Chapter 6 homeworkChapter 6 homeworkAnswer to Fugacity problem3. A Fugacity problem (first some additional theory )In a compartment, Ci, the rate of decay in moles m-3 year-1 for a given process, j, israteCi = kj[Ci]where kj = bio degradationphotolysis hydrolysis oxidationadvectionetc. rateCi= kB[Ci]+ kP[Ci]+ kH[Ci]+ kOX[Ci]+ kA[Ci] rateCi = [Ci]  kj = [Ci] kTin moles per year the total rate in a compartment is rateTi = [Ci] kT Vi if the system is at steady state in each compartment, the total input ratefor all the compartments in moles/ year will equal the amount reacted in moles/yearI = Ci]kT Vi) = Zi fi kT Vi) and I = fi Zi kT Vi); why?sofi= I /Zi kT Vi)Assume three compartments air, water and sediment (sed) in equilibrium with one another and a total input rate of 1000 moles/yeargoes into the entire system. From the following rate constant data (years-1), calculate for each compartment the fugacity, the resulting concentration (moles/m3), the total decay rate (moles/year) and the total #moles in each compartment. Use Z values and volumes from the class example. From the total decay rate, where does most of the degradation take place?; where does most of the mass end up?Rate constants (years-1)Biodeg. photolysis hydrolysis oxidation advection kTiAir 0 130 0 0 50water 100 100 100 0 200Sediment 0.1 0 0 0 0AnswerThe key to the problem is estimating the fugacity in one compartment; then you know it in the others, since we said the system was at equilibrium.We derived f = I/(Zi kTi Vi)This means the equilibrium fugacity is equal to the total input I, into the system, divided bythe sum of the individual products  Zi x kT x ViTo get kTi , the total rate constant in each cell, I, add up the individual rates for each cell orcompartment, Enter this in a tableNext enter the volume and the Zi values for each compartment in the table.Find the product of Zi x kT x Vi and then its sum.This sum divided into I equals f as per f = I/(Zi kTi Vi) The mass/ year reacted in each compartment, called the totali = the mass in each compartment multiplied by its total rate constant, kT; e.g. d mass/dt = kTj x mass in each compartment and this equals kTi x Ci xViktjViZVi x Zi x kijf = I/(ZikTiVi)Ci= f xZitotal ratekTi x Ci xVitotal mass Ci x Viyears-1m3atmospheres moles/year molesAir 200 1x101040 8 x10131.16x10-114.65x10-10929.15 4.65Water 500 1x1061x1045.1x10121.16x10-111.16x10-758.07 0.12Sediment 0.1 1x1041x1091 x10121.16x10-111.16x10-211.61 116.14total 8.61xqo13998.84120.91So, I get most of the mass reacting in the air and most of the mass accumulates in the