Question

Given that the atmosphere has a mass of 5.1x10¹⁸ kg and the mole fraction of methane is 1.8 ppm(V) in the atmosphere the gas-phase reaction with OH^. radicals (hydroxyl radicals) is the mechanism for removal of methane from the atmosphere. The first order rate coefficient for methane reaction from the atmosphere is 0.1 yr^-1. Emissions from the surface of the earth is the source of methane to the atmosphere.

a.) What is the mass of methane in the atmosphere?

b.) Using a mass balance, what is the expected emission rate in grams/year of methane to maintain a constant concentration of 1.8 ppm(V)?

c.) Methane concentrations are actually increasing by about 0.6% mass percent per year. Estimate the actual emission rate of methane in grams/year needed to both replace reacted methane and to increase the concentration in the atmosphere.

Answer 1

Given Mass of atmosphere = 5.1*10-18 kg Mole fraction of methane = 1.8 ppm (v) Parta- Mass of methane in atmosphere - Ppm (I is volume of the methane in 10^6 parts volume of the atmosphere. We know that the moles are proportional to the volume in the case of the gas. Thus 1 ppm V= 1 m^3 in 10^6 m^3 = 1 mole in 10^6 mole Atmosphere has mainly oxygen and nitrogen and thus the molecular mass of the air can be taken as 29 g/mole. Moles of air in the atmosphere = 5.1*10^18*1000 g/29 g/mole =1.75*10-20 moles Moles of methane = Moles of air ppm V/10^6 = 1.75*10*20/1.8*10^6 =9.72 €10^13 moles Mass = moles *MW = 9.72*10^13 * 16 gb MW is the molecular weight - molecular weight of methane = 16 g/mole) = 155*10^14 g = 1.55*10^11 kg Part b Using the mass balance, we can write

Paan Paragraph Styles Rate of methane coming to atmosphere by generation-rate of methane going out by degradation = rate of accumulation of methane in atmosphere In the case when we want no accumulation of the methane. Rate of accumulation becomes and thus we can write Rate of methane generated-moles of methane reacted (degraded) Since it is first order reaction - Rate of methane reacted = K*CA CA is the concentration of methane Thus rate of methane reacted=0.1 year"-1 *155-11 =155*10^10 kg/year Moles of methane generated = moles of methane reacted = 1.55*10^10 kg /year Part Given we have a accumulation of the 0.6 weight % in one year Taking one year as the basis Moles generated - Moles reacted = Moles accumulated Moles generated = Moles accumulated + Moles reacted =(0.6/100)*1.55*10^11 +0.1*1.55*10^11 kg/year =1643*10^11 kg /year Please note that here we are taking in account moles that are getting reacted as well as that are getting accumulated within a year. Hope this helps Thanks