2.2 Assume that you work for the United Nations and that a group of scientists c

Question

2.2 Assume that you work for the United Nations and that a group of scientists comes to you with a proposal to reduce the concentration of CO2 in the atmosphere. The proposal is to fertilize the surface waters of the ocean to stimulate the fixation of about 7.5 × 1016 grams of carbon, i.e., about 10% of the CO2 in the atmosphere. They rationalize that the uptake of this much carbon from the surface waters of the ocean will cause an equivalent amount of inorganic carbon to enter the ocean from the atmosphere (recall Lake 227) and thereby reduce the concentration of CO2 in the atmosphere by about 10%. They propose to fertilize with either nitrogen (N) , phosphorus (P) , or iron (Fe). Assuming that the organic matter contains C, N, and P in the Redfield ratio, calculate the amount of N and P that would be required to stimulate the fixation of 7.5 × 1016 grams of carbon. Compare these amounts of N and P with present global production of N and P for fertilizer use, 7 × 107 tonnes of N and 3.5 × 107 tonnes of P per year. How many years of fertilizer production would it take the world to produce enough N and P to stimulate the uptake of 7.5 × 1016 grams of carbon by marine phytoplankton? The ratio of carbon to iron in marine phytoplankton is about 104 grams of C per gram of Fe. How many grams of iron would be required to stimulate the uptake of 7.5 × 1016 grams of carbon? Global iron production is presently about 5.5 × 1014 g per year. How many days would it take the world to produce enough Fe to stimulate the uptake of 7.5 × 1016 grams of carbon by marine phytoplankton?

Explanation / Answer

Part 1: Amount of N and P in Redfield ratio

We know that ratio of C:N:P is 106:16:1 in Redfield ratio

Production of C= 7.5*10^16 gram

Production of N= 16/106 *(7.5*10^16)

= 1.13*10^16 gram

Production of P= (1/106)*7.5*10^16 gram

=.070*10^1gram

Part 2: Global production of N= 7*10^7 tonnes

Require production of N 1.13*10^16 grams

We know that 1tonnes= 1000000 gram

7*10^7 tonnes= 7*10^13 gram

Production status of N as compared to required production

= Global production- Required Production

= 7*10^13-1.13*10^16 gram

=-1.123*10^16 gram

Ratio of required production to global production

= required production/global production

= 1.13*10^16/7*10^13

= 161:1

Similarly Production of P

= 3.5*10^14-7.1*10^14 gram

= -6.75*10^14 gram

Ratio of P production

= 7.1*10^14/3.5*10^14

= 20.28:1

This will take 20.28 (21) years for P production and 161.4(162) years for production of N to produce enough to meet the requirement.

Iron Production

Ratio of C:Fe= 1:104

So for 7.5*1016 grams the required Fe is

= (7.5*10^16)*104

= 7.8*10^18 grams

Days required to produce 7.8*10^18 grams iron

Yearly Production

= 5.5*10^14 grams

No of days =((7.5*10^18)/(5.5*10^14))*365

= 5176363.636

=5176364 days

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