 107.640.(d) Write the relation between ∆Gro and Kp as follows: ∆Gro = -RTlnKp Substitute the corresponding value sin the above equation and calculate the value of ∆Gro. ∆Gro = -RTlnKp= -8.314 J/molK×298 K×ln107.64= -11592.0 J/mol Therefore, the required solution is -11592.0 J/mol.